Analogs of the Richard Stanley Amer. Math. Monthly Problem 11610 for ALL pairs of words of length, 3, in an alphabet of, 6 letters. By Shalosh B. Ekhad Proposition Number , 1, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 1] as those of, [1, 1, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 5 4 3 2 -t - (6 t - 1) (t + 1) (6 t + 11 t + 12 t + 9 t + 4 t - 1) P 2 5 4 3 2 2 - (6 t - 1) (5 t + 5 t - 1) (6 t + 11 t + 12 t + 9 t + 4 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 5 4 3 2 a(n + 8) = (-3900480 + 960 n - 5480 n - 286943 n - 2076379 n - 5276758 n) a(n + 7)/((n + 8) %1) - 2 5 4 3 2 (-6172704 + 1920 n - 11600 n - 538378 n - 3706388 n - 8971635 n) a(n + 6)/((n + 8) %1) - 5 4 3 2 (-11644992 + 1088 n - 6680 n - 389537 n - 3325453 n - 10663848 n) a(n + 5)/((n + 8) %1) + 4 5 4 3 2 (-10354272 + 3584 n - 23552 n - 921534 n - 6002953 n - 14226866 n) a(n + 4)/((n + 8) %1) + 5 4 3 2 (-91691136 + 29120 n - 206120 n - 7426999 n - 48114645 n - 117300972 n) a(n + 3)/((n + 8) %1) + ( 5 4 3 2 -105249600 + 36416 n - 275864 n - 9200297 n - 133804326 n - 56990773 n ) a(n + 2)/((n + 8) %1) + 30 (n + 2) 4 3 2 (1024 n - 10304 n - 236856 n - 1020355 n - 1133872) a(n + 1)/((n + 8) %1 4 3 2 180 (n + 1) (64 n - 600 n - 15473 n - 73449 n - 96080) a(n) ) + -------------------------------------------------------------- (n + 8) %1 4 3 2 %1 := 64 n - 856 n - 13289 n - 44559 n - 37440 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1275, a[5] = 7595, a[6] = 45243, a[7] = 269517, a[8] = 1605575 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 859 15588067\ 4.15 6 (1/n) |1 + ---- + --------| | 32 n 2 | \ 6144 n / Note that everything is rigorous except the constant in front, 4.15 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t^2-(6*t-1)*(t+1)*(6*t^5+11*t^4+12*t^3+9*t^2+4*t-1)*P-(6*t-1)*(5*t^2+5*t-1)*(6 *t^5+11*t^4+12*t^3+9*t^2+4*t-1)*P^2, [-180*(n+1)*(64*n^4-600*n^3-15473*n^2-\ 73449*n-96080)/(n+8)/(64*n^4-856*n^3-13289*n^2-44559*n-37440)-30*(n+2)*(1024*n^ 4-10304*n^3-236856*n^2-1020355*n-1133872)/(n+8)/(64*n^4-856*n^3-13289*n^2-44559 *n-37440)*N-(-105249600+36416*n^5-275864*n^4-9200297*n^3-133804326*n-56990773*n ^2)/(n+8)/(64*n^4-856*n^3-13289*n^2-44559*n-37440)*N^2-(-91691136+29120*n^5-\ 206120*n^4-7426999*n^3-48114645*n^2-117300972*n)/(n+8)/(64*n^4-856*n^3-13289*n^ 2-44559*n-37440)*N^3-4*(-10354272+3584*n^5-23552*n^4-921534*n^3-6002953*n^2-\ 14226866*n)/(n+8)/(64*n^4-856*n^3-13289*n^2-44559*n-37440)*N^4+(-11644992+1088* n^5-6680*n^4-389537*n^3-3325453*n^2-10663848*n)/(n+8)/(64*n^4-856*n^3-13289*n^2 -44559*n-37440)*N^5+2*(-6172704+1920*n^5-11600*n^4-538378*n^3-3706388*n^2-\ 8971635*n)/(n+8)/(64*n^4-856*n^3-13289*n^2-44559*n-37440)*N^6-(-3900480+960*n^5 -5480*n^4-286943*n^3-2076379*n^2-5276758*n)/(n+8)/(64*n^4-856*n^3-13289*n^2-\ 44559*n-37440)*N^7+N^8, [6, 36, 214, 1275, 7595, 45243, 269517, 1605575]], 4.15 *6^n*(1/n)^(1/2)*(1+859/32/n+15588067/6144/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1275, 7595, 45243, 269517, 1605575, 9564996, 56983449, 339486440, 2022580880, 12050335365, 71796313640, 427774546709, 2548810334214, 15186926429409, 90492388857286, 539217545161530, 3213111362880896, 19146850454109113, 114098180641991805, 679938922634089233, 4052013628034808309, 24148029891916357791, 143913762109677998651, 857694772664274920364, 5111790139159989594750, 30466536441223095566672, 181586254043065071032079, 1082312490220849123080936, 6451076413952815256244846, 38452228650981600043330954, 229203190527056627418807720, 1366248230260145100952843521, 8144198964873647930837006814, 48548626122476289985116378396, 289411189026407211300036382391, 1725295508102748470956858334341, 10285406563192865634138846215982, 61318176146803583670250042997020, 365566830956863543406080378534239, 2179486205497404386186558380485384, 12994250837764206068053781530885269, 77474386017081274994422303295286180, 461928555863118200107337379915277257, 2754236584941685081486735616620397286, 16422430538054995844632312695828653806, 97922699341237457290562884495199504721, 583900793630823047333988650502186411988] Proposition Number , 2, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 1] as those of, [1, 2, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 3 4 5 6 18 t + 8 t + 140 t + 112 t + 120 t + 1 - 10 t 3 2 3 2 + (6 t - 1) (20 t + 2 t + 5 t - 1) (10 t + 8 t + 9 t - 2) P 3 2 3 2 2 + (6 t - 1) (5 t - 1) (5 t + 4 t + 5 t - 1) (20 t + 2 t + 5 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 3 2 4 (-380546 + 325 n - 2580 n - 88389 n) a(n + 8) a(n + 9) = ------------------------------------------------- (n + 8) %1 3 2 2 (4485 n - 36202 n - 4890384 - 1169839 n) a(n + 7) - ----------------------------------------------------- (n + 8) %1 3 2 (-5643874 n + 22425 n - 179125 n - 23489176) a(n + 6) + ------------------------------------------------------- (n + 8) %1 2 3 2 (-26531 n - 1310075 n + 4680 n - 7272084) a(n + 5) - ------------------------------------------------------ (n + 8) %1 2 3 (-323805 n - 9789688 n + 37375 n - 42831312) a(n + 4) + ------------------------------------------------------- (n + 8) %1 3 2 (-109115616 + 122005 n - 1032271 n - 27738648 n) a(n + 3) - ----------------------------------------------------------- (n + 8) %1 3 2 10 (-15983084 + 30550 n - 324225 n - 6396227 n) a(n + 2) - ---------------------------------------------------------- (n + 8) %1 3 2 100 (-1822494 + 2990 n - 30743 n - 644117 n) a(n + 1) - ------------------------------------------------------- (n + 8) %1 2 1500 (2 n + 5) (65 n - 893 n - 11210) a(n) - ------------------------------------------- (n + 8) %1 2 %1 := 65 n - 1023 n - 10252 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1273, a[5] = 7577, a[6] = 45095, a[7] = 268386, a[8] = 1597383, a[9] = 9507595 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 1443 13288537\ 3.84 6 (1/n) |1 + ---- + --------| | 56 n 2 | \ 6272 n / Note that everything is rigorous except the constant in front, 3.84 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format 18*t^2+8*t^3+140*t^4+112*t^5+120*t^6+1-10*t+(6*t-1)*(20*t^3+2*t^2+5*t-1)*(10*t^ 3+8*t^2+9*t-2)*P+(6*t-1)*(5*t-1)*(5*t^3+4*t^2+5*t-1)*(20*t^3+2*t^2+5*t-1)*P^2, [1500*(2*n+5)*(65*n^2-893*n-11210)/(n+8)/(65*n^2-1023*n-10252)+100*(-1822494+ 2990*n^3-30743*n^2-644117*n)/(n+8)/(65*n^2-1023*n-10252)*N+10*(-15983084+30550* n^3-324225*n^2-6396227*n)/(n+8)/(65*n^2-1023*n-10252)*N^2+(-109115616+122005*n^ 3-1032271*n^2-27738648*n)/(n+8)/(65*n^2-1023*n-10252)*N^3-(-323805*n^2-9789688* n+37375*n^3-42831312)/(n+8)/(65*n^2-1023*n-10252)*N^4+2*(-26531*n^2-1310075*n+ 4680*n^3-7272084)/(n+8)/(65*n^2-1023*n-10252)*N^5-(-5643874*n+22425*n^3-179125* n^2-23489176)/(n+8)/(65*n^2-1023*n-10252)*N^6+2*(4485*n^3-36202*n^2-4890384-\ 1169839*n)/(n+8)/(65*n^2-1023*n-10252)*N^7-4*(-380546+325*n^3-2580*n^2-88389*n) /(n+8)/(65*n^2-1023*n-10252)*N^8+N^9, [6, 36, 214, 1273, 7577, 45095, 268386, 1597383, 9507595]], 3.84*6^n*(1/n)^(1/2)*(1+1443/56/n+13288537/6272/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1273, 7577, 45095, 268386, 1597383, 9507595, 56590566, 336844684, 2005061248, 11935423671, 71049381678, 422955856163, 2517920796469, 14989991140579, 89242757794198, 531320887104805, 3163392647535088, 18834826422749082, 112145656293186105, 667752631318133133, 3976134608452199057, 23676574575443590960, 140990219323407456917, 839598059029661374904, 4999956094810383588232, 29776474792899788489419, 177334279808303787254675, 1056147156341077374384921, 6290258523359428573998128, 37464925732141011655339024, 223148296041649889907209054, 1329151814587540019936595629, 7917132498935664718460369910, 47159972691246302824917968045, 280925704721727804229896985629, 1673484561694122322311379093768, 9969290079431908895104395946108, 59390781633851288866949836810640, 353823054666947680666424315602241, 2107975143161804305715231666132620, 12559059178688862358786839348246405, 74827456560323444482476522229201009, 445838025377243508726974045621792522, 2656473370514950628745463325614305358, 15828729382565770176385781657370307780, 94318928553069217471465705820552576991, 562035656854794955972936354256229138736] Proposition Number , 3, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 1] as those of, [1, 2, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 5 4 3 2 -t - (6 t - 1) (6 t + 7 t + 12 t + 9 t + 4 t - 1) P 2 5 4 3 2 2 + (6 t - 1) (t - 6 t + 1) (6 t + 7 t + 12 t + 9 t + 4 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (21511078 + 663359 n + 8085378 n) a(n + 8) a(n + 9) = ------------------------------------------- (n + 9) (30393 n + 123649) 2 2 (2571502 n + 29025484 n + 72725633) a(n + 7) - ----------------------------------------------- (n + 9) (30393 n + 123649) 2 (157155909 n + 364767620 + 15220190 n ) a(n + 6) + ------------------------------------------------ (n + 9) (30393 n + 123649) 2 31 (162662 n + 1363761 n + 2390263) a(n + 5) - --------------------------------------------- (n + 9) (30393 n + 123649) 2 2 (10486798 n + 88821980 n + 164655247) a(n + 4) - ------------------------------------------------- (n + 9) (30393 n + 123649) 2 (51750957 n + 439777283 n + 949836558) a(n + 3) - ------------------------------------------------ (n + 9) (30393 n + 123649) 2 (14892425 n + 108462577 n + 192106728) a(n + 2) - ------------------------------------------------ (n + 9) (30393 n + 123649) 2 18 (19766397 + 1831496 n + 13946483 n) a(n + 1) - ------------------------------------------------ (n + 9) (30393 n + 123649) 36 (177071 n + 997012) (n + 2) a(n) + ----------------------------------- (n + 9) (30393 n + 123649) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1273, a[5] = 7574, a[6] = 45062, a[7] = 268110, a[8] = 1595250, a[9] = 9492004 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 2559 46105945\ 3.84 6 (1/n) |1 + ----- + --------| | 112 n 2| \ 25088 n / Note that everything is rigorous except the constant in front, 3.84 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t^2-(6*t-1)*(6*t^5+7*t^4+12*t^3+9*t^2+4*t-1)*P+(6*t-1)*(t^2-6*t+1)*(6*t^5+7*t^ 4+12*t^3+9*t^2+4*t-1)*P^2, [-36*(177071*n+997012)*(n+2)/(n+9)/(30393*n+123649)+ 18*(19766397+1831496*n^2+13946483*n)/(n+9)/(30393*n+123649)*N+(14892425*n^2+ 108462577*n+192106728)/(n+9)/(30393*n+123649)*N^2+(51750957*n^2+439777283*n+ 949836558)/(n+9)/(30393*n+123649)*N^3+2*(10486798*n^2+88821980*n+164655247)/(n+ 9)/(30393*n+123649)*N^4+31*(162662*n^2+1363761*n+2390263)/(n+9)/(30393*n+123649 )*N^5-(157155909*n+364767620+15220190*n^2)/(n+9)/(30393*n+123649)*N^6+2*( 2571502*n^2+29025484*n+72725633)/(n+9)/(30393*n+123649)*N^7-(21511078+663359*n^ 2+8085378*n)/(n+9)/(30393*n+123649)*N^8+N^9, [6, 36, 214, 1273, 7574, 45062, 268110, 1595250, 9492004]], 3.84*6^n*(1/n)^(1/2)*(1+2559/112/n+46105945/25088/n ^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1273, 7574, 45062, 268110, 1595250, 9492004, 56480766, 336090880, 1999983336, 11901714640, 70828199789, 421518262298, 2508650231539, 14930603614930, 88864470039872, 528923041681082, 3148258350786655, 18739663575515922, 111549278836861835, 664026324390995716, 3952914131613161272, 23532226415863102560, 140094862845936045405, 834055509931806367690, 4965708856099303212084, 29565218960324252086444, 176033163393026985224111, 1048145154133613619195574, 6241111013065373298177688, 37163441026291141584218312, 221301042811405350941881302, 1317845582554896758912023124, 7848002033422966226635756764, 46737685132227116231612408056, 278348440193175026916791104887, 1657768483963320398752557580630, 9873530006147684345491975259166, 58807740917032404938676062335884, 350275697718551763903120276224919, 2086406709111542637885459114415690, 12428003764245188557321882167242656, 74031612888889947903306528991904900, 441007995280444883091753341614411358, 2627175699045958411452207123542237020, 15651110684891073479864618035129269196, 93242644427357589530219845265672471312, 555517008491465768015201840816261992337] Proposition Number , 4, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 1] as those of, [1, 2, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 5 4 3 2 -t - (6 t - 1) (6 t + 7 t + 12 t + 9 t + 4 t - 1) P 2 5 4 3 2 2 + (6 t - 1) (t - 6 t + 1) (6 t + 7 t + 12 t + 9 t + 4 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (21511078 + 663359 n + 8085378 n) a(n + 8) a(n + 9) = ------------------------------------------- (n + 9) (30393 n + 123649) 2 2 (2571502 n + 29025484 n + 72725633) a(n + 7) - ----------------------------------------------- (n + 9) (30393 n + 123649) 2 (157155909 n + 364767620 + 15220190 n ) a(n + 6) + ------------------------------------------------ (n + 9) (30393 n + 123649) 2 31 (162662 n + 1363761 n + 2390263) a(n + 5) - --------------------------------------------- (n + 9) (30393 n + 123649) 2 2 (10486798 n + 88821980 n + 164655247) a(n + 4) - ------------------------------------------------- (n + 9) (30393 n + 123649) 2 (51750957 n + 439777283 n + 949836558) a(n + 3) - ------------------------------------------------ (n + 9) (30393 n + 123649) 2 (14892425 n + 108462577 n + 192106728) a(n + 2) - ------------------------------------------------ (n + 9) (30393 n + 123649) 2 18 (19766397 + 1831496 n + 13946483 n) a(n + 1) - ------------------------------------------------ (n + 9) (30393 n + 123649) 36 (177071 n + 997012) (n + 2) a(n) + ----------------------------------- (n + 9) (30393 n + 123649) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1273, a[5] = 7574, a[6] = 45062, a[7] = 268110, a[8] = 1595250, a[9] = 9492004 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 2559 46105945\ 3.84 6 (1/n) |1 + ----- + --------| | 112 n 2| \ 25088 n / Note that everything is rigorous except the constant in front, 3.84 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t^2-(6*t-1)*(6*t^5+7*t^4+12*t^3+9*t^2+4*t-1)*P+(6*t-1)*(t^2-6*t+1)*(6*t^5+7*t^ 4+12*t^3+9*t^2+4*t-1)*P^2, [-36*(177071*n+997012)*(n+2)/(n+9)/(30393*n+123649)+ 18*(19766397+1831496*n^2+13946483*n)/(n+9)/(30393*n+123649)*N+(14892425*n^2+ 108462577*n+192106728)/(n+9)/(30393*n+123649)*N^2+(51750957*n^2+439777283*n+ 949836558)/(n+9)/(30393*n+123649)*N^3+2*(10486798*n^2+88821980*n+164655247)/(n+ 9)/(30393*n+123649)*N^4+31*(162662*n^2+1363761*n+2390263)/(n+9)/(30393*n+123649 )*N^5-(157155909*n+364767620+15220190*n^2)/(n+9)/(30393*n+123649)*N^6+2*( 2571502*n^2+29025484*n+72725633)/(n+9)/(30393*n+123649)*N^7-(21511078+663359*n^ 2+8085378*n)/(n+9)/(30393*n+123649)*N^8+N^9, [6, 36, 214, 1273, 7574, 45062, 268110, 1595250, 9492004]], 3.84*6^n*(1/n)^(1/2)*(1+2559/112/n+46105945/25088/n ^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1273, 7574, 45062, 268110, 1595250, 9492004, 56480766, 336090880, 1999983336, 11901714640, 70828199789, 421518262298, 2508650231539, 14930603614930, 88864470039872, 528923041681082, 3148258350786655, 18739663575515922, 111549278836861835, 664026324390995716, 3952914131613161272, 23532226415863102560, 140094862845936045405, 834055509931806367690, 4965708856099303212084, 29565218960324252086444, 176033163393026985224111, 1048145154133613619195574, 6241111013065373298177688, 37163441026291141584218312, 221301042811405350941881302, 1317845582554896758912023124, 7848002033422966226635756764, 46737685132227116231612408056, 278348440193175026916791104887, 1657768483963320398752557580630, 9873530006147684345491975259166, 58807740917032404938676062335884, 350275697718551763903120276224919, 2086406709111542637885459114415690, 12428003764245188557321882167242656, 74031612888889947903306528991904900, 441007995280444883091753341614411358, 2627175699045958411452207123542237020, 15651110684891073479864618035129269196, 93242644427357589530219845265672471312, 555517008491465768015201840816261992337] Proposition Number , 5, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 1] as those of, [2, 1, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 2 3 4 8 6 7 9 1 + 199 t + 7 t + 37 t + 126 t + 112 t + 233 t - 9 t + 202 t + 48 t + 3 2 (6 t - 1) (8 t + 4 t + 5 t - 1) 6 5 4 3 2 (8 t + 16 t + 24 t + 24 t + 15 t + 7 t - 2) P + (6 t - 1) 3 2 4 3 2 (4 t + 4 t + 4 t - 1) (4 t + 4 t + 4 t + 5 t - 1) 3 2 2 (8 t + 4 t + 5 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence a(n + 13) = ( 2 3 5 4 -208413678 - 85459321 n - 214063928 n - 8005890 n + 33440 n + 207337 n ) 2 a(n + 12)/((n + 12) %1) - (-1265918076 - 502820244 n - 1252732550 n 3 5 4 - 48070599 n + 209440 n + 1261717 n ) a(n + 11)/((n + 12) %1) + ( 2 3 4 -2523917586 - 867529511 n - 2167452244 n - 86075466 n + 2316915 n 5 + 396000 n ) a(n + 10)/((n + 12) %1) + 2 ( 2 3 5 4 47850054 - 495992354 n - 238422423 n - 17800122 n + 80960 n + 740778 n ) 2 a(n + 9)/((n + 12) %1) + 4 (-1434412500 - 1525268623 n - 575686810 n 3 5 4 - 66476344 n + 299200 n + 763550 n ) a(n + 8)/((n + 12) %1) - 8 ( 2 3 5 -1628844216 - 1667436683 n - 693421261 n - 73003621 n + 373120 n 4 + 1837596 n ) a(n + 7)/((n + 12) %1) - 16 (-1888056996 - 2359960853 n 2 3 5 4 - 982526088 n - 105234017 n + 529760 n + 2273303 n ) a(n + 6)/((n + 12) 2 3 %1) - 16 (-4027964166 - 4889147900 n - 1972289701 n - 225829457 n 5 4 + 1133440 n + 3721612 n ) a(n + 5)/((n + 12) %1) - 16 (-4979934918 2 3 5 4 - 6071558864 n - 2422596453 n - 284987738 n + 1430880 n + 4128919 n ) 3 2 a(n + 4)/((n + 12) %1) - 32 (-2352708870 - 136206476 n - 1119654805 n 5 4 - 2844066472 n + 686400 n + 1679410 n ) a(n + 3)/((n + 12) %1) - 64 ( 2 3 5 -864224526 - 393745295 n - 1013642701 n - 49188412 n + 248160 n 4 + 527863 n ) a(n + 2)/((n + 12) %1) - 128 ( 3 2 5 4 -204096750 - 11790649 n - 91261252 n - 237941763 n + 59840 n + 103782 n ) a(n + 1)/((n + 12) %1) 4 3 2 1536 (n + 3) (1760 n - 2797 n - 338876 n - 1581327 n - 2204118) a(n) - ----------------------------------------------------------------------- (n + 12) %1 4 3 2 %1 := 1760 n - 9837 n - 319925 n - 919006 n - 957110 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1273, a[5] = 7573, a[6] = 45053, a[7] = 268036, a[8] = 1594687, a[9] = 9487926, a[10] = 56452163, a[11] = 335894846, a[12] = 1998662959, a[13] = 11892941089 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 271 7481629\ 3.75 6 (1/n) |1 + ---- + -------| | 11 n 2| \ 3872 n / Note that everything is rigorous except the constant in front, 3.75 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format 1+199*t^5+7*t^2+37*t^3+126*t^4+112*t^8+233*t^6-9*t+202*t^7+48*t^9+(6*t-1)*(8*t^ 3+4*t^2+5*t-1)*(8*t^6+16*t^5+24*t^4+24*t^3+15*t^2+7*t-2)*P+(6*t-1)*(4*t^3+4*t^2 +4*t-1)*(4*t^4+4*t^3+4*t^2+5*t-1)*(8*t^3+4*t^2+5*t-1)*P^2, [1536*(n+3)*(1760*n^ 4-2797*n^3-338876*n^2-1581327*n-2204118)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-\ 919006*n-957110)+128*(-204096750-11790649*n^3-91261252*n^2-237941763*n+59840*n^ 5+103782*n^4)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-919006*n-957110)*N+64*(-\ 864224526-393745295*n^2-1013642701*n-49188412*n^3+248160*n^5+527863*n^4)/(n+12) /(1760*n^4-9837*n^3-319925*n^2-919006*n-957110)*N^2+32*(-2352708870-136206476*n ^3-1119654805*n^2-2844066472*n+686400*n^5+1679410*n^4)/(n+12)/(1760*n^4-9837*n^ 3-319925*n^2-919006*n-957110)*N^3+16*(-4979934918-6071558864*n-2422596453*n^2-\ 284987738*n^3+1430880*n^5+4128919*n^4)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-\ 919006*n-957110)*N^4+16*(-4027964166-4889147900*n-1972289701*n^2-225829457*n^3+ 1133440*n^5+3721612*n^4)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-919006*n-957110)* N^5+16*(-1888056996-2359960853*n-982526088*n^2-105234017*n^3+529760*n^5+2273303 *n^4)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-919006*n-957110)*N^6+8*(-1628844216-\ 1667436683*n-693421261*n^2-73003621*n^3+373120*n^5+1837596*n^4)/(n+12)/(1760*n^ 4-9837*n^3-319925*n^2-919006*n-957110)*N^7-4*(-1434412500-1525268623*n-\ 575686810*n^2-66476344*n^3+299200*n^5+763550*n^4)/(n+12)/(1760*n^4-9837*n^3-\ 319925*n^2-919006*n-957110)*N^8-2*(47850054-495992354*n-238422423*n^2-17800122* n^3+80960*n^5+740778*n^4)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-919006*n-957110) *N^9-(-2523917586-867529511*n^2-2167452244*n-86075466*n^3+2316915*n^4+396000*n^ 5)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-919006*n-957110)*N^10+(-1265918076-\ 502820244*n^2-1252732550*n-48070599*n^3+209440*n^5+1261717*n^4)/(n+12)/(1760*n^ 4-9837*n^3-319925*n^2-919006*n-957110)*N^11-(-208413678-85459321*n^2-214063928* n-8005890*n^3+33440*n^5+207337*n^4)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-919006 *n-957110)*N^12+N^13, [6, 36, 214, 1273, 7573, 45053, 268036, 1594687, 9487926, 56452163, 335894846, 1998662959, 11892941089]], 3.75*6^n*(1/n)^(1/2)*(1+271/11/ n+7481629/3872/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1273, 7573, 45053, 268036, 1594687, 9487926, 56452163, 335894846, 1998662959, 11892941089, 70770531987, 421142576672, 2506220883021, 14914992890145, 88764697038861, 528288337795698, 3144237196010719, 18714279531259255, 111389552886262979, 663024153808359008, 3946642446275922777, 23493069319360564256, 139850906044999524317, 832538556696794682710, 4956293021457469968469, 29506869755699052089195, 175672123224746539862225, 1045914308799017599335932, 6227344611719853607122951, 37078591858561949148826678, 220778663215719780598816699, 1314632897575651306052068374, 7828263163567916082082249303, 46616520772288437795208585575, 277605335638514358589174224707, 1653214732938488708506272985496, 9845646135842713682690825326537, 58637124938973777987314050727430, 349232452923131372130012846753253, 2080031883108474902671476612600534, 12389074109465427506658338938936953, 73794018410588146665952691216632391, 439558728913037302136557167417357189, 2618340266850621290000506984192198204, 15597273054090546082081965057667011011, 92914750963605862383282307882322533473, 553520929521016796369646420444959132443] Proposition Number , 6, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 1] as those of, [2, 1, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 3 2 -t - (6 t - 1) (t + 1) (2 t + 5 t + 5 t - 1) P 3 2 4 3 2 2 + (6 t - 1) (2 t + 5 t + 5 t - 1) (t + t - 5 t - 5 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 3 (-1582782 - 411573 n - 4010 n + 2325 n ) a(n + 9) a(n + 10) = -------------------------------------------------- (n + 10) %1 2 3 2 (-2807719 - 759514 n - 8795 n + 4650 n ) a(n + 8) - ---------------------------------------------------- (n + 10) %1 2 3 2 (-1258715 - 278394 n - 6374 n + 1395 n ) a(n + 7) - ---------------------------------------------------- (n + 10) %1 2 3 (-19237936 - 5218845 n - 75934 n + 35495 n ) a(n + 6) + ------------------------------------------------------ (n + 10) %1 2 3 2 (-17179595 - 5104556 n - 148370 n + 37975 n ) a(n + 5) + --------------------------------------------------------- (n + 10) %1 2 3 2 (-12680691 - 4349991 n - 186998 n + 35340 n ) a(n + 4) + --------------------------------------------------------- (n + 10) %1 2 3 (-5265254 - 2556857 n - 168808 n + 23715 n ) a(n + 3) + ------------------------------------------------------ (n + 10) %1 2 3 (-3682790 - 976227 n - 21950 n + 6975 n ) a(n + 2) - --------------------------------------------------- (n + 10) %1 3 2 4 (-629041 + 2015 n - 11773 n - 238116 n) a(n + 1) - ---------------------------------------------------- (n + 10) %1 2 12 (n + 3) (155 n - 1466 n - 13220) a(n) - ----------------------------------------- (n + 10) %1 2 %1 := 155 n - 1776 n - 11599 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1273, a[5] = 7572, a[6] = 45041, a[7] = 267929, a[8] = 1593843, a[9] = 9481685, a[10] = 56407868 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 3833 103420369\ 3.79 6 (1/n) |1 + ----- + ---------| | 172 n 2 | \ 59168 n / Note that everything is rigorous except the constant in front, 3.79 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t^2-(6*t-1)*(t+1)*(2*t^3+5*t^2+5*t-1)*P+(6*t-1)*(2*t^3+5*t^2+5*t-1)*(t^4+t^3-5 *t^2-5*t+1)*P^2, [12*(n+3)*(155*n^2-1466*n-13220)/(n+10)/(155*n^2-1776*n-11599) +4*(-629041+2015*n^3-11773*n^2-238116*n)/(n+10)/(155*n^2-1776*n-11599)*N+(-\ 3682790-976227*n-21950*n^2+6975*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^2-(-\ 5265254-2556857*n-168808*n^2+23715*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^3-2*(-\ 12680691-4349991*n-186998*n^2+35340*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^4-2*(-\ 17179595-5104556*n-148370*n^2+37975*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^5-(-\ 19237936-5218845*n-75934*n^2+35495*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^6+2*(-\ 1258715-278394*n-6374*n^2+1395*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^7+2*(-\ 2807719-759514*n-8795*n^2+4650*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^8-(-1582782 -411573*n-4010*n^2+2325*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^9+N^10, [6, 36, 214, 1273, 7572, 45041, 267929, 1593843, 9481685, 56407868]], 3.79*6^n*(1/n)^(1 /2)*(1+3833/172/n+103420369/59168/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1273, 7572, 45041, 267929, 1593843, 9481685, 56407868, 335589246, 1996597773, 11879203848, 70680286435, 420555664820, 2502435546783, 14890749111356, 88610345507530, 527310665321892, 3138072163062357, 18675556237137043, 111147171676637102, 661511712339396123, 3937231154609263917, 23434653734463592512, 139489147824652923342, 830302895060018001914, 4942502893308353232860, 29421957129880376879802, 175150115123589035260346, 1042709989715135350811353, 6207702209213205662786286, 36958338932950844023526812, 220043343040269273819995531, 1310141608754561115333149613, 7800859425634664862798629485, 46449480665861665170879283165, 276588082228913309896267632850, 1647025196865859027686565660971, 9808016610944577293011095782954, 58408533235568509504444080222774, 347844831086131207800631021100237, 2071614498635243943817749327933023, 12338047845797002409309649719166674, 73484892793814214858394089870135504, 437687125276797581259023251314683450, 2607015154203959410648331688441067866, 15528782258081082723492147706522004727, 92500757550297218186499732427921899461, 551019801123658115216634578409787565816] Proposition Number , 7, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 1] as those of, [2, 2, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 4 2 3 5 6 1 - 48 t - 2 t - 17 t - 39 t - 34 t - 12 t 2 3 2 - 2 (6 t - 1) (t + 1) (t + t + 1) (8 t + 10 t + 4 t - 1) P 3 2 4 3 2 2 - (6 t - 1) (8 t + 10 t + 4 t - 1) (4 t + 9 t + 9 t + 4 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (47 + 6 n) a(n + 9) (164 + 19 n) a(n + 8) a(n + 10) = --------------------- - --------------------- n + 8 n + 8 (936 + 139 n) a(n + 7) 2 (-27 + 22 n) a(n + 6) - ---------------------- - ----------------------- n + 8 n + 8 2 (3073 + 496 n) a(n + 5) 2 (7373 + 1424 n) a(n + 4) + ------------------------- + -------------------------- n + 8 n + 8 12 (1517 + 337 n) a(n + 3) 4 (3359 + 846 n) a(n + 2) + -------------------------- + ------------------------- n + 8 n + 8 16 (361 + 104 n) a(n + 1) 384 (n + 3) a(n) + ------------------------- + ---------------- n + 8 n + 8 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1274, a[5] = 7584, a[6] = 45148, a[7] = 268774, a[8] = 1600098, a[9] = 9526118, a[10] = 56714672 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 7171 269642137\ 3.52 6 (1/n) |1 + ----- + ---------| | 200 n 2 | \ 80000 n / Note that everything is rigorous except the constant in front, 3.52 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format 1-48*t^4-2*t-17*t^2-39*t^3-34*t^5-12*t^6-2*(6*t-1)*(t+1)*(t^2+t+1)*(8*t^3+10*t^ 2+4*t-1)*P-(6*t-1)*(8*t^3+10*t^2+4*t-1)*(4*t^4+9*t^3+9*t^2+4*t-1)*P^2, [-384*(n +3)/(n+8)-16*(361+104*n)/(n+8)*N-4*(3359+846*n)/(n+8)*N^2-12*(1517+337*n)/(n+8) *N^3-2*(7373+1424*n)/(n+8)*N^4-2*(3073+496*n)/(n+8)*N^5+2*(-27+22*n)/(n+8)*N^6+ (936+139*n)/(n+8)*N^7+(164+19*n)/(n+8)*N^8-2*(47+6*n)/(n+8)*N^9+N^10, [6, 36, 214, 1274, 7584, 45148, 268774, 1600098, 9526118, 56714672]], 3.52*6^n*(1/n)^(1 /2)*(1+7171/200/n+269642137/80000/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1274, 7584, 45148, 268774, 1600098, 9526118, 56714672, 337664204, 2010410558, 11970013460, 71271300400, 424370221110, 2526883996828, 15046514442602, 89597670082178, 533540935311598, 3177232565227472, 18920843990682384, 112678810668496438, 671049045943956680, 3996469847271244150, 23801761286465442710, 141759422534886323108, 844316096947432278272, 5028847944867518323738, 29953132512217006226088, 178412922333428979668002, 1062724487784044951390034, 6330316106129670349772030, 37708602756823419622034580, 224628999775481267880630120, 1338139993091783067400442866, 7971639076538227328857100252, 47490206468859197963117206426, 282924688044680226162643015266, 1685574664670728365808710949406, 10042353585769620083196065318932, 59831982001285290688068553917312, 356485311776353484651838357263674, 2124028124985455186642712768531716, 12655789987922071955912886682762466, 75409946184821854129964222478533542, 449343411535545394327601505099889942, 2677555627571803760065501349379139040, 15955448009724998540205210267858964836, 95080154939266410212403948168209604846, 566605954944728764403465519296647500730] Proposition Number , 8, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 1] as those of, [2, 2, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 3 2 -t - (6 t - 1) (t + 1) (2 t + 5 t + 5 t - 1) P 3 2 4 3 2 2 + (6 t - 1) (2 t + 5 t + 5 t - 1) (t + t - 5 t - 5 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 3 (-1582782 - 411573 n - 4010 n + 2325 n ) a(n + 9) a(n + 10) = -------------------------------------------------- (n + 10) %1 2 3 2 (-2807719 - 759514 n - 8795 n + 4650 n ) a(n + 8) - ---------------------------------------------------- (n + 10) %1 2 3 2 (-1258715 - 278394 n - 6374 n + 1395 n ) a(n + 7) - ---------------------------------------------------- (n + 10) %1 2 3 (-19237936 - 5218845 n - 75934 n + 35495 n ) a(n + 6) + ------------------------------------------------------ (n + 10) %1 2 3 2 (-17179595 - 5104556 n - 148370 n + 37975 n ) a(n + 5) + --------------------------------------------------------- (n + 10) %1 2 3 2 (-12680691 - 4349991 n - 186998 n + 35340 n ) a(n + 4) + --------------------------------------------------------- (n + 10) %1 2 3 (-5265254 - 2556857 n - 168808 n + 23715 n ) a(n + 3) + ------------------------------------------------------ (n + 10) %1 2 3 (-3682790 - 976227 n - 21950 n + 6975 n ) a(n + 2) - --------------------------------------------------- (n + 10) %1 3 2 4 (-629041 + 2015 n - 11773 n - 238116 n) a(n + 1) - ---------------------------------------------------- (n + 10) %1 2 12 (n + 3) (155 n - 1466 n - 13220) a(n) - ----------------------------------------- (n + 10) %1 2 %1 := 155 n - 1776 n - 11599 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1273, a[5] = 7572, a[6] = 45041, a[7] = 267929, a[8] = 1593843, a[9] = 9481685, a[10] = 56407868 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 3833 103420369\ 3.79 6 (1/n) |1 + ----- + ---------| | 172 n 2 | \ 59168 n / Note that everything is rigorous except the constant in front, 3.79 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t^2-(6*t-1)*(t+1)*(2*t^3+5*t^2+5*t-1)*P+(6*t-1)*(2*t^3+5*t^2+5*t-1)*(t^4+t^3-5 *t^2-5*t+1)*P^2, [12*(n+3)*(155*n^2-1466*n-13220)/(n+10)/(155*n^2-1776*n-11599) +4*(-629041+2015*n^3-11773*n^2-238116*n)/(n+10)/(155*n^2-1776*n-11599)*N+(-\ 3682790-976227*n-21950*n^2+6975*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^2-(-\ 5265254-2556857*n-168808*n^2+23715*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^3-2*(-\ 12680691-4349991*n-186998*n^2+35340*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^4-2*(-\ 17179595-5104556*n-148370*n^2+37975*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^5-(-\ 19237936-5218845*n-75934*n^2+35495*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^6+2*(-\ 1258715-278394*n-6374*n^2+1395*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^7+2*(-\ 2807719-759514*n-8795*n^2+4650*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^8-(-1582782 -411573*n-4010*n^2+2325*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^9+N^10, [6, 36, 214, 1273, 7572, 45041, 267929, 1593843, 9481685, 56407868]], 3.79*6^n*(1/n)^(1 /2)*(1+3833/172/n+103420369/59168/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1273, 7572, 45041, 267929, 1593843, 9481685, 56407868, 335589246, 1996597773, 11879203848, 70680286435, 420555664820, 2502435546783, 14890749111356, 88610345507530, 527310665321892, 3138072163062357, 18675556237137043, 111147171676637102, 661511712339396123, 3937231154609263917, 23434653734463592512, 139489147824652923342, 830302895060018001914, 4942502893308353232860, 29421957129880376879802, 175150115123589035260346, 1042709989715135350811353, 6207702209213205662786286, 36958338932950844023526812, 220043343040269273819995531, 1310141608754561115333149613, 7800859425634664862798629485, 46449480665861665170879283165, 276588082228913309896267632850, 1647025196865859027686565660971, 9808016610944577293011095782954, 58408533235568509504444080222774, 347844831086131207800631021100237, 2071614498635243943817749327933023, 12338047845797002409309649719166674, 73484892793814214858394089870135504, 437687125276797581259023251314683450, 2607015154203959410648331688441067866, 15528782258081082723492147706522004727, 92500757550297218186499732427921899461, 551019801123658115216634578409787565816] Proposition Number , 9, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 1] as those of, [2, 3, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 2 3 4 8 6 7 9 1 + 199 t + 7 t + 37 t + 126 t + 112 t + 233 t - 9 t + 202 t + 48 t + 3 2 (6 t - 1) (8 t + 4 t + 5 t - 1) 6 5 4 3 2 (8 t + 16 t + 24 t + 24 t + 15 t + 7 t - 2) P + (6 t - 1) 3 2 4 3 2 (4 t + 4 t + 4 t - 1) (4 t + 4 t + 4 t + 5 t - 1) 3 2 2 (8 t + 4 t + 5 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence a(n + 13) = ( 2 3 5 4 -208413678 - 85459321 n - 214063928 n - 8005890 n + 33440 n + 207337 n ) 2 a(n + 12)/((n + 12) %1) - (-1265918076 - 502820244 n - 1252732550 n 3 5 4 - 48070599 n + 209440 n + 1261717 n ) a(n + 11)/((n + 12) %1) + ( 2 3 4 -2523917586 - 867529511 n - 2167452244 n - 86075466 n + 2316915 n 5 + 396000 n ) a(n + 10)/((n + 12) %1) + 2 ( 2 3 5 4 47850054 - 495992354 n - 238422423 n - 17800122 n + 80960 n + 740778 n ) 2 a(n + 9)/((n + 12) %1) + 4 (-1434412500 - 1525268623 n - 575686810 n 3 5 4 - 66476344 n + 299200 n + 763550 n ) a(n + 8)/((n + 12) %1) - 8 ( 2 3 5 -1628844216 - 1667436683 n - 693421261 n - 73003621 n + 373120 n 4 + 1837596 n ) a(n + 7)/((n + 12) %1) - 16 (-1888056996 - 2359960853 n 2 3 5 4 - 982526088 n - 105234017 n + 529760 n + 2273303 n ) a(n + 6)/((n + 12) 2 3 %1) - 16 (-4027964166 - 4889147900 n - 1972289701 n - 225829457 n 5 4 + 1133440 n + 3721612 n ) a(n + 5)/((n + 12) %1) - 16 (-4979934918 2 3 5 4 - 6071558864 n - 2422596453 n - 284987738 n + 1430880 n + 4128919 n ) 3 2 a(n + 4)/((n + 12) %1) - 32 (-2352708870 - 136206476 n - 1119654805 n 5 4 - 2844066472 n + 686400 n + 1679410 n ) a(n + 3)/((n + 12) %1) - 64 ( 2 3 5 -864224526 - 393745295 n - 1013642701 n - 49188412 n + 248160 n 4 + 527863 n ) a(n + 2)/((n + 12) %1) - 128 ( 3 2 5 4 -204096750 - 11790649 n - 91261252 n - 237941763 n + 59840 n + 103782 n ) a(n + 1)/((n + 12) %1) 4 3 2 1536 (n + 3) (1760 n - 2797 n - 338876 n - 1581327 n - 2204118) a(n) - ----------------------------------------------------------------------- (n + 12) %1 4 3 2 %1 := 1760 n - 9837 n - 319925 n - 919006 n - 957110 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1273, a[5] = 7573, a[6] = 45053, a[7] = 268036, a[8] = 1594687, a[9] = 9487926, a[10] = 56452163, a[11] = 335894846, a[12] = 1998662959, a[13] = 11892941089 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 271 7481629\ 3.75 6 (1/n) |1 + ---- + -------| | 11 n 2| \ 3872 n / Note that everything is rigorous except the constant in front, 3.75 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format 1+199*t^5+7*t^2+37*t^3+126*t^4+112*t^8+233*t^6-9*t+202*t^7+48*t^9+(6*t-1)*(8*t^ 3+4*t^2+5*t-1)*(8*t^6+16*t^5+24*t^4+24*t^3+15*t^2+7*t-2)*P+(6*t-1)*(4*t^3+4*t^2 +4*t-1)*(4*t^4+4*t^3+4*t^2+5*t-1)*(8*t^3+4*t^2+5*t-1)*P^2, [1536*(n+3)*(1760*n^ 4-2797*n^3-338876*n^2-1581327*n-2204118)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-\ 919006*n-957110)+128*(-204096750-11790649*n^3-91261252*n^2-237941763*n+59840*n^ 5+103782*n^4)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-919006*n-957110)*N+64*(-\ 864224526-393745295*n^2-1013642701*n-49188412*n^3+248160*n^5+527863*n^4)/(n+12) /(1760*n^4-9837*n^3-319925*n^2-919006*n-957110)*N^2+32*(-2352708870-136206476*n ^3-1119654805*n^2-2844066472*n+686400*n^5+1679410*n^4)/(n+12)/(1760*n^4-9837*n^ 3-319925*n^2-919006*n-957110)*N^3+16*(-4979934918-6071558864*n-2422596453*n^2-\ 284987738*n^3+1430880*n^5+4128919*n^4)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-\ 919006*n-957110)*N^4+16*(-4027964166-4889147900*n-1972289701*n^2-225829457*n^3+ 1133440*n^5+3721612*n^4)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-919006*n-957110)* N^5+16*(-1888056996-2359960853*n-982526088*n^2-105234017*n^3+529760*n^5+2273303 *n^4)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-919006*n-957110)*N^6+8*(-1628844216-\ 1667436683*n-693421261*n^2-73003621*n^3+373120*n^5+1837596*n^4)/(n+12)/(1760*n^ 4-9837*n^3-319925*n^2-919006*n-957110)*N^7-4*(-1434412500-1525268623*n-\ 575686810*n^2-66476344*n^3+299200*n^5+763550*n^4)/(n+12)/(1760*n^4-9837*n^3-\ 319925*n^2-919006*n-957110)*N^8-2*(47850054-495992354*n-238422423*n^2-17800122* n^3+80960*n^5+740778*n^4)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-919006*n-957110) *N^9-(-2523917586-867529511*n^2-2167452244*n-86075466*n^3+2316915*n^4+396000*n^ 5)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-919006*n-957110)*N^10+(-1265918076-\ 502820244*n^2-1252732550*n-48070599*n^3+209440*n^5+1261717*n^4)/(n+12)/(1760*n^ 4-9837*n^3-319925*n^2-919006*n-957110)*N^11-(-208413678-85459321*n^2-214063928* n-8005890*n^3+33440*n^5+207337*n^4)/(n+12)/(1760*n^4-9837*n^3-319925*n^2-919006 *n-957110)*N^12+N^13, [6, 36, 214, 1273, 7573, 45053, 268036, 1594687, 9487926, 56452163, 335894846, 1998662959, 11892941089]], 3.75*6^n*(1/n)^(1/2)*(1+271/11/ n+7481629/3872/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1273, 7573, 45053, 268036, 1594687, 9487926, 56452163, 335894846, 1998662959, 11892941089, 70770531987, 421142576672, 2506220883021, 14914992890145, 88764697038861, 528288337795698, 3144237196010719, 18714279531259255, 111389552886262979, 663024153808359008, 3946642446275922777, 23493069319360564256, 139850906044999524317, 832538556696794682710, 4956293021457469968469, 29506869755699052089195, 175672123224746539862225, 1045914308799017599335932, 6227344611719853607122951, 37078591858561949148826678, 220778663215719780598816699, 1314632897575651306052068374, 7828263163567916082082249303, 46616520772288437795208585575, 277605335638514358589174224707, 1653214732938488708506272985496, 9845646135842713682690825326537, 58637124938973777987314050727430, 349232452923131372130012846753253, 2080031883108474902671476612600534, 12389074109465427506658338938936953, 73794018410588146665952691216632391, 439558728913037302136557167417357189, 2618340266850621290000506984192198204, 15597273054090546082081965057667011011, 92914750963605862383282307882322533473, 553520929521016796369646420444959132443] Proposition Number , 10, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 1] as those of, [2, 3, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 3 2 -t - (6 t - 1) (t + 1) (2 t + 5 t + 5 t - 1) P 3 2 4 3 2 2 + (6 t - 1) (2 t + 5 t + 5 t - 1) (t + t - 5 t - 5 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 3 (-1582782 - 411573 n - 4010 n + 2325 n ) a(n + 9) a(n + 10) = -------------------------------------------------- (n + 10) %1 2 3 2 (-2807719 - 759514 n - 8795 n + 4650 n ) a(n + 8) - ---------------------------------------------------- (n + 10) %1 2 3 2 (-1258715 - 278394 n - 6374 n + 1395 n ) a(n + 7) - ---------------------------------------------------- (n + 10) %1 2 3 (-19237936 - 5218845 n - 75934 n + 35495 n ) a(n + 6) + ------------------------------------------------------ (n + 10) %1 2 3 2 (-17179595 - 5104556 n - 148370 n + 37975 n ) a(n + 5) + --------------------------------------------------------- (n + 10) %1 2 3 2 (-12680691 - 4349991 n - 186998 n + 35340 n ) a(n + 4) + --------------------------------------------------------- (n + 10) %1 2 3 (-5265254 - 2556857 n - 168808 n + 23715 n ) a(n + 3) + ------------------------------------------------------ (n + 10) %1 2 3 (-3682790 - 976227 n - 21950 n + 6975 n ) a(n + 2) - --------------------------------------------------- (n + 10) %1 3 2 4 (-629041 + 2015 n - 11773 n - 238116 n) a(n + 1) - ---------------------------------------------------- (n + 10) %1 2 12 (n + 3) (155 n - 1466 n - 13220) a(n) - ----------------------------------------- (n + 10) %1 2 %1 := 155 n - 1776 n - 11599 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1273, a[5] = 7572, a[6] = 45041, a[7] = 267929, a[8] = 1593843, a[9] = 9481685, a[10] = 56407868 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 3833 103420369\ 3.79 6 (1/n) |1 + ----- + ---------| | 172 n 2 | \ 59168 n / Note that everything is rigorous except the constant in front, 3.79 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t^2-(6*t-1)*(t+1)*(2*t^3+5*t^2+5*t-1)*P+(6*t-1)*(2*t^3+5*t^2+5*t-1)*(t^4+t^3-5 *t^2-5*t+1)*P^2, [12*(n+3)*(155*n^2-1466*n-13220)/(n+10)/(155*n^2-1776*n-11599) +4*(-629041+2015*n^3-11773*n^2-238116*n)/(n+10)/(155*n^2-1776*n-11599)*N+(-\ 3682790-976227*n-21950*n^2+6975*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^2-(-\ 5265254-2556857*n-168808*n^2+23715*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^3-2*(-\ 12680691-4349991*n-186998*n^2+35340*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^4-2*(-\ 17179595-5104556*n-148370*n^2+37975*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^5-(-\ 19237936-5218845*n-75934*n^2+35495*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^6+2*(-\ 1258715-278394*n-6374*n^2+1395*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^7+2*(-\ 2807719-759514*n-8795*n^2+4650*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^8-(-1582782 -411573*n-4010*n^2+2325*n^3)/(n+10)/(155*n^2-1776*n-11599)*N^9+N^10, [6, 36, 214, 1273, 7572, 45041, 267929, 1593843, 9481685, 56407868]], 3.79*6^n*(1/n)^(1 /2)*(1+3833/172/n+103420369/59168/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1273, 7572, 45041, 267929, 1593843, 9481685, 56407868, 335589246, 1996597773, 11879203848, 70680286435, 420555664820, 2502435546783, 14890749111356, 88610345507530, 527310665321892, 3138072163062357, 18675556237137043, 111147171676637102, 661511712339396123, 3937231154609263917, 23434653734463592512, 139489147824652923342, 830302895060018001914, 4942502893308353232860, 29421957129880376879802, 175150115123589035260346, 1042709989715135350811353, 6207702209213205662786286, 36958338932950844023526812, 220043343040269273819995531, 1310141608754561115333149613, 7800859425634664862798629485, 46449480665861665170879283165, 276588082228913309896267632850, 1647025196865859027686565660971, 9808016610944577293011095782954, 58408533235568509504444080222774, 347844831086131207800631021100237, 2071614498635243943817749327933023, 12338047845797002409309649719166674, 73484892793814214858394089870135504, 437687125276797581259023251314683450, 2607015154203959410648331688441067866, 15528782258081082723492147706522004727, 92500757550297218186499732427921899461, 551019801123658115216634578409787565816] Proposition Number , 11, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [1, 1, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 12, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [1, 2, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 4 3 2 -1 + 36 t + 10 t - 96 t - 3 t - 27 t 4 3 2 + (6 t - 1) (10 t - 1) (6 t - 15 t - 3 t - 5 t + 1) P 4 3 2 2 + 5 t (6 t - 1) (5 t - 1) (6 t - 15 t - 3 t - 5 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 3 2 (12091 n + 102 n + 1913 n + 26016) a(n + 6) a(n + 7) = --------------------------------------------- (n + 8) %1 3 2 2 (70434 + 294 n + 5383 n + 33413 n) a(n + 5) - ----------------------------------------------- (n + 8) %1 3 2 2 (138435 + 68446 n + 642 n + 11381 n ) a(n + 4) + ------------------------------------------------- (n + 8) %1 3 2 3 (74338 + 40337 n + 426 n + 7169 n ) a(n + 3) - ----------------------------------------------- (n + 8) %1 2 3 3 (223854 + 115971 n + 20203 n + 1194 n ) a(n + 2) + --------------------------------------------------- (n + 8) %1 2 3 6 (90573 + 57307 n + 11066 n + 696 n ) a(n + 1) - ------------------------------------------------ (n + 8) %1 2 180 (n + 2) (6 n + 79 n + 274) a(n) + ------------------------------------ (n + 8) %1 2 %1 := 6 n + 67 n + 201 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1274, a[5] = 7587, a[6] = 45177, a[7] = 269018 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 1371 15898897\ 4.54 6 (1/n) |1 + ---- + --------| | 80 n 2| \ 12800 n / Note that everything is rigorous except the constant in front, 4.54 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+36*t^5+10*t-96*t^4-3*t^3-27*t^2+(6*t-1)*(10*t-1)*(6*t^4-15*t^3-3*t^2-5*t+1)* P+5*t*(6*t-1)*(5*t-1)*(6*t^4-15*t^3-3*t^2-5*t+1)*P^2, [-180*(n+2)*(6*n^2+79*n+ 274)/(n+8)/(6*n^2+67*n+201)+6*(90573+57307*n+11066*n^2+696*n^3)/(n+8)/(6*n^2+67 *n+201)*N-3*(223854+115971*n+20203*n^2+1194*n^3)/(n+8)/(6*n^2+67*n+201)*N^2+3*( 74338+40337*n+426*n^3+7169*n^2)/(n+8)/(6*n^2+67*n+201)*N^3-2*(138435+68446*n+ 642*n^3+11381*n^2)/(n+8)/(6*n^2+67*n+201)*N^4+2*(70434+294*n^3+5383*n^2+33413*n )/(n+8)/(6*n^2+67*n+201)*N^5-(12091*n+102*n^3+1913*n^2+26016)/(n+8)/(6*n^2+67*n +201)*N^6+N^7, [6, 36, 214, 1274, 7587, 45177, 269018]], 4.54*6^n*(1/n)^(1/2)*( 1+1371/80/n+15898897/12800/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1274, 7587, 45177, 269018, 1601998, 9540133, 56814563, 338359087, 2015156807, 12001973914, 71484081584, 425773695946, 2536069367486, 15106235014605, 89983761784755, 536024750856845, 3193142672387845, 19022367820957156, 113324441507294066, 675142347988993259, 4022349935133844219, 23964980369898993792, 142786457286758972982, 850765087361980991638, 5069264913966675627218, 30205984001201446410763, 179992183815276808789193, 1072573213756073583082282, 6391648357453023850544102, 38090038448833386964734469, 226998266925218711937049739, 1352839414245694111641356341, 8062737336238001861096056781, 48054195987959404079119539460, 286412942617301327810529253010, 1707129488134987531769972819065, 10175429980443113838545232996065, 60652895252486663044336834555186, 361545318768611088519659932789496, 2155193988240903817651480035384304, 12847611260253732750273009377592164, 76589772860409496982336103803561497, 456595394510136460450884212680941287, 2722103282211201468970309908616275883, 16228933304701759597364663308251969563, 96758166695889564338860836392668294768, 576896018006554253567546655291232427198] Proposition Number , 13, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [1, 2, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 4 3 2 -1 + (6 t - 1) (4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 9) a(n + 4) 36 (n + 4) a(n + 3) 2 (2 n + 7) a(n + 2) a(n + 5) = -------------------- - ------------------- - -------------------- n + 5 n + 5 n + 5 28 (n + 3) a(n + 1) 12 (2 n + 5) a(n) + ------------------- - ----------------- n + 5 n + 5 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1274, a[5] = 7584 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 123 695069\ 4.54 6 (1/n) |1 + --- + ------| | 8 n 2| \ 640 n / Note that everything is rigorous except the constant in front, 4.54 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^4-4*t^3+6*t-1)*P^2, [12*(2*n+5)/(n+5)-28*(n+3)/(n+5)*N+2*(2*n+7 )/(n+5)*N^2+36*(n+4)/(n+5)*N^3-6*(2*n+9)/(n+5)*N^4+N^5, [6, 36, 214, 1274, 7584 ]], 4.54*6^n*(1/n)^(1/2)*(1+123/8/n+695069/640/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1274, 7584, 45150, 268800, 1600350, 9528280, 56731980, 337796580, 2011390650, 11977094300, 71321483100, 424720386548, 2529296162338, 15062951835228, 89708646326892, 534284187403572, 3182175501625152, 18953512711237200, 112893530790348300, 672453330484995000, 4005612876110404170, 23861048142331583220, 142142438007251826120, 846782135072415899500, 5044676016546888618600, 30054431721780670339800, 179059509482029891889356, 1066841422807391131253336, 6356469083483105077080066, 37874383914544058480771364, 225677760224060424323294424, 1344762143470011374247297084, 8013379119978900118763991000, 47752859851268543163447483000, 284574865250604978279207619500, 1695926883690409044912722017400, 10107206027371092040674853628400, 60237715822665630876836367761400, 359020484906466215621579298638000, 2139849815612482240631377758550000, 12754418466742855168395745679311500, 76024103303096857220873192473997480, 453163788745094354129861703853808880, 2701296846354865449169372076269754280, 16102845075785700213455373389074341570, 95994439850663113030947308368011770620, 572272232474458929228546060400693657920] Proposition Number , 14, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [1, 2, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 4 3 2 -1 + (6 t - 1) (4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 9) a(n + 4) 36 (n + 4) a(n + 3) 2 (2 n + 7) a(n + 2) a(n + 5) = -------------------- - ------------------- - -------------------- n + 5 n + 5 n + 5 28 (n + 3) a(n + 1) 12 (2 n + 5) a(n) + ------------------- - ----------------- n + 5 n + 5 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1274, a[5] = 7584 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 123 695069\ 4.54 6 (1/n) |1 + --- + ------| | 8 n 2| \ 640 n / Note that everything is rigorous except the constant in front, 4.54 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^4-4*t^3+6*t-1)*P^2, [12*(2*n+5)/(n+5)-28*(n+3)/(n+5)*N+2*(2*n+7 )/(n+5)*N^2+36*(n+4)/(n+5)*N^3-6*(2*n+9)/(n+5)*N^4+N^5, [6, 36, 214, 1274, 7584 ]], 4.54*6^n*(1/n)^(1/2)*(1+123/8/n+695069/640/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1274, 7584, 45150, 268800, 1600350, 9528280, 56731980, 337796580, 2011390650, 11977094300, 71321483100, 424720386548, 2529296162338, 15062951835228, 89708646326892, 534284187403572, 3182175501625152, 18953512711237200, 112893530790348300, 672453330484995000, 4005612876110404170, 23861048142331583220, 142142438007251826120, 846782135072415899500, 5044676016546888618600, 30054431721780670339800, 179059509482029891889356, 1066841422807391131253336, 6356469083483105077080066, 37874383914544058480771364, 225677760224060424323294424, 1344762143470011374247297084, 8013379119978900118763991000, 47752859851268543163447483000, 284574865250604978279207619500, 1695926883690409044912722017400, 10107206027371092040674853628400, 60237715822665630876836367761400, 359020484906466215621579298638000, 2139849815612482240631377758550000, 12754418466742855168395745679311500, 76024103303096857220873192473997480, 453163788745094354129861703853808880, 2701296846354865449169372076269754280, 16102845075785700213455373389074341570, 95994439850663113030947308368011770620, 572272232474458929228546060400693657920] Proposition Number , 15, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [1, 3, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 4 3 2 -t - (6 t - 1) (6 t - t + 8 t - 2 t + 6 t - 1) P 5 4 3 2 2 - (6 t - 1) (5 t - 1) (6 t - t + 8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (66108 + 28671 n + 2660 n ) a(n + 7) a(n + 8) = ------------------------------------ (n + 8) (158 n + 499) 2 2 (78101 n + 7521 n + 183400) a(n + 6) - --------------------------------------- (n + 8) (158 n + 499) 2 2 (360690 + 15316 n + 151371 n) a(n + 5) + ----------------------------------------- (n + 8) (158 n + 499) 2 (348420 + 18082 n + 155303 n) a(n + 4) - --------------------------------------- (n + 8) (158 n + 499) 2 (714650 + 36732 n + 324931 n) a(n + 3) + --------------------------------------- (n + 8) (158 n + 499) 2 2 (31083 + 4243 n + 25770 n) a(n + 2) - -------------------------------------- (n + 8) (158 n + 499) 2 24 (13048 + 1081 n + 8539 n) a(n + 1) 360 (13 n + 124) (n + 1) a(n) + -------------------------------------- + ----------------------------- (n + 8) (158 n + 499) (n + 8) (158 n + 499) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7563, a[6] = 44970, a[7] = 267404, a[8] = 1590124 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 475 5518051\ 4.146 6 (1/n) |1 + ---- + -------| | 32 n 2| \ 6144 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(6*t^5-t^4+8*t^3-2*t^2+6*t-1)*P-(6*t-1)*(5*t-1)*(6*t^5-t^4+8*t^3-2*t ^2+6*t-1)*P^2, [-360*(13*n+124)*(n+1)/(n+8)/(158*n+499)-24*(13048+1081*n^2+8539 *n)/(n+8)/(158*n+499)*N+2*(31083+4243*n^2+25770*n)/(n+8)/(158*n+499)*N^2-( 714650+36732*n^2+324931*n)/(n+8)/(158*n+499)*N^3+(348420+18082*n^2+155303*n)/(n +8)/(158*n+499)*N^4-2*(360690+15316*n^2+151371*n)/(n+8)/(158*n+499)*N^5+2*( 78101*n+7521*n^2+183400)/(n+8)/(158*n+499)*N^6-(66108+28671*n+2660*n^2)/(n+8)/( 158*n+499)*N^7+N^8, [6, 36, 214, 1272, 7563, 44970, 267404, 1590124]], 4.146*6^ n*(1/n)^(1/2)*(1+475/32/n+5518051/6144/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7563, 44970, 267404, 1590124, 9456105, 56235657, 334448649, 1989139738, 11830941475, 70370635328, 418583185888, 2489947635948, 14812104305103, 88117345672707, 524232729504521, 3118924913068933, 18556828335074974, 110413100144521621, 656985008404103183, 3909383670056342758, 23263715890842936399, 138441981794281648509, 823899872354158707277, 4903418290178399535504, 29183763766743366407553, 173700665489812529798640, 1033902192595810255096604, 6154250822218346294159428, 36634364190883836804506855, 218082000321969613529426573, 1298280843234694862185370923, 7729209951988533431886244479, 46017090108955653699315319024, 273981195693660308255074524881, 1631322697586360889690055884423, 9713516457563042570852728723737, 57840297952541056988151299041116, 344430785101414334799002723685492, 2051118535045184752314129781624140, 12215095639197088829611745785961877, 72747865593330709904405188125229536, 433272242649538566015752751327707871, 2580587923471380933902626140778237911, 15370698012885394944391120104143238334, 91555746672844212946716627249337069537, 545374300536898773491123160416428473929] Proposition Number , 16, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [1, 3, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 17, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [1, 3, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 18, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [1, 3, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 19, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [2, 1, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 4 3 5 5 4 -1 - 10 t + 4 t + 2 t - 12 t - (6 t - 1) (16 t + 16 t - 6 t + 1) P 5 4 2 - 2 t (6 t - 1) (16 t + 16 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 2 4 (677 + 183 n + 12 n ) a(n + 5) 4 (1823 + 531 n + 36 n ) a(n + 4) a(n + 6) = -------------------------------- - --------------------------------- (4 n + 35) (n + 7) (4 n + 35) (n + 7) 2 544 a(n + 3) 16 (243 + 55 n + 4 n ) a(n + 2) - ------------------ - ------------------------------- (4 n + 35) (n + 7) (4 n + 35) (n + 7) 2 8 (1227 + 530 n + 40 n ) a(n + 1) 96 (4 n + 39) (n + 3) a(n) + --------------------------------- + -------------------------- (4 n + 35) (n + 7) (4 n + 35) (n + 7) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1274, a[5] = 7588, a[6] = 45192 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 865 7316509\ 4.70 6 (1/n) |1 + ---- + -------| | 56 n 2| \ 6272 n / Note that everything is rigorous except the constant in front, 4.70 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-10*t^4+4*t+2*t^3-12*t^5-(6*t-1)*(16*t^5+16*t^4-6*t+1)*P-2*t*(6*t-1)*(16*t^5+ 16*t^4-6*t+1)*P^2, [-96*(4*n+39)*(n+3)/(4*n+35)/(n+7)-8*(1227+530*n+40*n^2)/(4* n+35)/(n+7)*N+16*(243+55*n+4*n^2)/(4*n+35)/(n+7)*N^2+544/(4*n+35)/(n+7)*N^3+4*( 1823+531*n+36*n^2)/(4*n+35)/(n+7)*N^4-4*(677+183*n+12*n^2)/(4*n+35)/(n+7)*N^5+N ^6, [6, 36, 214, 1274, 7588, 45192]], 4.70*6^n*(1/n)^(1/2)*(1+865/56/n+7316509/ 6272/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1274, 7588, 45192, 269160, 1603152, 9548856, 56877504, 338799136, 2018164672, 12022185792, 71618099776, 426652865024, 2541786354176, 15143139281152, 90220509517312, 537535447731584, 3202737955780096, 19083066192150528, 113707038141702144, 677546280148208640, 4037411335876684800, 24059103023846098944, 143373290447068483584, 854416143404005994496, 5091936689767327997952, 30346519381367791595520, 180861905863503387721728, 1077947545588869527715840, 6424812300913209082675200, 38294423382645787070791680, 228256356299357117122904064, 1360574925559165333554724864, 8110250408799006005062795264, 48345745593923636755602997248, 288200307215831716226200240128, 1718077594358562201613972897792, 10242435686464454572415233949696, 61062675931784205669260321357824, 364049561564601256992565325463552, 2170487371394529172746153724477440, 12940947114715692502542155635228672, 77159052567365772491533484184567808, 460065546973805886604790340258889728, 2743244476432008392883732557115424768, 16357663422102782562824388262831849472, 97541616724256654433488129553213685760, 581661787153240399842072614695155531776] Proposition Number , 20, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [2, 1, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 4 3 2 -t - (6 t - 1) (6 t - t + 8 t - 2 t + 6 t - 1) P 5 4 3 2 2 - (6 t - 1) (5 t - 1) (6 t - t + 8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (66108 + 28671 n + 2660 n ) a(n + 7) a(n + 8) = ------------------------------------ (n + 8) (158 n + 499) 2 2 (78101 n + 7521 n + 183400) a(n + 6) - --------------------------------------- (n + 8) (158 n + 499) 2 2 (360690 + 15316 n + 151371 n) a(n + 5) + ----------------------------------------- (n + 8) (158 n + 499) 2 (348420 + 18082 n + 155303 n) a(n + 4) - --------------------------------------- (n + 8) (158 n + 499) 2 (714650 + 36732 n + 324931 n) a(n + 3) + --------------------------------------- (n + 8) (158 n + 499) 2 2 (31083 + 4243 n + 25770 n) a(n + 2) - -------------------------------------- (n + 8) (158 n + 499) 2 24 (13048 + 1081 n + 8539 n) a(n + 1) 360 (13 n + 124) (n + 1) a(n) + -------------------------------------- + ----------------------------- (n + 8) (158 n + 499) (n + 8) (158 n + 499) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7563, a[6] = 44970, a[7] = 267404, a[8] = 1590124 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 475 5518051\ 4.146 6 (1/n) |1 + ---- + -------| | 32 n 2| \ 6144 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(6*t^5-t^4+8*t^3-2*t^2+6*t-1)*P-(6*t-1)*(5*t-1)*(6*t^5-t^4+8*t^3-2*t ^2+6*t-1)*P^2, [-360*(13*n+124)*(n+1)/(n+8)/(158*n+499)-24*(13048+1081*n^2+8539 *n)/(n+8)/(158*n+499)*N+2*(31083+4243*n^2+25770*n)/(n+8)/(158*n+499)*N^2-( 714650+36732*n^2+324931*n)/(n+8)/(158*n+499)*N^3+(348420+18082*n^2+155303*n)/(n +8)/(158*n+499)*N^4-2*(360690+15316*n^2+151371*n)/(n+8)/(158*n+499)*N^5+2*( 78101*n+7521*n^2+183400)/(n+8)/(158*n+499)*N^6-(66108+28671*n+2660*n^2)/(n+8)/( 158*n+499)*N^7+N^8, [6, 36, 214, 1272, 7563, 44970, 267404, 1590124]], 4.146*6^ n*(1/n)^(1/2)*(1+475/32/n+5518051/6144/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7563, 44970, 267404, 1590124, 9456105, 56235657, 334448649, 1989139738, 11830941475, 70370635328, 418583185888, 2489947635948, 14812104305103, 88117345672707, 524232729504521, 3118924913068933, 18556828335074974, 110413100144521621, 656985008404103183, 3909383670056342758, 23263715890842936399, 138441981794281648509, 823899872354158707277, 4903418290178399535504, 29183763766743366407553, 173700665489812529798640, 1033902192595810255096604, 6154250822218346294159428, 36634364190883836804506855, 218082000321969613529426573, 1298280843234694862185370923, 7729209951988533431886244479, 46017090108955653699315319024, 273981195693660308255074524881, 1631322697586360889690055884423, 9713516457563042570852728723737, 57840297952541056988151299041116, 344430785101414334799002723685492, 2051118535045184752314129781624140, 12215095639197088829611745785961877, 72747865593330709904405188125229536, 433272242649538566015752751327707871, 2580587923471380933902626140778237911, 15370698012885394944391120104143238334, 91555746672844212946716627249337069537, 545374300536898773491123160416428473929] Proposition Number , 21, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [2, 1, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 3 2 -1 + (6 t - 1) (4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 11) a(n + 5) 36 (n + 5) a(n + 4) 2 (2 n + 9) a(n + 3) a(n + 6) = --------------------- - ------------------- - -------------------- n + 6 n + 6 n + 6 24 (n + 4) a(n + 2) 2 (2 n + 7) a(n + 1) 24 (n + 3) a(n) + ------------------- + -------------------- - --------------- n + 6 n + 6 n + 6 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7562, a[6] = 44958 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 521 2502529\ 4.205 6 (1/n) |1 + ---- + -------| | 40 n 2| \ 3200 n / Note that everything is rigorous except the constant in front, 4.205 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^5-4*t^3+6*t-1)*P^2, [24*(n+3)/(n+6)-2*(2*n+7)/(n+6)*N-24*(n+4)/ (n+6)*N^2+2*(2*n+9)/(n+6)*N^3+36*(n+5)/(n+6)*N^4-6*(2*n+11)/(n+6)*N^5+N^6, [6, 36, 214, 1272, 7562, 44958]], 4.205*6^n*(1/n)^(1/2)*(1+521/40/n+2502529/3200/n^ 2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7562, 44958, 267300, 1589316, 9450196, 56194086, 334163976, 1987228726, 11818307256, 70288111376, 418049425304, 2486523268404, 14790285061524, 87979129336868, 523361595527232, 3113458587734110, 18522660336804468, 110200263338247084, 655663298780619876, 3901198605504457830, 23213154398977342440, 138130359633331334748, 821983264478393999764, 4891652813438850746832, 29111665836085553119332, 173259571618490266150768, 1031207646797687254066248, 6137813457870869748529148, 36534223227564741192094536, 217472658430937710414174884, 1294577337077895162316177296, 7706724699788043479343760612, 45880712769880308338900727384, 273154832326134868125779903736, 1626319971156341349181021068352, 9683256394543314610491985962798, 57657412536593910625062421487180, 343326320640053256691648058786504, 2044453480460431778198363148838404, 12174902716459651798539044567315652, 72505649567614945427959319917982692, 431813508858897006790224811251688428, 2571808203855980343902893285555965280, 15317886647816722444541262208671130286, 91238259470573690493118524002219522676, 543466704651662865564987092697981902116] Proposition Number , 22, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [2, 2, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 4 3 3 2 -1 + 6 t - t + 12 t - 2 t + 2 t (6 t - 1) (8 t - 2 t + 6 t - 1) P 2 3 2 2 + (6 t - 1) (4 t - t + 1) (8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence (98 + 13 n) a(n + 7) 2 (189 + 26 n) a(n + 6) a(n + 8) = -------------------- - ----------------------- n + 8 n + 8 2 (319 + 40 n) a(n + 5) 2 (515 + 56 n) a(n + 4) + ----------------------- - ----------------------- n + 8 n + 8 4 (17 + 21 n) a(n + 3) 4 (123 + 62 n) a(n + 2) - ---------------------- + ----------------------- n + 8 n + 8 16 (53 + 16 n) a(n + 1) 384 (n + 3) a(n) - ----------------------- + ---------------- n + 8 n + 8 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7564, a[6] = 44982, a[7] = 267510, a[8] = 1590960 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 1811 30474361\ 4.266 6 (1/n) |1 + ----- + --------| | 136 n 2| \ 36992 n / Note that everything is rigorous except the constant in front, 4.266 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+6*t^2-t+12*t^4-2*t^3+2*t*(6*t-1)*(8*t^3-2*t^2+6*t-1)*P+(6*t-1)*(4*t^2-t+1)*( 8*t^3-2*t^2+6*t-1)*P^2, [-384*(n+3)/(n+8)+16*(53+16*n)/(n+8)*N-4*(123+62*n)/(n+ 8)*N^2+4*(17+21*n)/(n+8)*N^3+2*(515+56*n)/(n+8)*N^4-2*(319+40*n)/(n+8)*N^5+2*( 189+26*n)/(n+8)*N^6-(98+13*n)/(n+8)*N^7+N^8, [6, 36, 214, 1272, 7564, 44982, 267510, 1590960]], 4.266*6^n*(1/n)^(1/2)*(1+1811/136/n+30474361/36992/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7564, 44982, 267510, 1590960, 9462304, 56279824, 334754740, 1991218422, 11844840490, 70462437914, 419183549742, 2493841783056, 14837189545148, 88277993824228, 525256344529584, 3125418503868136, 18597863189912640, 110671524361870956, 658607517979402308, 3919542687612002598, 23327167582804676058, 138837405019825914486, 826359082608264642994, 4918684091984746497586, 29278365853496908110682, 174285991447733509340302, 1037518480568323458299822, 6176563047215064400299456, 36771856692794288245706172, 218928270022255688105292880, 1303483988146515002030433268, 7761168048728232251269689180, 46213191845093350822617773756, 275183440000330223507481674312, 1638687104909021652440222956736, 9758591664940169274223117265304, 58115982150651571900089785812320, 346115700201681931796498777955984, 2061409416962276939039306251901872, 12277908816166423734366705287412252, 73131031238216130129010557201408484, 435608243706209161655584765074557708, 2594821770283355790905159092446544980, 15457383340939303467946023312412958086, 92083405830508669102507402359742482818, 548584675492787626673637920117515577510] Proposition Number , 23, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [2, 2, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 3 2 -1 + (6 t - 1) (4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 11) a(n + 5) 36 (n + 5) a(n + 4) 2 (2 n + 9) a(n + 3) a(n + 6) = --------------------- - ------------------- - -------------------- n + 6 n + 6 n + 6 24 (n + 4) a(n + 2) 2 (2 n + 7) a(n + 1) 24 (n + 3) a(n) + ------------------- + -------------------- - --------------- n + 6 n + 6 n + 6 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7562, a[6] = 44958 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 521 2502529\ 4.205 6 (1/n) |1 + ---- + -------| | 40 n 2| \ 3200 n / Note that everything is rigorous except the constant in front, 4.205 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^5-4*t^3+6*t-1)*P^2, [24*(n+3)/(n+6)-2*(2*n+7)/(n+6)*N-24*(n+4)/ (n+6)*N^2+2*(2*n+9)/(n+6)*N^3+36*(n+5)/(n+6)*N^4-6*(2*n+11)/(n+6)*N^5+N^6, [6, 36, 214, 1272, 7562, 44958]], 4.205*6^n*(1/n)^(1/2)*(1+521/40/n+2502529/3200/n^ 2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7562, 44958, 267300, 1589316, 9450196, 56194086, 334163976, 1987228726, 11818307256, 70288111376, 418049425304, 2486523268404, 14790285061524, 87979129336868, 523361595527232, 3113458587734110, 18522660336804468, 110200263338247084, 655663298780619876, 3901198605504457830, 23213154398977342440, 138130359633331334748, 821983264478393999764, 4891652813438850746832, 29111665836085553119332, 173259571618490266150768, 1031207646797687254066248, 6137813457870869748529148, 36534223227564741192094536, 217472658430937710414174884, 1294577337077895162316177296, 7706724699788043479343760612, 45880712769880308338900727384, 273154832326134868125779903736, 1626319971156341349181021068352, 9683256394543314610491985962798, 57657412536593910625062421487180, 343326320640053256691648058786504, 2044453480460431778198363148838404, 12174902716459651798539044567315652, 72505649567614945427959319917982692, 431813508858897006790224811251688428, 2571808203855980343902893285555965280, 15317886647816722444541262208671130286, 91238259470573690493118524002219522676, 543466704651662865564987092697981902116] Proposition Number , 24, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [2, 3, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 4 3 3 2 -1 + 6 t - t + 12 t - 2 t + 2 t (6 t - 1) (8 t - 2 t + 6 t - 1) P 2 3 2 2 + (6 t - 1) (4 t - t + 1) (8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence (98 + 13 n) a(n + 7) 2 (189 + 26 n) a(n + 6) a(n + 8) = -------------------- - ----------------------- n + 8 n + 8 2 (319 + 40 n) a(n + 5) 2 (515 + 56 n) a(n + 4) + ----------------------- - ----------------------- n + 8 n + 8 4 (17 + 21 n) a(n + 3) 4 (123 + 62 n) a(n + 2) - ---------------------- + ----------------------- n + 8 n + 8 16 (53 + 16 n) a(n + 1) 384 (n + 3) a(n) - ----------------------- + ---------------- n + 8 n + 8 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7564, a[6] = 44982, a[7] = 267510, a[8] = 1590960 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 1811 30474361\ 4.266 6 (1/n) |1 + ----- + --------| | 136 n 2| \ 36992 n / Note that everything is rigorous except the constant in front, 4.266 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+6*t^2-t+12*t^4-2*t^3+2*t*(6*t-1)*(8*t^3-2*t^2+6*t-1)*P+(6*t-1)*(4*t^2-t+1)*( 8*t^3-2*t^2+6*t-1)*P^2, [-384*(n+3)/(n+8)+16*(53+16*n)/(n+8)*N-4*(123+62*n)/(n+ 8)*N^2+4*(17+21*n)/(n+8)*N^3+2*(515+56*n)/(n+8)*N^4-2*(319+40*n)/(n+8)*N^5+2*( 189+26*n)/(n+8)*N^6-(98+13*n)/(n+8)*N^7+N^8, [6, 36, 214, 1272, 7564, 44982, 267510, 1590960]], 4.266*6^n*(1/n)^(1/2)*(1+1811/136/n+30474361/36992/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7564, 44982, 267510, 1590960, 9462304, 56279824, 334754740, 1991218422, 11844840490, 70462437914, 419183549742, 2493841783056, 14837189545148, 88277993824228, 525256344529584, 3125418503868136, 18597863189912640, 110671524361870956, 658607517979402308, 3919542687612002598, 23327167582804676058, 138837405019825914486, 826359082608264642994, 4918684091984746497586, 29278365853496908110682, 174285991447733509340302, 1037518480568323458299822, 6176563047215064400299456, 36771856692794288245706172, 218928270022255688105292880, 1303483988146515002030433268, 7761168048728232251269689180, 46213191845093350822617773756, 275183440000330223507481674312, 1638687104909021652440222956736, 9758591664940169274223117265304, 58115982150651571900089785812320, 346115700201681931796498777955984, 2061409416962276939039306251901872, 12277908816166423734366705287412252, 73131031238216130129010557201408484, 435608243706209161655584765074557708, 2594821770283355790905159092446544980, 15457383340939303467946023312412958086, 92083405830508669102507402359742482818, 548584675492787626673637920117515577510] Proposition Number , 25, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [2, 3, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 4 3 2 -t - (6 t - 1) (6 t - t + 8 t - 2 t + 6 t - 1) P 5 4 3 2 2 - (6 t - 1) (5 t - 1) (6 t - t + 8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (66108 + 28671 n + 2660 n ) a(n + 7) a(n + 8) = ------------------------------------ (n + 8) (158 n + 499) 2 2 (78101 n + 7521 n + 183400) a(n + 6) - --------------------------------------- (n + 8) (158 n + 499) 2 2 (360690 + 15316 n + 151371 n) a(n + 5) + ----------------------------------------- (n + 8) (158 n + 499) 2 (348420 + 18082 n + 155303 n) a(n + 4) - --------------------------------------- (n + 8) (158 n + 499) 2 (714650 + 36732 n + 324931 n) a(n + 3) + --------------------------------------- (n + 8) (158 n + 499) 2 2 (31083 + 4243 n + 25770 n) a(n + 2) - -------------------------------------- (n + 8) (158 n + 499) 2 24 (13048 + 1081 n + 8539 n) a(n + 1) 360 (13 n + 124) (n + 1) a(n) + -------------------------------------- + ----------------------------- (n + 8) (158 n + 499) (n + 8) (158 n + 499) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7563, a[6] = 44970, a[7] = 267404, a[8] = 1590124 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 475 5518051\ 4.146 6 (1/n) |1 + ---- + -------| | 32 n 2| \ 6144 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(6*t^5-t^4+8*t^3-2*t^2+6*t-1)*P-(6*t-1)*(5*t-1)*(6*t^5-t^4+8*t^3-2*t ^2+6*t-1)*P^2, [-360*(13*n+124)*(n+1)/(n+8)/(158*n+499)-24*(13048+1081*n^2+8539 *n)/(n+8)/(158*n+499)*N+2*(31083+4243*n^2+25770*n)/(n+8)/(158*n+499)*N^2-( 714650+36732*n^2+324931*n)/(n+8)/(158*n+499)*N^3+(348420+18082*n^2+155303*n)/(n +8)/(158*n+499)*N^4-2*(360690+15316*n^2+151371*n)/(n+8)/(158*n+499)*N^5+2*( 78101*n+7521*n^2+183400)/(n+8)/(158*n+499)*N^6-(66108+28671*n+2660*n^2)/(n+8)/( 158*n+499)*N^7+N^8, [6, 36, 214, 1272, 7563, 44970, 267404, 1590124]], 4.146*6^ n*(1/n)^(1/2)*(1+475/32/n+5518051/6144/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7563, 44970, 267404, 1590124, 9456105, 56235657, 334448649, 1989139738, 11830941475, 70370635328, 418583185888, 2489947635948, 14812104305103, 88117345672707, 524232729504521, 3118924913068933, 18556828335074974, 110413100144521621, 656985008404103183, 3909383670056342758, 23263715890842936399, 138441981794281648509, 823899872354158707277, 4903418290178399535504, 29183763766743366407553, 173700665489812529798640, 1033902192595810255096604, 6154250822218346294159428, 36634364190883836804506855, 218082000321969613529426573, 1298280843234694862185370923, 7729209951988533431886244479, 46017090108955653699315319024, 273981195693660308255074524881, 1631322697586360889690055884423, 9713516457563042570852728723737, 57840297952541056988151299041116, 344430785101414334799002723685492, 2051118535045184752314129781624140, 12215095639197088829611745785961877, 72747865593330709904405188125229536, 433272242649538566015752751327707871, 2580587923471380933902626140778237911, 15370698012885394944391120104143238334, 91555746672844212946716627249337069537, 545374300536898773491123160416428473929] Proposition Number , 26, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [2, 3, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 3 2 -1 + (6 t - 1) (4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 11) a(n + 5) 36 (n + 5) a(n + 4) 2 (2 n + 9) a(n + 3) a(n + 6) = --------------------- - ------------------- - -------------------- n + 6 n + 6 n + 6 24 (n + 4) a(n + 2) 2 (2 n + 7) a(n + 1) 24 (n + 3) a(n) + ------------------- + -------------------- - --------------- n + 6 n + 6 n + 6 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7562, a[6] = 44958 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 521 2502529\ 4.205 6 (1/n) |1 + ---- + -------| | 40 n 2| \ 3200 n / Note that everything is rigorous except the constant in front, 4.205 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^5-4*t^3+6*t-1)*P^2, [24*(n+3)/(n+6)-2*(2*n+7)/(n+6)*N-24*(n+4)/ (n+6)*N^2+2*(2*n+9)/(n+6)*N^3+36*(n+5)/(n+6)*N^4-6*(2*n+11)/(n+6)*N^5+N^6, [6, 36, 214, 1272, 7562, 44958]], 4.205*6^n*(1/n)^(1/2)*(1+521/40/n+2502529/3200/n^ 2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7562, 44958, 267300, 1589316, 9450196, 56194086, 334163976, 1987228726, 11818307256, 70288111376, 418049425304, 2486523268404, 14790285061524, 87979129336868, 523361595527232, 3113458587734110, 18522660336804468, 110200263338247084, 655663298780619876, 3901198605504457830, 23213154398977342440, 138130359633331334748, 821983264478393999764, 4891652813438850746832, 29111665836085553119332, 173259571618490266150768, 1031207646797687254066248, 6137813457870869748529148, 36534223227564741192094536, 217472658430937710414174884, 1294577337077895162316177296, 7706724699788043479343760612, 45880712769880308338900727384, 273154832326134868125779903736, 1626319971156341349181021068352, 9683256394543314610491985962798, 57657412536593910625062421487180, 343326320640053256691648058786504, 2044453480460431778198363148838404, 12174902716459651798539044567315652, 72505649567614945427959319917982692, 431813508858897006790224811251688428, 2571808203855980343902893285555965280, 15317886647816722444541262208671130286, 91238259470573690493118524002219522676, 543466704651662865564987092697981902116] Proposition Number , 27, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [2, 3, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 3 2 -1 + (6 t - 1) (4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 11) a(n + 5) 36 (n + 5) a(n + 4) 2 (2 n + 9) a(n + 3) a(n + 6) = --------------------- - ------------------- - -------------------- n + 6 n + 6 n + 6 24 (n + 4) a(n + 2) 2 (2 n + 7) a(n + 1) 24 (n + 3) a(n) + ------------------- + -------------------- - --------------- n + 6 n + 6 n + 6 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7562, a[6] = 44958 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 521 2502529\ 4.205 6 (1/n) |1 + ---- + -------| | 40 n 2| \ 3200 n / Note that everything is rigorous except the constant in front, 4.205 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^5-4*t^3+6*t-1)*P^2, [24*(n+3)/(n+6)-2*(2*n+7)/(n+6)*N-24*(n+4)/ (n+6)*N^2+2*(2*n+9)/(n+6)*N^3+36*(n+5)/(n+6)*N^4-6*(2*n+11)/(n+6)*N^5+N^6, [6, 36, 214, 1272, 7562, 44958]], 4.205*6^n*(1/n)^(1/2)*(1+521/40/n+2502529/3200/n^ 2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7562, 44958, 267300, 1589316, 9450196, 56194086, 334163976, 1987228726, 11818307256, 70288111376, 418049425304, 2486523268404, 14790285061524, 87979129336868, 523361595527232, 3113458587734110, 18522660336804468, 110200263338247084, 655663298780619876, 3901198605504457830, 23213154398977342440, 138130359633331334748, 821983264478393999764, 4891652813438850746832, 29111665836085553119332, 173259571618490266150768, 1031207646797687254066248, 6137813457870869748529148, 36534223227564741192094536, 217472658430937710414174884, 1294577337077895162316177296, 7706724699788043479343760612, 45880712769880308338900727384, 273154832326134868125779903736, 1626319971156341349181021068352, 9683256394543314610491985962798, 57657412536593910625062421487180, 343326320640053256691648058786504, 2044453480460431778198363148838404, 12174902716459651798539044567315652, 72505649567614945427959319917982692, 431813508858897006790224811251688428, 2571808203855980343902893285555965280, 15317886647816722444541262208671130286, 91238259470573690493118524002219522676, 543466704651662865564987092697981902116] Proposition Number , 28, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 1, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 4 3 2 -1 + (6 t - 1) (4 t + 4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 11) a(n + 5) 36 (n + 5) a(n + 4) 2 (2 n + 9) a(n + 3) a(n + 6) = --------------------- - ------------------- - -------------------- n + 6 n + 6 n + 6 28 (n + 4) a(n + 2) 10 (2 n + 7) a(n + 1) 24 (n + 3) a(n) + ------------------- - --------------------- - --------------- n + 6 n + 6 n + 6 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1274, a[5] = 7586, a[6] = 45174 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 3701 125779657\ 4.62 6 (1/n) |1 + ----- + ---------| | 232 n 2| \ 107648 n / Note that everything is rigorous except the constant in front, 4.62 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^5+4*t^4-4*t^3+6*t-1)*P^2, [24*(n+3)/(n+6)+10*(2*n+7)/(n+6)*N-28 *(n+4)/(n+6)*N^2+2*(2*n+9)/(n+6)*N^3+36*(n+5)/(n+6)*N^4-6*(2*n+11)/(n+6)*N^5+N^ 6, [6, 36, 214, 1274, 7586, 45174]], 4.62*6^n*(1/n)^(1/2)*(1+3701/232/n+ 125779657/107648/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1274, 7586, 45174, 269016, 1602066, 9541036, 56822922, 338426604, 2015664882, 12005632796, 71509652576, 425948563360, 2537245649422, 15114045838240, 90035090124008, 536359172458228, 3195305959512790, 19036276502706456, 113413398579155280, 675708712094024004, 4025941426800456534, 23987674780262445720, 142929411148228938888, 851663028265702911220, 5074890858617532135852, 30241151616220336851612, 180211555256316867029008, 1073939008197485956324712, 6400136787494499428921918, 38142708864115947940487148, 227324596998767712733140708, 1354858460185836106007473932, 8075213375886397422404125308, 48131195338820182114988513652, 286887634373184438565852910172, 1710052845102745422237618164312, 10193415619531022859937266195750, 60763448000195858315599363875924, 362224266552942188115098291309140, 2159360262969131008033124651652436, 12873157317291103949380618728712036, 76746297407953442846848295310168372, 457553780339917053051835763053284604, 2727967540922906660129241123298560712, 16264793692719367859307819450026963606, 96977325664015780179890475009606345476, 578234640886295060286262718300233040228] Proposition Number , 29, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 1, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 30, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 1, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -t - (6 t - 1) (2 t - t + 6 t - 1) P 3 3 2 2 + (6 t - 1) (t - 5 t + 1) (2 t - t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (181619 + 58106 n + 4335 n ) a(n + 8) a(n + 9) = ------------------------------------- (n + 9) (255 n + 1213) 2 2 (162967 n + 12495 n + 510920) a(n + 7) - ----------------------------------------- (n + 9) (255 n + 1213) 2 (2078799 + 668402 n + 53295 n ) a(n + 6) + ---------------------------------------- (n + 9) (255 n + 1213) 2 (978061 + 369416 n + 34935 n ) a(n + 5) - --------------------------------------- (n + 9) (255 n + 1213) 2 (1272103 + 535940 n + 54825 n ) a(n + 4) + ---------------------------------------- (n + 9) (255 n + 1213) 2 2 (54361 + 32413 n + 4080 n ) a(n + 3) - -------------------------------------- (n + 9) (255 n + 1213) 2 (-59025 + 5948 n + 2805 n ) a(n + 2) + ------------------------------------ (n + 9) (255 n + 1213) 2 4 (10612 + 510 n + 4721 n) a(n + 1) 12 (255 n + 1468) (n + 3) a(n) + ------------------------------------ - ------------------------------ (n + 9) (255 n + 1213) (n + 9) (255 n + 1213) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7561, a[6] = 44946, a[7] = 267191, a[8] = 1588444, a[9] = 9443681 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 533 250411\ 4.090 6 (1/n) |1 + ---- + ------| | 37 n 2| \ 296 n / Note that everything is rigorous except the constant in front, 4.090 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(2*t^3-t^2+6*t-1)*P+(6*t-1)*(t^3-5*t+1)*(2*t^3-t^2+6*t-1)*P^2, [12*( 255*n+1468)*(n+3)/(n+9)/(255*n+1213)-4*(10612+510*n^2+4721*n)/(n+9)/(255*n+1213 )*N-(-59025+5948*n+2805*n^2)/(n+9)/(255*n+1213)*N^2+2*(54361+32413*n+4080*n^2)/ (n+9)/(255*n+1213)*N^3-(1272103+535940*n+54825*n^2)/(n+9)/(255*n+1213)*N^4+( 978061+369416*n+34935*n^2)/(n+9)/(255*n+1213)*N^5-(2078799+668402*n+53295*n^2)/ (n+9)/(255*n+1213)*N^6+2*(162967*n+12495*n^2+510920)/(n+9)/(255*n+1213)*N^7-( 181619+58106*n+4335*n^2)/(n+9)/(255*n+1213)*N^8+N^9, [6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681]], 4.090*6^n*(1/n)^(1/2)*(1+533/37/n+250411/296 /n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681, 56147447, 333839726, 1985021900, 11803528804, 70190395149, 417409912200, 2482373125000, 14763542516248, 87807840206271, 522270131681000, 3106534936881197, 18478913451779103, 109924812719279688, 653934314413092628, 3890376060255274350, 23145580582447686799, 137709399424316701632, 819366249097547144735, 4875414094260698558059, 29011077965235340163136, 172637489329251700785438, 1027366051270731010220680, 6114122403744102323868426, 36388305371016650507374899, 216574977829647209364169136, 1289060897599775360173879365, 7672859745592032079419777473, 45673018330306205202754795796, 271882186417215625776682800094, 1618528454514806077450756476952, 9635592502719568051760339887815, 57366052693492200870421861560944, 341546563981446716149056370054411, 2033589251105051752536426587647311, 12108626157269399233715395997287936, 72101578336902016901637316741723080, 429351406841302402155847671638976200, 2556814236950587384105926921537635575, 15226622364564768960091967830743804494, 90683033750708720782107507423223180637, 540090469688866397874901220747414911673] Proposition Number , 31, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 1, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 32, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 2, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 3 2 -1 + (6 t - 1) (4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 11) a(n + 5) 36 (n + 5) a(n + 4) 2 (2 n + 9) a(n + 3) a(n + 6) = --------------------- - ------------------- - -------------------- n + 6 n + 6 n + 6 24 (n + 4) a(n + 2) 2 (2 n + 7) a(n + 1) 24 (n + 3) a(n) + ------------------- + -------------------- - --------------- n + 6 n + 6 n + 6 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7562, a[6] = 44958 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 521 2502529\ 4.205 6 (1/n) |1 + ---- + -------| | 40 n 2| \ 3200 n / Note that everything is rigorous except the constant in front, 4.205 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^5-4*t^3+6*t-1)*P^2, [24*(n+3)/(n+6)-2*(2*n+7)/(n+6)*N-24*(n+4)/ (n+6)*N^2+2*(2*n+9)/(n+6)*N^3+36*(n+5)/(n+6)*N^4-6*(2*n+11)/(n+6)*N^5+N^6, [6, 36, 214, 1272, 7562, 44958]], 4.205*6^n*(1/n)^(1/2)*(1+521/40/n+2502529/3200/n^ 2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7562, 44958, 267300, 1589316, 9450196, 56194086, 334163976, 1987228726, 11818307256, 70288111376, 418049425304, 2486523268404, 14790285061524, 87979129336868, 523361595527232, 3113458587734110, 18522660336804468, 110200263338247084, 655663298780619876, 3901198605504457830, 23213154398977342440, 138130359633331334748, 821983264478393999764, 4891652813438850746832, 29111665836085553119332, 173259571618490266150768, 1031207646797687254066248, 6137813457870869748529148, 36534223227564741192094536, 217472658430937710414174884, 1294577337077895162316177296, 7706724699788043479343760612, 45880712769880308338900727384, 273154832326134868125779903736, 1626319971156341349181021068352, 9683256394543314610491985962798, 57657412536593910625062421487180, 343326320640053256691648058786504, 2044453480460431778198363148838404, 12174902716459651798539044567315652, 72505649567614945427959319917982692, 431813508858897006790224811251688428, 2571808203855980343902893285555965280, 15317886647816722444541262208671130286, 91238259470573690493118524002219522676, 543466704651662865564987092697981902116] Proposition Number , 33, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 2, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -t - (6 t - 1) (2 t - t + 6 t - 1) P 3 3 2 2 + (6 t - 1) (t - 5 t + 1) (2 t - t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (181619 + 58106 n + 4335 n ) a(n + 8) a(n + 9) = ------------------------------------- (n + 9) (255 n + 1213) 2 2 (162967 n + 12495 n + 510920) a(n + 7) - ----------------------------------------- (n + 9) (255 n + 1213) 2 (2078799 + 668402 n + 53295 n ) a(n + 6) + ---------------------------------------- (n + 9) (255 n + 1213) 2 (978061 + 369416 n + 34935 n ) a(n + 5) - --------------------------------------- (n + 9) (255 n + 1213) 2 (1272103 + 535940 n + 54825 n ) a(n + 4) + ---------------------------------------- (n + 9) (255 n + 1213) 2 2 (54361 + 32413 n + 4080 n ) a(n + 3) - -------------------------------------- (n + 9) (255 n + 1213) 2 (-59025 + 5948 n + 2805 n ) a(n + 2) + ------------------------------------ (n + 9) (255 n + 1213) 2 4 (10612 + 510 n + 4721 n) a(n + 1) 12 (255 n + 1468) (n + 3) a(n) + ------------------------------------ - ------------------------------ (n + 9) (255 n + 1213) (n + 9) (255 n + 1213) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7561, a[6] = 44946, a[7] = 267191, a[8] = 1588444, a[9] = 9443681 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 533 250411\ 4.090 6 (1/n) |1 + ---- + ------| | 37 n 2| \ 296 n / Note that everything is rigorous except the constant in front, 4.090 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(2*t^3-t^2+6*t-1)*P+(6*t-1)*(t^3-5*t+1)*(2*t^3-t^2+6*t-1)*P^2, [12*( 255*n+1468)*(n+3)/(n+9)/(255*n+1213)-4*(10612+510*n^2+4721*n)/(n+9)/(255*n+1213 )*N-(-59025+5948*n+2805*n^2)/(n+9)/(255*n+1213)*N^2+2*(54361+32413*n+4080*n^2)/ (n+9)/(255*n+1213)*N^3-(1272103+535940*n+54825*n^2)/(n+9)/(255*n+1213)*N^4+( 978061+369416*n+34935*n^2)/(n+9)/(255*n+1213)*N^5-(2078799+668402*n+53295*n^2)/ (n+9)/(255*n+1213)*N^6+2*(162967*n+12495*n^2+510920)/(n+9)/(255*n+1213)*N^7-( 181619+58106*n+4335*n^2)/(n+9)/(255*n+1213)*N^8+N^9, [6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681]], 4.090*6^n*(1/n)^(1/2)*(1+533/37/n+250411/296 /n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681, 56147447, 333839726, 1985021900, 11803528804, 70190395149, 417409912200, 2482373125000, 14763542516248, 87807840206271, 522270131681000, 3106534936881197, 18478913451779103, 109924812719279688, 653934314413092628, 3890376060255274350, 23145580582447686799, 137709399424316701632, 819366249097547144735, 4875414094260698558059, 29011077965235340163136, 172637489329251700785438, 1027366051270731010220680, 6114122403744102323868426, 36388305371016650507374899, 216574977829647209364169136, 1289060897599775360173879365, 7672859745592032079419777473, 45673018330306205202754795796, 271882186417215625776682800094, 1618528454514806077450756476952, 9635592502719568051760339887815, 57366052693492200870421861560944, 341546563981446716149056370054411, 2033589251105051752536426587647311, 12108626157269399233715395997287936, 72101578336902016901637316741723080, 429351406841302402155847671638976200, 2556814236950587384105926921537635575, 15226622364564768960091967830743804494, 90683033750708720782107507423223180637, 540090469688866397874901220747414911673] Proposition Number , 34, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 2, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 35, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 3, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 36, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 3, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 37, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 4, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 3 2 -1 + (6 t - 1) (4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 11) a(n + 5) 36 (n + 5) a(n + 4) 2 (2 n + 9) a(n + 3) a(n + 6) = --------------------- - ------------------- - -------------------- n + 6 n + 6 n + 6 24 (n + 4) a(n + 2) 2 (2 n + 7) a(n + 1) 24 (n + 3) a(n) + ------------------- + -------------------- - --------------- n + 6 n + 6 n + 6 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7562, a[6] = 44958 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 521 2502529\ 4.205 6 (1/n) |1 + ---- + -------| | 40 n 2| \ 3200 n / Note that everything is rigorous except the constant in front, 4.205 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^5-4*t^3+6*t-1)*P^2, [24*(n+3)/(n+6)-2*(2*n+7)/(n+6)*N-24*(n+4)/ (n+6)*N^2+2*(2*n+9)/(n+6)*N^3+36*(n+5)/(n+6)*N^4-6*(2*n+11)/(n+6)*N^5+N^6, [6, 36, 214, 1272, 7562, 44958]], 4.205*6^n*(1/n)^(1/2)*(1+521/40/n+2502529/3200/n^ 2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7562, 44958, 267300, 1589316, 9450196, 56194086, 334163976, 1987228726, 11818307256, 70288111376, 418049425304, 2486523268404, 14790285061524, 87979129336868, 523361595527232, 3113458587734110, 18522660336804468, 110200263338247084, 655663298780619876, 3901198605504457830, 23213154398977342440, 138130359633331334748, 821983264478393999764, 4891652813438850746832, 29111665836085553119332, 173259571618490266150768, 1031207646797687254066248, 6137813457870869748529148, 36534223227564741192094536, 217472658430937710414174884, 1294577337077895162316177296, 7706724699788043479343760612, 45880712769880308338900727384, 273154832326134868125779903736, 1626319971156341349181021068352, 9683256394543314610491985962798, 57657412536593910625062421487180, 343326320640053256691648058786504, 2044453480460431778198363148838404, 12174902716459651798539044567315652, 72505649567614945427959319917982692, 431813508858897006790224811251688428, 2571808203855980343902893285555965280, 15317886647816722444541262208671130286, 91238259470573690493118524002219522676, 543466704651662865564987092697981902116] Proposition Number , 38, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 4, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 39, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 4, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -t - (6 t - 1) (2 t - t + 6 t - 1) P 3 3 2 2 + (6 t - 1) (t - 5 t + 1) (2 t - t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (181619 + 58106 n + 4335 n ) a(n + 8) a(n + 9) = ------------------------------------- (n + 9) (255 n + 1213) 2 2 (162967 n + 12495 n + 510920) a(n + 7) - ----------------------------------------- (n + 9) (255 n + 1213) 2 (2078799 + 668402 n + 53295 n ) a(n + 6) + ---------------------------------------- (n + 9) (255 n + 1213) 2 (978061 + 369416 n + 34935 n ) a(n + 5) - --------------------------------------- (n + 9) (255 n + 1213) 2 (1272103 + 535940 n + 54825 n ) a(n + 4) + ---------------------------------------- (n + 9) (255 n + 1213) 2 2 (54361 + 32413 n + 4080 n ) a(n + 3) - -------------------------------------- (n + 9) (255 n + 1213) 2 (-59025 + 5948 n + 2805 n ) a(n + 2) + ------------------------------------ (n + 9) (255 n + 1213) 2 4 (10612 + 510 n + 4721 n) a(n + 1) 12 (255 n + 1468) (n + 3) a(n) + ------------------------------------ - ------------------------------ (n + 9) (255 n + 1213) (n + 9) (255 n + 1213) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7561, a[6] = 44946, a[7] = 267191, a[8] = 1588444, a[9] = 9443681 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 533 250411\ 4.090 6 (1/n) |1 + ---- + ------| | 37 n 2| \ 296 n / Note that everything is rigorous except the constant in front, 4.090 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(2*t^3-t^2+6*t-1)*P+(6*t-1)*(t^3-5*t+1)*(2*t^3-t^2+6*t-1)*P^2, [12*( 255*n+1468)*(n+3)/(n+9)/(255*n+1213)-4*(10612+510*n^2+4721*n)/(n+9)/(255*n+1213 )*N-(-59025+5948*n+2805*n^2)/(n+9)/(255*n+1213)*N^2+2*(54361+32413*n+4080*n^2)/ (n+9)/(255*n+1213)*N^3-(1272103+535940*n+54825*n^2)/(n+9)/(255*n+1213)*N^4+( 978061+369416*n+34935*n^2)/(n+9)/(255*n+1213)*N^5-(2078799+668402*n+53295*n^2)/ (n+9)/(255*n+1213)*N^6+2*(162967*n+12495*n^2+510920)/(n+9)/(255*n+1213)*N^7-( 181619+58106*n+4335*n^2)/(n+9)/(255*n+1213)*N^8+N^9, [6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681]], 4.090*6^n*(1/n)^(1/2)*(1+533/37/n+250411/296 /n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681, 56147447, 333839726, 1985021900, 11803528804, 70190395149, 417409912200, 2482373125000, 14763542516248, 87807840206271, 522270131681000, 3106534936881197, 18478913451779103, 109924812719279688, 653934314413092628, 3890376060255274350, 23145580582447686799, 137709399424316701632, 819366249097547144735, 4875414094260698558059, 29011077965235340163136, 172637489329251700785438, 1027366051270731010220680, 6114122403744102323868426, 36388305371016650507374899, 216574977829647209364169136, 1289060897599775360173879365, 7672859745592032079419777473, 45673018330306205202754795796, 271882186417215625776682800094, 1618528454514806077450756476952, 9635592502719568051760339887815, 57366052693492200870421861560944, 341546563981446716149056370054411, 2033589251105051752536426587647311, 12108626157269399233715395997287936, 72101578336902016901637316741723080, 429351406841302402155847671638976200, 2556814236950587384105926921537635575, 15226622364564768960091967830743804494, 90683033750708720782107507423223180637, 540090469688866397874901220747414911673] Proposition Number , 40, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 4, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 41, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 1, 2] as those of, [3, 4, 5] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 42, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [1, 2, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 4 3 2 -t - (6 t - 1) (6 t - t + 8 t - 2 t + 6 t - 1) P 5 4 3 2 2 - (6 t - 1) (5 t - 1) (6 t - t + 8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (66108 + 28671 n + 2660 n ) a(n + 7) a(n + 8) = ------------------------------------ (n + 8) (158 n + 499) 2 2 (78101 n + 7521 n + 183400) a(n + 6) - --------------------------------------- (n + 8) (158 n + 499) 2 2 (360690 + 15316 n + 151371 n) a(n + 5) + ----------------------------------------- (n + 8) (158 n + 499) 2 (348420 + 18082 n + 155303 n) a(n + 4) - --------------------------------------- (n + 8) (158 n + 499) 2 (714650 + 36732 n + 324931 n) a(n + 3) + --------------------------------------- (n + 8) (158 n + 499) 2 2 (31083 + 4243 n + 25770 n) a(n + 2) - -------------------------------------- (n + 8) (158 n + 499) 2 24 (13048 + 1081 n + 8539 n) a(n + 1) 360 (13 n + 124) (n + 1) a(n) + -------------------------------------- + ----------------------------- (n + 8) (158 n + 499) (n + 8) (158 n + 499) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7563, a[6] = 44970, a[7] = 267404, a[8] = 1590124 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 475 5518051\ 4.146 6 (1/n) |1 + ---- + -------| | 32 n 2| \ 6144 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(6*t^5-t^4+8*t^3-2*t^2+6*t-1)*P-(6*t-1)*(5*t-1)*(6*t^5-t^4+8*t^3-2*t ^2+6*t-1)*P^2, [-360*(13*n+124)*(n+1)/(n+8)/(158*n+499)-24*(13048+1081*n^2+8539 *n)/(n+8)/(158*n+499)*N+2*(31083+4243*n^2+25770*n)/(n+8)/(158*n+499)*N^2-( 714650+36732*n^2+324931*n)/(n+8)/(158*n+499)*N^3+(348420+18082*n^2+155303*n)/(n +8)/(158*n+499)*N^4-2*(360690+15316*n^2+151371*n)/(n+8)/(158*n+499)*N^5+2*( 78101*n+7521*n^2+183400)/(n+8)/(158*n+499)*N^6-(66108+28671*n+2660*n^2)/(n+8)/( 158*n+499)*N^7+N^8, [6, 36, 214, 1272, 7563, 44970, 267404, 1590124]], 4.146*6^ n*(1/n)^(1/2)*(1+475/32/n+5518051/6144/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7563, 44970, 267404, 1590124, 9456105, 56235657, 334448649, 1989139738, 11830941475, 70370635328, 418583185888, 2489947635948, 14812104305103, 88117345672707, 524232729504521, 3118924913068933, 18556828335074974, 110413100144521621, 656985008404103183, 3909383670056342758, 23263715890842936399, 138441981794281648509, 823899872354158707277, 4903418290178399535504, 29183763766743366407553, 173700665489812529798640, 1033902192595810255096604, 6154250822218346294159428, 36634364190883836804506855, 218082000321969613529426573, 1298280843234694862185370923, 7729209951988533431886244479, 46017090108955653699315319024, 273981195693660308255074524881, 1631322697586360889690055884423, 9713516457563042570852728723737, 57840297952541056988151299041116, 344430785101414334799002723685492, 2051118535045184752314129781624140, 12215095639197088829611745785961877, 72747865593330709904405188125229536, 433272242649538566015752751327707871, 2580587923471380933902626140778237911, 15370698012885394944391120104143238334, 91555746672844212946716627249337069537, 545374300536898773491123160416428473929] Proposition Number , 43, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [1, 3, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 3 2 1 - 24 t + 4 t - 7 t - 2 (6 t - 1) (20 t - 4 t + 6 t - 1) P 3 2 2 - (6 t - 1) (5 t - 1) (20 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence (62 + 17 n) a(n + 4) 10 (33 + 10 n) a(n + 3) a(n + 5) = -------------------- - ----------------------- n + 4 n + 4 2 (355 + 122 n) a(n + 2) 10 (79 + 34 n) a(n + 1) 600 (n + 2) a(n) + ------------------------ - ----------------------- + ---------------- n + 4 n + 4 n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7566 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 137 137521\ 4.15 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.15 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format 1-24*t^3+4*t^2-7*t-2*(6*t-1)*(20*t^3-4*t^2+6*t-1)*P-(6*t-1)*(5*t-1)*(20*t^3-4*t ^2+6*t-1)*P^2, [-600*(n+2)/(n+4)+10*(79+34*n)/(n+4)*N-2*(355+122*n)/(n+4)*N^2+ 10*(33+10*n)/(n+4)*N^3-(62+17*n)/(n+4)*N^4+N^5, [6, 36, 214, 1272, 7566]], 4.15 *6^n*(1/n)^(1/2)*(1+137/8/n+137521/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7566, 45006, 267712, 1592504, 9473522, 56358426, 335291394, 1994813350, 11868571656, 70617275508, 420184242444, 2500258199352, 14878057910106, 88536817193598, 526887422413054, 3135653104555922, 18661838698439306, 111070078580931458, 661082924863651894, 3934875523185191462, 23421906246438697616, 139421465218560806356, 829952415564046620792, 4940749912370886101964, 29413632221628646101372, 175113864111261292439156, 1042577794611934715470752, 6207438906245767957504248, 36960042117802649306879778, 220073858290682533643737534, 1310449921626239942437516202, 7803480454544717499159952694, 46469947755726732727379308446, 276739985446278452681821227006, 1648114982562681628375702746898, 9815647110688146037371617138514, 58460990701648487274406585440402, 348200331488500792843772020316226, 2073996090350373818439707132853498, 12353852113688223528853135694703954, 73588939064017410770510041825653222, 438367481672026753907815130769533766, 2611438064108820242574258865081764918, 15557388797009748632717063502330258534, 92684950117514908752380049547054910832, 552201063848208832474277716680953408076] Proposition Number , 44, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [1, 3, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 4 3 2 -t - (6 t - 1) (6 t - t + 8 t - 2 t + 6 t - 1) P 5 4 3 2 2 - (6 t - 1) (5 t - 1) (6 t - t + 8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (66108 + 28671 n + 2660 n ) a(n + 7) a(n + 8) = ------------------------------------ (n + 8) (158 n + 499) 2 2 (78101 n + 7521 n + 183400) a(n + 6) - --------------------------------------- (n + 8) (158 n + 499) 2 2 (360690 + 15316 n + 151371 n) a(n + 5) + ----------------------------------------- (n + 8) (158 n + 499) 2 (348420 + 18082 n + 155303 n) a(n + 4) - --------------------------------------- (n + 8) (158 n + 499) 2 (714650 + 36732 n + 324931 n) a(n + 3) + --------------------------------------- (n + 8) (158 n + 499) 2 2 (31083 + 4243 n + 25770 n) a(n + 2) - -------------------------------------- (n + 8) (158 n + 499) 2 24 (13048 + 1081 n + 8539 n) a(n + 1) 360 (13 n + 124) (n + 1) a(n) + -------------------------------------- + ----------------------------- (n + 8) (158 n + 499) (n + 8) (158 n + 499) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7563, a[6] = 44970, a[7] = 267404, a[8] = 1590124 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 475 5518051\ 4.146 6 (1/n) |1 + ---- + -------| | 32 n 2| \ 6144 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(6*t^5-t^4+8*t^3-2*t^2+6*t-1)*P-(6*t-1)*(5*t-1)*(6*t^5-t^4+8*t^3-2*t ^2+6*t-1)*P^2, [-360*(13*n+124)*(n+1)/(n+8)/(158*n+499)-24*(13048+1081*n^2+8539 *n)/(n+8)/(158*n+499)*N+2*(31083+4243*n^2+25770*n)/(n+8)/(158*n+499)*N^2-( 714650+36732*n^2+324931*n)/(n+8)/(158*n+499)*N^3+(348420+18082*n^2+155303*n)/(n +8)/(158*n+499)*N^4-2*(360690+15316*n^2+151371*n)/(n+8)/(158*n+499)*N^5+2*( 78101*n+7521*n^2+183400)/(n+8)/(158*n+499)*N^6-(66108+28671*n+2660*n^2)/(n+8)/( 158*n+499)*N^7+N^8, [6, 36, 214, 1272, 7563, 44970, 267404, 1590124]], 4.146*6^ n*(1/n)^(1/2)*(1+475/32/n+5518051/6144/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7563, 44970, 267404, 1590124, 9456105, 56235657, 334448649, 1989139738, 11830941475, 70370635328, 418583185888, 2489947635948, 14812104305103, 88117345672707, 524232729504521, 3118924913068933, 18556828335074974, 110413100144521621, 656985008404103183, 3909383670056342758, 23263715890842936399, 138441981794281648509, 823899872354158707277, 4903418290178399535504, 29183763766743366407553, 173700665489812529798640, 1033902192595810255096604, 6154250822218346294159428, 36634364190883836804506855, 218082000321969613529426573, 1298280843234694862185370923, 7729209951988533431886244479, 46017090108955653699315319024, 273981195693660308255074524881, 1631322697586360889690055884423, 9713516457563042570852728723737, 57840297952541056988151299041116, 344430785101414334799002723685492, 2051118535045184752314129781624140, 12215095639197088829611745785961877, 72747865593330709904405188125229536, 433272242649538566015752751327707871, 2580587923471380933902626140778237911, 15370698012885394944391120104143238334, 91555746672844212946716627249337069537, 545374300536898773491123160416428473929] Proposition Number , 45, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [1, 3, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 4 3 2 -t - (6 t - 1) (6 t - t + 8 t - 2 t + 6 t - 1) P 5 4 3 2 2 - (6 t - 1) (5 t - 1) (6 t - t + 8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (66108 + 28671 n + 2660 n ) a(n + 7) a(n + 8) = ------------------------------------ (n + 8) (158 n + 499) 2 2 (78101 n + 7521 n + 183400) a(n + 6) - --------------------------------------- (n + 8) (158 n + 499) 2 2 (360690 + 15316 n + 151371 n) a(n + 5) + ----------------------------------------- (n + 8) (158 n + 499) 2 (348420 + 18082 n + 155303 n) a(n + 4) - --------------------------------------- (n + 8) (158 n + 499) 2 (714650 + 36732 n + 324931 n) a(n + 3) + --------------------------------------- (n + 8) (158 n + 499) 2 2 (31083 + 4243 n + 25770 n) a(n + 2) - -------------------------------------- (n + 8) (158 n + 499) 2 24 (13048 + 1081 n + 8539 n) a(n + 1) 360 (13 n + 124) (n + 1) a(n) + -------------------------------------- + ----------------------------- (n + 8) (158 n + 499) (n + 8) (158 n + 499) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7563, a[6] = 44970, a[7] = 267404, a[8] = 1590124 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 475 5518051\ 4.146 6 (1/n) |1 + ---- + -------| | 32 n 2| \ 6144 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(6*t^5-t^4+8*t^3-2*t^2+6*t-1)*P-(6*t-1)*(5*t-1)*(6*t^5-t^4+8*t^3-2*t ^2+6*t-1)*P^2, [-360*(13*n+124)*(n+1)/(n+8)/(158*n+499)-24*(13048+1081*n^2+8539 *n)/(n+8)/(158*n+499)*N+2*(31083+4243*n^2+25770*n)/(n+8)/(158*n+499)*N^2-( 714650+36732*n^2+324931*n)/(n+8)/(158*n+499)*N^3+(348420+18082*n^2+155303*n)/(n +8)/(158*n+499)*N^4-2*(360690+15316*n^2+151371*n)/(n+8)/(158*n+499)*N^5+2*( 78101*n+7521*n^2+183400)/(n+8)/(158*n+499)*N^6-(66108+28671*n+2660*n^2)/(n+8)/( 158*n+499)*N^7+N^8, [6, 36, 214, 1272, 7563, 44970, 267404, 1590124]], 4.146*6^ n*(1/n)^(1/2)*(1+475/32/n+5518051/6144/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7563, 44970, 267404, 1590124, 9456105, 56235657, 334448649, 1989139738, 11830941475, 70370635328, 418583185888, 2489947635948, 14812104305103, 88117345672707, 524232729504521, 3118924913068933, 18556828335074974, 110413100144521621, 656985008404103183, 3909383670056342758, 23263715890842936399, 138441981794281648509, 823899872354158707277, 4903418290178399535504, 29183763766743366407553, 173700665489812529798640, 1033902192595810255096604, 6154250822218346294159428, 36634364190883836804506855, 218082000321969613529426573, 1298280843234694862185370923, 7729209951988533431886244479, 46017090108955653699315319024, 273981195693660308255074524881, 1631322697586360889690055884423, 9713516457563042570852728723737, 57840297952541056988151299041116, 344430785101414334799002723685492, 2051118535045184752314129781624140, 12215095639197088829611745785961877, 72747865593330709904405188125229536, 433272242649538566015752751327707871, 2580587923471380933902626140778237911, 15370698012885394944391120104143238334, 91555746672844212946716627249337069537, 545374300536898773491123160416428473929] Proposition Number , 46, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [2, 1, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 4 3 3 2 -1 - 4 t + t - 12 t - 10 t - 2 t (6 t - 1) (8 t + 10 t + 4 t - 1) P 2 3 2 2 - (6 t - 1) (4 t + t - 1) (8 t + 10 t + 4 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence (96 + 13 n) a(n + 7) 4 (71 + 11 n) a(n + 6) a(n + 8) = -------------------- - ---------------------- n + 8 n + 8 2 (109 + 8 n) a(n + 5) 2 (563 + 88 n) a(n + 4) - ---------------------- + ----------------------- n + 8 n + 8 4 (-81 + 5 n) a(n + 3) 4 (209 + 50 n) a(n + 2) 16 (1 + 8 n) a(n + 1) - ---------------------- - ----------------------- + --------------------- n + 8 n + 8 n + 8 384 (n + 3) a(n) + ---------------- n + 8 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1276, a[5] = 7604, a[6] = 45318, a[7] = 270090, a[8] = 1609740 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 1879 32088481\ 4.88 6 (1/n) |1 + ----- + --------| | 104 n 2| \ 21632 n / Note that everything is rigorous except the constant in front, 4.88 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-4*t^2+t-12*t^4-10*t^3-2*t*(6*t-1)*(8*t^3+10*t^2+4*t-1)*P-(6*t-1)*(4*t^2+t-1) *(8*t^3+10*t^2+4*t-1)*P^2, [-384*(n+3)/(n+8)-16*(1+8*n)/(n+8)*N+4*(209+50*n)/(n +8)*N^2+4*(-81+5*n)/(n+8)*N^3-2*(563+88*n)/(n+8)*N^4+2*(109+8*n)/(n+8)*N^5+4*( 71+11*n)/(n+8)*N^6-(96+13*n)/(n+8)*N^7+N^8, [6, 36, 214, 1276, 7604, 45318, 270090, 1609740]], 4.88*6^n*(1/n)^(1/2)*(1+1879/104/n+32088481/21632/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1276, 7604, 45318, 270090, 1609740, 9594304, 57184904, 340847060, 2031645502, 12110066510, 72186392734, 430302987082, 2565095777300, 15291250708492, 91157545783412, 543441030958520, 3239831367358744, 19315348939009344, 115157671611719724, 686583479129464812, 4093586497733810286, 24407580711915536622, 145531038383283692058, 867754039916218231798, 5174254345095267194174, 30853824480864175603334, 183984115283796625831370, 1097139206507071077191602, 6542642829214648407665508, 39017078694394456605779068, 232683900470560976708010848, 1387675617380377009228863876, 8275983267565408281979849940, 49358421481852536234294293108, 294383130523087671586823872528, 1755798049429805394937820685392, 10472398921294237902990742275080, 62463706454131679111924346306128, 372579768344016090131529705416912, 2222392259259689280926020864563696, 13256597735996571779784965436311292, 79077574551274342916162212596386796, 471720214054529122632728772295939284, 2814009131309456850173365296194157212, 16787126787607583426400048029133277630, 100146799608309862144342250634688450902, 597458300922148340925030067470073469658] Proposition Number , 47, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [2, 1, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 5 4 3 2 -t (t + 1 - t) - (6 t - 1) (2 t + 3 t + 8 t - 2 t + 6 t - 1) P 3 2 5 4 3 2 2 + (6 t - 1) (t - t - 5 t + 1) (2 t + 3 t + 8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (2318710 n + 163862 n + 7291607) a(n + 9) a(n + 10) = ------------------------------------------ (n + 10) (9719 n + 44048) 2 (39961537 + 12638986 n + 919606 n ) a(n + 8) - -------------------------------------------- (n + 10) (9719 n + 44048) 2 (23506434 n + 1782626 n + 73752647) a(n + 7) + --------------------------------------------- (n + 10) (9719 n + 44048) 2 (7071606 n + 655188 n + 18136093) a(n + 6) - ------------------------------------------- (n + 10) (9719 n + 44048) 2 (50440393 + 15987706 n + 1338274 n ) a(n + 5) + --------------------------------------------- (n + 10) (9719 n + 44048) 2 (57973629 + 19186318 n + 1498138 n ) a(n + 4) + --------------------------------------------- (n + 10) (9719 n + 44048) 2 (14372093 + 5923310 n + 482302 n ) a(n + 3) + ------------------------------------------- (n + 10) (9719 n + 44048) 2 (-1159919 + 3315 n + 41676 n) a(n + 2) + --------------------------------------- (n + 10) (9719 n + 44048) 2 4 (965549 + 358904 n + 30518 n ) a(n + 1) - ----------------------------------------- (n + 10) (9719 n + 44048) 12 (1361 n + 15856) (n + 3) a(n) - -------------------------------- (n + 10) (9719 n + 44048) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1274, a[5] = 7585, a[6] = 45159, a[7] = 268871, a[8] = 1600874, a[9] = 9531993, a[10] = 56757538 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 2159 37934977\ 4.47 6 (1/n) |1 + ----- + --------| | 124 n 2| \ 30752 n / Note that everything is rigorous except the constant in front, 4.47 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t*(t^2+1-t)-(6*t-1)*(2*t^5+3*t^4+8*t^3-2*t^2+6*t-1)*P+(6*t-1)*(t^3-t^2-5*t+1)* (2*t^5+3*t^4+8*t^3-2*t^2+6*t-1)*P^2, [12*(1361*n+15856)*(n+3)/(n+10)/(9719*n+ 44048)+4*(965549+358904*n+30518*n^2)/(n+10)/(9719*n+44048)*N-(-1159919+3315*n^2 +41676*n)/(n+10)/(9719*n+44048)*N^2-(14372093+5923310*n+482302*n^2)/(n+10)/( 9719*n+44048)*N^3-(57973629+19186318*n+1498138*n^2)/(n+10)/(9719*n+44048)*N^4-( 50440393+15987706*n+1338274*n^2)/(n+10)/(9719*n+44048)*N^5+(7071606*n+655188*n^ 2+18136093)/(n+10)/(9719*n+44048)*N^6-(23506434*n+1782626*n^2+73752647)/(n+10)/ (9719*n+44048)*N^7+(39961537+12638986*n+919606*n^2)/(n+10)/(9719*n+44048)*N^8-( 2318710*n+163862*n^2+7291607)/(n+10)/(9719*n+44048)*N^9+N^10, [6, 36, 214, 1274 , 7585, 45159, 268871, 1600874, 9531993, 56757538]], 4.47*6^n*(1/n)^(1/2)*(1+ 2159/124/n+37934977/30752/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1274, 7585, 45159, 268871, 1600874, 9531993, 56757538, 337968828, 2012532976, 11984574579, 71369966255, 425031977935, 2531284498095, 15075563142236, 89788217866860, 534783961152718, 3185301901831095, 18973000723855156, 113014620339367937, 673203571271044585, 4010249136713644998, 23889631270013404661, 142318276629486667606, 847861722731956889976, 5051292201855217772990, 30094910043229332221525, 179306769891016028203079, 1068349592732643743252879, 6365655601286527783808648, 37930268693578976809052139, 226017316205563407800853334, 1346822936764874166880065320, 8025872782303157004008952804, 47828526049959534278577958799, 285032683577703407648772691911, 1698694359383326595665231216719, 10123920486533680897438413593051, 60338579686273816678091942649568, 359628660619797694722645488735708, 2143514080758438484309130346510278, 12776479305521032435244233394806901, 76156826354083839030920568396200102, 453961729817467757478277892256740765, 2706090928171639191927889509832783949, 16131629625088180574289860142992867873, 96167159305320481378905857178353810210, 573307991392973961616755566311866620680] Proposition Number , 48, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [2, 3, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 5 4 2 1 - 18 t - 12 t - 7 t + 2 t + 3 t 2 3 2 - 2 (6 t - 1) (t + 1) (8 t - 2 t + 6 t - 1) P 3 2 3 2 2 - (6 t - 1) (4 t - t + 5 t - 1) (8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence (130 + 17 n) a(n + 8) (734 + 101 n) a(n + 7) a(n + 9) = --------------------- - ---------------------- n + 8 n + 8 20 (87 + 13 n) a(n + 6) 6 (403 + 72 n) a(n + 5) + ----------------------- - ----------------------- n + 8 n + 8 2 (2221 + 432 n) a(n + 4) 4 (633 + 149 n) a(n + 3) + ------------------------- - ------------------------ n + 8 n + 8 4 (987 + 254 n) a(n + 2) 16 (53 + 16 n) a(n + 1) 384 (n + 3) a(n) + ------------------------ - ----------------------- + ---------------- n + 8 n + 8 n + 8 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7562, a[6] = 44958, a[7] = 267298, a[8] = 1589288, a[9] = 9449910 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 2473 44608609\ 4.035 6 (1/n) |1 + ----- + --------| | 152 n 2| \ 46208 n / Note that everything is rigorous except the constant in front, 4.035 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format 1-18*t^3-12*t^5-7*t+2*t^4+3*t^2-2*(6*t-1)*(t^2+1)*(8*t^3-2*t^2+6*t-1)*P-(6*t-1) *(4*t^3-t^2+5*t-1)*(8*t^3-2*t^2+6*t-1)*P^2, [-384*(n+3)/(n+8)+16*(53+16*n)/(n+8 )*N-4*(987+254*n)/(n+8)*N^2+4*(633+149*n)/(n+8)*N^3-2*(2221+432*n)/(n+8)*N^4+6* (403+72*n)/(n+8)*N^5-20*(87+13*n)/(n+8)*N^6+(734+101*n)/(n+8)*N^7-(130+17*n)/(n +8)*N^8+N^9, [6, 36, 214, 1272, 7562, 44958, 267298, 1589288, 9449910]], 4.035* 6^n*(1/n)^(1/2)*(1+2473/152/n+44608609/46208/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7562, 44958, 267298, 1589288, 9449910, 56191546, 334143124, 1987066058, 11817083468, 70279152406, 417985226274, 2486071089240, 14787145203950, 87957586155998, 523215286617796, 3112473682666518, 18516081254182386, 110156613723673548, 655375447127053202, 3899310585596868326, 23200831064069829986, 138050277742555443344, 821464944665619228190, 4888310331065522981042, 29090183581909732294604, 173121931258471387410442, 1030328281216062597216898, 6132210182146699265392264, 36498607253013540699307102, 217246792330903989889581694, 1293148026925461208746346600, 7697697966390080985559827954, 45823812091351817474272986598, 272796788583469042338377432692, 1624070754054147740413469030038, 9669149080201283605337849997146, 57569061298757579039981648524732, 342773771514333480925817898643578, 2041002432073373258861226402819654, 12153375761499747480002889082296228, 72371529773627208080420289962066422, 430978853430744035818690407986362738, 2566619615754676257279877011107211956, 15285665520218796796907225459568979978, 91038364300131494078122984590089400716, 542227756846253791374764958224304638054] Proposition Number , 49, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [2, 3, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -t - (6 t - 1) (2 t - t + 6 t - 1) P 3 3 2 2 + (6 t - 1) (t - 5 t + 1) (2 t - t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (181619 + 58106 n + 4335 n ) a(n + 8) a(n + 9) = ------------------------------------- (n + 9) (255 n + 1213) 2 2 (162967 n + 12495 n + 510920) a(n + 7) - ----------------------------------------- (n + 9) (255 n + 1213) 2 (2078799 + 668402 n + 53295 n ) a(n + 6) + ---------------------------------------- (n + 9) (255 n + 1213) 2 (978061 + 369416 n + 34935 n ) a(n + 5) - --------------------------------------- (n + 9) (255 n + 1213) 2 (1272103 + 535940 n + 54825 n ) a(n + 4) + ---------------------------------------- (n + 9) (255 n + 1213) 2 2 (54361 + 32413 n + 4080 n ) a(n + 3) - -------------------------------------- (n + 9) (255 n + 1213) 2 (-59025 + 5948 n + 2805 n ) a(n + 2) + ------------------------------------ (n + 9) (255 n + 1213) 2 4 (10612 + 510 n + 4721 n) a(n + 1) 12 (255 n + 1468) (n + 3) a(n) + ------------------------------------ - ------------------------------ (n + 9) (255 n + 1213) (n + 9) (255 n + 1213) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7561, a[6] = 44946, a[7] = 267191, a[8] = 1588444, a[9] = 9443681 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 533 250411\ 4.090 6 (1/n) |1 + ---- + ------| | 37 n 2| \ 296 n / Note that everything is rigorous except the constant in front, 4.090 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(2*t^3-t^2+6*t-1)*P+(6*t-1)*(t^3-5*t+1)*(2*t^3-t^2+6*t-1)*P^2, [12*( 255*n+1468)*(n+3)/(n+9)/(255*n+1213)-4*(10612+510*n^2+4721*n)/(n+9)/(255*n+1213 )*N-(-59025+5948*n+2805*n^2)/(n+9)/(255*n+1213)*N^2+2*(54361+32413*n+4080*n^2)/ (n+9)/(255*n+1213)*N^3-(1272103+535940*n+54825*n^2)/(n+9)/(255*n+1213)*N^4+( 978061+369416*n+34935*n^2)/(n+9)/(255*n+1213)*N^5-(2078799+668402*n+53295*n^2)/ (n+9)/(255*n+1213)*N^6+2*(162967*n+12495*n^2+510920)/(n+9)/(255*n+1213)*N^7-( 181619+58106*n+4335*n^2)/(n+9)/(255*n+1213)*N^8+N^9, [6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681]], 4.090*6^n*(1/n)^(1/2)*(1+533/37/n+250411/296 /n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681, 56147447, 333839726, 1985021900, 11803528804, 70190395149, 417409912200, 2482373125000, 14763542516248, 87807840206271, 522270131681000, 3106534936881197, 18478913451779103, 109924812719279688, 653934314413092628, 3890376060255274350, 23145580582447686799, 137709399424316701632, 819366249097547144735, 4875414094260698558059, 29011077965235340163136, 172637489329251700785438, 1027366051270731010220680, 6114122403744102323868426, 36388305371016650507374899, 216574977829647209364169136, 1289060897599775360173879365, 7672859745592032079419777473, 45673018330306205202754795796, 271882186417215625776682800094, 1618528454514806077450756476952, 9635592502719568051760339887815, 57366052693492200870421861560944, 341546563981446716149056370054411, 2033589251105051752536426587647311, 12108626157269399233715395997287936, 72101578336902016901637316741723080, 429351406841302402155847671638976200, 2556814236950587384105926921537635575, 15226622364564768960091967830743804494, 90683033750708720782107507423223180637, 540090469688866397874901220747414911673] Proposition Number , 50, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [3, 1, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -t - (6 t - 1) (2 t - t + 6 t - 1) P 3 3 2 2 + (6 t - 1) (t - 5 t + 1) (2 t - t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (181619 + 58106 n + 4335 n ) a(n + 8) a(n + 9) = ------------------------------------- (n + 9) (255 n + 1213) 2 2 (162967 n + 12495 n + 510920) a(n + 7) - ----------------------------------------- (n + 9) (255 n + 1213) 2 (2078799 + 668402 n + 53295 n ) a(n + 6) + ---------------------------------------- (n + 9) (255 n + 1213) 2 (978061 + 369416 n + 34935 n ) a(n + 5) - --------------------------------------- (n + 9) (255 n + 1213) 2 (1272103 + 535940 n + 54825 n ) a(n + 4) + ---------------------------------------- (n + 9) (255 n + 1213) 2 2 (54361 + 32413 n + 4080 n ) a(n + 3) - -------------------------------------- (n + 9) (255 n + 1213) 2 (-59025 + 5948 n + 2805 n ) a(n + 2) + ------------------------------------ (n + 9) (255 n + 1213) 2 4 (10612 + 510 n + 4721 n) a(n + 1) 12 (255 n + 1468) (n + 3) a(n) + ------------------------------------ - ------------------------------ (n + 9) (255 n + 1213) (n + 9) (255 n + 1213) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7561, a[6] = 44946, a[7] = 267191, a[8] = 1588444, a[9] = 9443681 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 533 250411\ 4.090 6 (1/n) |1 + ---- + ------| | 37 n 2| \ 296 n / Note that everything is rigorous except the constant in front, 4.090 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(2*t^3-t^2+6*t-1)*P+(6*t-1)*(t^3-5*t+1)*(2*t^3-t^2+6*t-1)*P^2, [12*( 255*n+1468)*(n+3)/(n+9)/(255*n+1213)-4*(10612+510*n^2+4721*n)/(n+9)/(255*n+1213 )*N-(-59025+5948*n+2805*n^2)/(n+9)/(255*n+1213)*N^2+2*(54361+32413*n+4080*n^2)/ (n+9)/(255*n+1213)*N^3-(1272103+535940*n+54825*n^2)/(n+9)/(255*n+1213)*N^4+( 978061+369416*n+34935*n^2)/(n+9)/(255*n+1213)*N^5-(2078799+668402*n+53295*n^2)/ (n+9)/(255*n+1213)*N^6+2*(162967*n+12495*n^2+510920)/(n+9)/(255*n+1213)*N^7-( 181619+58106*n+4335*n^2)/(n+9)/(255*n+1213)*N^8+N^9, [6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681]], 4.090*6^n*(1/n)^(1/2)*(1+533/37/n+250411/296 /n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681, 56147447, 333839726, 1985021900, 11803528804, 70190395149, 417409912200, 2482373125000, 14763542516248, 87807840206271, 522270131681000, 3106534936881197, 18478913451779103, 109924812719279688, 653934314413092628, 3890376060255274350, 23145580582447686799, 137709399424316701632, 819366249097547144735, 4875414094260698558059, 29011077965235340163136, 172637489329251700785438, 1027366051270731010220680, 6114122403744102323868426, 36388305371016650507374899, 216574977829647209364169136, 1289060897599775360173879365, 7672859745592032079419777473, 45673018330306205202754795796, 271882186417215625776682800094, 1618528454514806077450756476952, 9635592502719568051760339887815, 57366052693492200870421861560944, 341546563981446716149056370054411, 2033589251105051752536426587647311, 12108626157269399233715395997287936, 72101578336902016901637316741723080, 429351406841302402155847671638976200, 2556814236950587384105926921537635575, 15226622364564768960091967830743804494, 90683033750708720782107507423223180637, 540090469688866397874901220747414911673] Proposition Number , 51, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [3, 2, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 5 4 2 1 - 18 t - 12 t - 7 t + 2 t + 3 t 2 3 2 - 2 (6 t - 1) (t + 1) (8 t - 2 t + 6 t - 1) P 3 2 3 2 2 - (6 t - 1) (4 t - t + 5 t - 1) (8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence (130 + 17 n) a(n + 8) (734 + 101 n) a(n + 7) a(n + 9) = --------------------- - ---------------------- n + 8 n + 8 20 (87 + 13 n) a(n + 6) 6 (403 + 72 n) a(n + 5) + ----------------------- - ----------------------- n + 8 n + 8 2 (2221 + 432 n) a(n + 4) 4 (633 + 149 n) a(n + 3) + ------------------------- - ------------------------ n + 8 n + 8 4 (987 + 254 n) a(n + 2) 16 (53 + 16 n) a(n + 1) 384 (n + 3) a(n) + ------------------------ - ----------------------- + ---------------- n + 8 n + 8 n + 8 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7562, a[6] = 44958, a[7] = 267298, a[8] = 1589288, a[9] = 9449910 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 2473 44608609\ 4.035 6 (1/n) |1 + ----- + --------| | 152 n 2| \ 46208 n / Note that everything is rigorous except the constant in front, 4.035 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format 1-18*t^3-12*t^5-7*t+2*t^4+3*t^2-2*(6*t-1)*(t^2+1)*(8*t^3-2*t^2+6*t-1)*P-(6*t-1) *(4*t^3-t^2+5*t-1)*(8*t^3-2*t^2+6*t-1)*P^2, [-384*(n+3)/(n+8)+16*(53+16*n)/(n+8 )*N-4*(987+254*n)/(n+8)*N^2+4*(633+149*n)/(n+8)*N^3-2*(2221+432*n)/(n+8)*N^4+6* (403+72*n)/(n+8)*N^5-20*(87+13*n)/(n+8)*N^6+(734+101*n)/(n+8)*N^7-(130+17*n)/(n +8)*N^8+N^9, [6, 36, 214, 1272, 7562, 44958, 267298, 1589288, 9449910]], 4.035* 6^n*(1/n)^(1/2)*(1+2473/152/n+44608609/46208/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7562, 44958, 267298, 1589288, 9449910, 56191546, 334143124, 1987066058, 11817083468, 70279152406, 417985226274, 2486071089240, 14787145203950, 87957586155998, 523215286617796, 3112473682666518, 18516081254182386, 110156613723673548, 655375447127053202, 3899310585596868326, 23200831064069829986, 138050277742555443344, 821464944665619228190, 4888310331065522981042, 29090183581909732294604, 173121931258471387410442, 1030328281216062597216898, 6132210182146699265392264, 36498607253013540699307102, 217246792330903989889581694, 1293148026925461208746346600, 7697697966390080985559827954, 45823812091351817474272986598, 272796788583469042338377432692, 1624070754054147740413469030038, 9669149080201283605337849997146, 57569061298757579039981648524732, 342773771514333480925817898643578, 2041002432073373258861226402819654, 12153375761499747480002889082296228, 72371529773627208080420289962066422, 430978853430744035818690407986362738, 2566619615754676257279877011107211956, 15285665520218796796907225459568979978, 91038364300131494078122984590089400716, 542227756846253791374764958224304638054] Proposition Number , 52, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [3, 2, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -t - (6 t - 1) (2 t - t + 6 t - 1) P 3 3 2 2 + (6 t - 1) (t - 5 t + 1) (2 t - t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (181619 + 58106 n + 4335 n ) a(n + 8) a(n + 9) = ------------------------------------- (n + 9) (255 n + 1213) 2 2 (162967 n + 12495 n + 510920) a(n + 7) - ----------------------------------------- (n + 9) (255 n + 1213) 2 (2078799 + 668402 n + 53295 n ) a(n + 6) + ---------------------------------------- (n + 9) (255 n + 1213) 2 (978061 + 369416 n + 34935 n ) a(n + 5) - --------------------------------------- (n + 9) (255 n + 1213) 2 (1272103 + 535940 n + 54825 n ) a(n + 4) + ---------------------------------------- (n + 9) (255 n + 1213) 2 2 (54361 + 32413 n + 4080 n ) a(n + 3) - -------------------------------------- (n + 9) (255 n + 1213) 2 (-59025 + 5948 n + 2805 n ) a(n + 2) + ------------------------------------ (n + 9) (255 n + 1213) 2 4 (10612 + 510 n + 4721 n) a(n + 1) 12 (255 n + 1468) (n + 3) a(n) + ------------------------------------ - ------------------------------ (n + 9) (255 n + 1213) (n + 9) (255 n + 1213) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7561, a[6] = 44946, a[7] = 267191, a[8] = 1588444, a[9] = 9443681 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 533 250411\ 4.090 6 (1/n) |1 + ---- + ------| | 37 n 2| \ 296 n / Note that everything is rigorous except the constant in front, 4.090 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(2*t^3-t^2+6*t-1)*P+(6*t-1)*(t^3-5*t+1)*(2*t^3-t^2+6*t-1)*P^2, [12*( 255*n+1468)*(n+3)/(n+9)/(255*n+1213)-4*(10612+510*n^2+4721*n)/(n+9)/(255*n+1213 )*N-(-59025+5948*n+2805*n^2)/(n+9)/(255*n+1213)*N^2+2*(54361+32413*n+4080*n^2)/ (n+9)/(255*n+1213)*N^3-(1272103+535940*n+54825*n^2)/(n+9)/(255*n+1213)*N^4+( 978061+369416*n+34935*n^2)/(n+9)/(255*n+1213)*N^5-(2078799+668402*n+53295*n^2)/ (n+9)/(255*n+1213)*N^6+2*(162967*n+12495*n^2+510920)/(n+9)/(255*n+1213)*N^7-( 181619+58106*n+4335*n^2)/(n+9)/(255*n+1213)*N^8+N^9, [6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681]], 4.090*6^n*(1/n)^(1/2)*(1+533/37/n+250411/296 /n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681, 56147447, 333839726, 1985021900, 11803528804, 70190395149, 417409912200, 2482373125000, 14763542516248, 87807840206271, 522270131681000, 3106534936881197, 18478913451779103, 109924812719279688, 653934314413092628, 3890376060255274350, 23145580582447686799, 137709399424316701632, 819366249097547144735, 4875414094260698558059, 29011077965235340163136, 172637489329251700785438, 1027366051270731010220680, 6114122403744102323868426, 36388305371016650507374899, 216574977829647209364169136, 1289060897599775360173879365, 7672859745592032079419777473, 45673018330306205202754795796, 271882186417215625776682800094, 1618528454514806077450756476952, 9635592502719568051760339887815, 57366052693492200870421861560944, 341546563981446716149056370054411, 2033589251105051752536426587647311, 12108626157269399233715395997287936, 72101578336902016901637316741723080, 429351406841302402155847671638976200, 2556814236950587384105926921537635575, 15226622364564768960091967830743804494, 90683033750708720782107507423223180637, 540090469688866397874901220747414911673] Proposition Number , 53, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [3, 4, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 5 4 2 1 - 18 t - 12 t - 7 t + 2 t + 3 t 2 3 2 - 2 (6 t - 1) (t + 1) (8 t - 2 t + 6 t - 1) P 3 2 3 2 2 - (6 t - 1) (4 t - t + 5 t - 1) (8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence (130 + 17 n) a(n + 8) (734 + 101 n) a(n + 7) a(n + 9) = --------------------- - ---------------------- n + 8 n + 8 20 (87 + 13 n) a(n + 6) 6 (403 + 72 n) a(n + 5) + ----------------------- - ----------------------- n + 8 n + 8 2 (2221 + 432 n) a(n + 4) 4 (633 + 149 n) a(n + 3) + ------------------------- - ------------------------ n + 8 n + 8 4 (987 + 254 n) a(n + 2) 16 (53 + 16 n) a(n + 1) 384 (n + 3) a(n) + ------------------------ - ----------------------- + ---------------- n + 8 n + 8 n + 8 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7562, a[6] = 44958, a[7] = 267298, a[8] = 1589288, a[9] = 9449910 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 2473 44608609\ 4.035 6 (1/n) |1 + ----- + --------| | 152 n 2| \ 46208 n / Note that everything is rigorous except the constant in front, 4.035 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format 1-18*t^3-12*t^5-7*t+2*t^4+3*t^2-2*(6*t-1)*(t^2+1)*(8*t^3-2*t^2+6*t-1)*P-(6*t-1) *(4*t^3-t^2+5*t-1)*(8*t^3-2*t^2+6*t-1)*P^2, [-384*(n+3)/(n+8)+16*(53+16*n)/(n+8 )*N-4*(987+254*n)/(n+8)*N^2+4*(633+149*n)/(n+8)*N^3-2*(2221+432*n)/(n+8)*N^4+6* (403+72*n)/(n+8)*N^5-20*(87+13*n)/(n+8)*N^6+(734+101*n)/(n+8)*N^7-(130+17*n)/(n +8)*N^8+N^9, [6, 36, 214, 1272, 7562, 44958, 267298, 1589288, 9449910]], 4.035* 6^n*(1/n)^(1/2)*(1+2473/152/n+44608609/46208/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7562, 44958, 267298, 1589288, 9449910, 56191546, 334143124, 1987066058, 11817083468, 70279152406, 417985226274, 2486071089240, 14787145203950, 87957586155998, 523215286617796, 3112473682666518, 18516081254182386, 110156613723673548, 655375447127053202, 3899310585596868326, 23200831064069829986, 138050277742555443344, 821464944665619228190, 4888310331065522981042, 29090183581909732294604, 173121931258471387410442, 1030328281216062597216898, 6132210182146699265392264, 36498607253013540699307102, 217246792330903989889581694, 1293148026925461208746346600, 7697697966390080985559827954, 45823812091351817474272986598, 272796788583469042338377432692, 1624070754054147740413469030038, 9669149080201283605337849997146, 57569061298757579039981648524732, 342773771514333480925817898643578, 2041002432073373258861226402819654, 12153375761499747480002889082296228, 72371529773627208080420289962066422, 430978853430744035818690407986362738, 2566619615754676257279877011107211956, 15285665520218796796907225459568979978, 91038364300131494078122984590089400716, 542227756846253791374764958224304638054] Proposition Number , 54, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 1] as those of, [3, 4, 5] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -t - (6 t - 1) (2 t - t + 6 t - 1) P 3 3 2 2 + (6 t - 1) (t - 5 t + 1) (2 t - t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 2 (181619 + 58106 n + 4335 n ) a(n + 8) a(n + 9) = ------------------------------------- (n + 9) (255 n + 1213) 2 2 (162967 n + 12495 n + 510920) a(n + 7) - ----------------------------------------- (n + 9) (255 n + 1213) 2 (2078799 + 668402 n + 53295 n ) a(n + 6) + ---------------------------------------- (n + 9) (255 n + 1213) 2 (978061 + 369416 n + 34935 n ) a(n + 5) - --------------------------------------- (n + 9) (255 n + 1213) 2 (1272103 + 535940 n + 54825 n ) a(n + 4) + ---------------------------------------- (n + 9) (255 n + 1213) 2 2 (54361 + 32413 n + 4080 n ) a(n + 3) - -------------------------------------- (n + 9) (255 n + 1213) 2 (-59025 + 5948 n + 2805 n ) a(n + 2) + ------------------------------------ (n + 9) (255 n + 1213) 2 4 (10612 + 510 n + 4721 n) a(n + 1) 12 (255 n + 1468) (n + 3) a(n) + ------------------------------------ - ------------------------------ (n + 9) (255 n + 1213) (n + 9) (255 n + 1213) subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7561, a[6] = 44946, a[7] = 267191, a[8] = 1588444, a[9] = 9443681 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 533 250411\ 4.090 6 (1/n) |1 + ---- + ------| | 37 n 2| \ 296 n / Note that everything is rigorous except the constant in front, 4.090 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -t-(6*t-1)*(2*t^3-t^2+6*t-1)*P+(6*t-1)*(t^3-5*t+1)*(2*t^3-t^2+6*t-1)*P^2, [12*( 255*n+1468)*(n+3)/(n+9)/(255*n+1213)-4*(10612+510*n^2+4721*n)/(n+9)/(255*n+1213 )*N-(-59025+5948*n+2805*n^2)/(n+9)/(255*n+1213)*N^2+2*(54361+32413*n+4080*n^2)/ (n+9)/(255*n+1213)*N^3-(1272103+535940*n+54825*n^2)/(n+9)/(255*n+1213)*N^4+( 978061+369416*n+34935*n^2)/(n+9)/(255*n+1213)*N^5-(2078799+668402*n+53295*n^2)/ (n+9)/(255*n+1213)*N^6+2*(162967*n+12495*n^2+510920)/(n+9)/(255*n+1213)*N^7-( 181619+58106*n+4335*n^2)/(n+9)/(255*n+1213)*N^8+N^9, [6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681]], 4.090*6^n*(1/n)^(1/2)*(1+533/37/n+250411/296 /n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7561, 44946, 267191, 1588444, 9443681, 56147447, 333839726, 1985021900, 11803528804, 70190395149, 417409912200, 2482373125000, 14763542516248, 87807840206271, 522270131681000, 3106534936881197, 18478913451779103, 109924812719279688, 653934314413092628, 3890376060255274350, 23145580582447686799, 137709399424316701632, 819366249097547144735, 4875414094260698558059, 29011077965235340163136, 172637489329251700785438, 1027366051270731010220680, 6114122403744102323868426, 36388305371016650507374899, 216574977829647209364169136, 1289060897599775360173879365, 7672859745592032079419777473, 45673018330306205202754795796, 271882186417215625776682800094, 1618528454514806077450756476952, 9635592502719568051760339887815, 57366052693492200870421861560944, 341546563981446716149056370054411, 2033589251105051752536426587647311, 12108626157269399233715395997287936, 72101578336902016901637316741723080, 429351406841302402155847671638976200, 2556814236950587384105926921537635575, 15226622364564768960091967830743804494, 90683033750708720782107507423223180637, 540090469688866397874901220747414911673] Proposition Number , 55, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [1, 2, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 56, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [1, 3, 2] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 57, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [1, 3, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 58, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [1, 4, 3] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 59, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [1, 4, 5] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 60, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [2, 1, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 61, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [2, 3, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 4 3 t 7 t 5 4 -12 t + 2 t + ---- + --- - 1 + (6 t - 1) (4 t - 20 t + 6 t - 1) P 2 2 5 4 2 t (t - 5) (6 t - 1) (4 t - 20 t + 6 t - 1) P - ----------------------------------------------- = 0 2 Not only that, a(n) safisfies the following linear recurrence 2 3 (1305152 + 573960 n + 66779 n + 1769 n ) a(n + 6) a(n + 7) = 1/5 -------------------------------------------------- (n + 8) %1 2 3 (1826242 + 844414 n + 103767 n + 2784 n ) a(n + 5) - 2/5 --------------------------------------------------- (n + 8) %1 2 3 (62190 + 19440 n + 106997 n + 522 n ) a(n + 4) + 2/5 ----------------------------------------------- (n + 8) %1 2 3 (538343 + 26275 n + 236572 n + 725 n ) a(n + 3) - 4/5 ------------------------------------------------ (n + 8) %1 2 3 (4420164 + 331825 n + 2315208 n + 9280 n ) a(n + 2) + 2/5 ---------------------------------------------------- (n + 8) %1 2 3 (1524546 + 125583 n + 840611 n + 3538 n ) a(n + 1) - 2/5 --------------------------------------------------- (n + 8) %1 2 (n + 3) (29 n + 935 n + 3810) a(n) + 24/5 ----------------------------------- (n + 8) %1 2 %1 := 29 n + 877 n + 2904 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1274, a[5] = 7586, a[6] = 45168, a[7] = 268950 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 3485 117740569\ 4.62 6 (1/n) |1 + ----- + ---------| | 232 n 2| \ 107648 n / Note that everything is rigorous except the constant in front, 4.62 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -12*t^4+2*t^3+1/2*t^2+7/2*t-1+(6*t-1)*(4*t^5-20*t^4+6*t-1)*P-1/2*t*(t-5)*(6*t-1 )*(4*t^5-20*t^4+6*t-1)*P^2, [-24/5*(n+3)*(29*n^2+935*n+3810)/(n+8)/(29*n^2+877* n+2904)+2/5*(1524546+125583*n^2+840611*n+3538*n^3)/(n+8)/(29*n^2+877*n+2904)*N-\ 2/5*(4420164+331825*n^2+2315208*n+9280*n^3)/(n+8)/(29*n^2+877*n+2904)*N^2+4/5*( 538343+26275*n^2+236572*n+725*n^3)/(n+8)/(29*n^2+877*n+2904)*N^3-2/5*(62190+ 19440*n^2+106997*n+522*n^3)/(n+8)/(29*n^2+877*n+2904)*N^4+2/5*(1826242+844414*n +103767*n^2+2784*n^3)/(n+8)/(29*n^2+877*n+2904)*N^5-1/5*(1305152+573960*n+66779 *n^2+1769*n^3)/(n+8)/(29*n^2+877*n+2904)*N^6+N^7, [6, 36, 214, 1274, 7586, 45168, 268950]], 4.62*6^n*(1/n)^(1/2)*(1+3485/232/n+117740569/107648/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1274, 7586, 45168, 268950, 1601496, 9536592, 56790294, 338195992, 2014077676, 11994921016, 71438442252, 425480789794, 2534202556852, 15094407466486, 89909209514486, 535556929227892, 3190218631607314, 19004155672744632, 113211364717543536, 674442255257250024, 4018026560558333052, 23938344029102174436, 142622701540179135564, 849760324338961531068, 5063111142008568461232, 30168358045311346273920, 179762487962709866910948, 1071173034556886760153990, 6383124828601056780158868, 38038218516273014927750634, 226683601953053546686553094, 1350930864723058685887145206, 8051173859596034254405316914, 47984207378904574044763841110, 285989746187752267662399033026, 1704572951020208352483081061788, 10159999633486437564180735449214, 60559842508737018503151831322988, 360984622627053920030446816888892, 2151818127355681261910556589382908, 12827301133659004302363264854036532, 76467670716020811120371139580671692, 455861849096345077135360536865520832, 2717699423636646758508488231175421784, 16202512190430932345683299846409018744, 96599754610816591713581163043455504396, 575946830070141011661612629620785080896] Proposition Number , 62, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [2, 3, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 4 3 2 -1 + (6 t - 1) (4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 9) a(n + 4) 36 (n + 4) a(n + 3) 2 (2 n + 7) a(n + 2) a(n + 5) = -------------------- - ------------------- - -------------------- n + 5 n + 5 n + 5 28 (n + 3) a(n + 1) 12 (2 n + 5) a(n) + ------------------- - ----------------- n + 5 n + 5 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1274, a[5] = 7584 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 123 695069\ 4.54 6 (1/n) |1 + --- + ------| | 8 n 2| \ 640 n / Note that everything is rigorous except the constant in front, 4.54 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^4-4*t^3+6*t-1)*P^2, [12*(2*n+5)/(n+5)-28*(n+3)/(n+5)*N+2*(2*n+7 )/(n+5)*N^2+36*(n+4)/(n+5)*N^3-6*(2*n+9)/(n+5)*N^4+N^5, [6, 36, 214, 1274, 7584 ]], 4.54*6^n*(1/n)^(1/2)*(1+123/8/n+695069/640/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1274, 7584, 45150, 268800, 1600350, 9528280, 56731980, 337796580, 2011390650, 11977094300, 71321483100, 424720386548, 2529296162338, 15062951835228, 89708646326892, 534284187403572, 3182175501625152, 18953512711237200, 112893530790348300, 672453330484995000, 4005612876110404170, 23861048142331583220, 142142438007251826120, 846782135072415899500, 5044676016546888618600, 30054431721780670339800, 179059509482029891889356, 1066841422807391131253336, 6356469083483105077080066, 37874383914544058480771364, 225677760224060424323294424, 1344762143470011374247297084, 8013379119978900118763991000, 47752859851268543163447483000, 284574865250604978279207619500, 1695926883690409044912722017400, 10107206027371092040674853628400, 60237715822665630876836367761400, 359020484906466215621579298638000, 2139849815612482240631377758550000, 12754418466742855168395745679311500, 76024103303096857220873192473997480, 453163788745094354129861703853808880, 2701296846354865449169372076269754280, 16102845075785700213455373389074341570, 95994439850663113030947308368011770620, 572272232474458929228546060400693657920] Proposition Number , 63, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [2, 4, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 3 2 -1 + (6 t - 1) (4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 11) a(n + 5) 36 (n + 5) a(n + 4) 2 (2 n + 9) a(n + 3) a(n + 6) = --------------------- - ------------------- - -------------------- n + 6 n + 6 n + 6 24 (n + 4) a(n + 2) 2 (2 n + 7) a(n + 1) 24 (n + 3) a(n) + ------------------- + -------------------- - --------------- n + 6 n + 6 n + 6 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7562, a[6] = 44958 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 521 2502529\ 4.205 6 (1/n) |1 + ---- + -------| | 40 n 2| \ 3200 n / Note that everything is rigorous except the constant in front, 4.205 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^5-4*t^3+6*t-1)*P^2, [24*(n+3)/(n+6)-2*(2*n+7)/(n+6)*N-24*(n+4)/ (n+6)*N^2+2*(2*n+9)/(n+6)*N^3+36*(n+5)/(n+6)*N^4-6*(2*n+11)/(n+6)*N^5+N^6, [6, 36, 214, 1272, 7562, 44958]], 4.205*6^n*(1/n)^(1/2)*(1+521/40/n+2502529/3200/n^ 2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7562, 44958, 267300, 1589316, 9450196, 56194086, 334163976, 1987228726, 11818307256, 70288111376, 418049425304, 2486523268404, 14790285061524, 87979129336868, 523361595527232, 3113458587734110, 18522660336804468, 110200263338247084, 655663298780619876, 3901198605504457830, 23213154398977342440, 138130359633331334748, 821983264478393999764, 4891652813438850746832, 29111665836085553119332, 173259571618490266150768, 1031207646797687254066248, 6137813457870869748529148, 36534223227564741192094536, 217472658430937710414174884, 1294577337077895162316177296, 7706724699788043479343760612, 45880712769880308338900727384, 273154832326134868125779903736, 1626319971156341349181021068352, 9683256394543314610491985962798, 57657412536593910625062421487180, 343326320640053256691648058786504, 2044453480460431778198363148838404, 12174902716459651798539044567315652, 72505649567614945427959319917982692, 431813508858897006790224811251688428, 2571808203855980343902893285555965280, 15317886647816722444541262208671130286, 91238259470573690493118524002219522676, 543466704651662865564987092697981902116] Proposition Number , 64, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [2, 4, 5] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 65, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [3, 2, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 4 3 3 2 -1 + 6 t - t + 12 t - 2 t + 2 t (6 t - 1) (8 t - 2 t + 6 t - 1) P 2 3 2 2 + (6 t - 1) (4 t - t + 1) (8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence (98 + 13 n) a(n + 7) 2 (189 + 26 n) a(n + 6) a(n + 8) = -------------------- - ----------------------- n + 8 n + 8 2 (319 + 40 n) a(n + 5) 2 (515 + 56 n) a(n + 4) + ----------------------- - ----------------------- n + 8 n + 8 4 (17 + 21 n) a(n + 3) 4 (123 + 62 n) a(n + 2) - ---------------------- + ----------------------- n + 8 n + 8 16 (53 + 16 n) a(n + 1) 384 (n + 3) a(n) - ----------------------- + ---------------- n + 8 n + 8 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7564, a[6] = 44982, a[7] = 267510, a[8] = 1590960 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 1811 30474361\ 4.266 6 (1/n) |1 + ----- + --------| | 136 n 2| \ 36992 n / Note that everything is rigorous except the constant in front, 4.266 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+6*t^2-t+12*t^4-2*t^3+2*t*(6*t-1)*(8*t^3-2*t^2+6*t-1)*P+(6*t-1)*(4*t^2-t+1)*( 8*t^3-2*t^2+6*t-1)*P^2, [-384*(n+3)/(n+8)+16*(53+16*n)/(n+8)*N-4*(123+62*n)/(n+ 8)*N^2+4*(17+21*n)/(n+8)*N^3+2*(515+56*n)/(n+8)*N^4-2*(319+40*n)/(n+8)*N^5+2*( 189+26*n)/(n+8)*N^6-(98+13*n)/(n+8)*N^7+N^8, [6, 36, 214, 1272, 7564, 44982, 267510, 1590960]], 4.266*6^n*(1/n)^(1/2)*(1+1811/136/n+30474361/36992/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7564, 44982, 267510, 1590960, 9462304, 56279824, 334754740, 1991218422, 11844840490, 70462437914, 419183549742, 2493841783056, 14837189545148, 88277993824228, 525256344529584, 3125418503868136, 18597863189912640, 110671524361870956, 658607517979402308, 3919542687612002598, 23327167582804676058, 138837405019825914486, 826359082608264642994, 4918684091984746497586, 29278365853496908110682, 174285991447733509340302, 1037518480568323458299822, 6176563047215064400299456, 36771856692794288245706172, 218928270022255688105292880, 1303483988146515002030433268, 7761168048728232251269689180, 46213191845093350822617773756, 275183440000330223507481674312, 1638687104909021652440222956736, 9758591664940169274223117265304, 58115982150651571900089785812320, 346115700201681931796498777955984, 2061409416962276939039306251901872, 12277908816166423734366705287412252, 73131031238216130129010557201408484, 435608243706209161655584765074557708, 2594821770283355790905159092446544980, 15457383340939303467946023312412958086, 92083405830508669102507402359742482818, 548584675492787626673637920117515577510] Proposition Number , 66, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [3, 2, 4] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 3 2 -1 + (6 t - 1) (4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 11) a(n + 5) 36 (n + 5) a(n + 4) 2 (2 n + 9) a(n + 3) a(n + 6) = --------------------- - ------------------- - -------------------- n + 6 n + 6 n + 6 24 (n + 4) a(n + 2) 2 (2 n + 7) a(n + 1) 24 (n + 3) a(n) + ------------------- + -------------------- - --------------- n + 6 n + 6 n + 6 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7562, a[6] = 44958 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 521 2502529\ 4.205 6 (1/n) |1 + ---- + -------| | 40 n 2| \ 3200 n / Note that everything is rigorous except the constant in front, 4.205 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^5-4*t^3+6*t-1)*P^2, [24*(n+3)/(n+6)-2*(2*n+7)/(n+6)*N-24*(n+4)/ (n+6)*N^2+2*(2*n+9)/(n+6)*N^3+36*(n+5)/(n+6)*N^4-6*(2*n+11)/(n+6)*N^5+N^6, [6, 36, 214, 1272, 7562, 44958]], 4.205*6^n*(1/n)^(1/2)*(1+521/40/n+2502529/3200/n^ 2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7562, 44958, 267300, 1589316, 9450196, 56194086, 334163976, 1987228726, 11818307256, 70288111376, 418049425304, 2486523268404, 14790285061524, 87979129336868, 523361595527232, 3113458587734110, 18522660336804468, 110200263338247084, 655663298780619876, 3901198605504457830, 23213154398977342440, 138130359633331334748, 821983264478393999764, 4891652813438850746832, 29111665836085553119332, 173259571618490266150768, 1031207646797687254066248, 6137813457870869748529148, 36534223227564741192094536, 217472658430937710414174884, 1294577337077895162316177296, 7706724699788043479343760612, 45880712769880308338900727384, 273154832326134868125779903736, 1626319971156341349181021068352, 9683256394543314610491985962798, 57657412536593910625062421487180, 343326320640053256691648058786504, 2044453480460431778198363148838404, 12174902716459651798539044567315652, 72505649567614945427959319917982692, 431813508858897006790224811251688428, 2571808203855980343902893285555965280, 15317886647816722444541262208671130286, 91238259470573690493118524002219522676, 543466704651662865564987092697981902116] Proposition Number , 67, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [3, 4, 1] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 2 4 3 3 2 -1 + 6 t - t + 12 t - 2 t + 2 t (6 t - 1) (8 t - 2 t + 6 t - 1) P 2 3 2 2 + (6 t - 1) (4 t - t + 1) (8 t - 2 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence (98 + 13 n) a(n + 7) 2 (189 + 26 n) a(n + 6) a(n + 8) = -------------------- - ----------------------- n + 8 n + 8 2 (319 + 40 n) a(n + 5) 2 (515 + 56 n) a(n + 4) + ----------------------- - ----------------------- n + 8 n + 8 4 (17 + 21 n) a(n + 3) 4 (123 + 62 n) a(n + 2) - ---------------------- + ----------------------- n + 8 n + 8 16 (53 + 16 n) a(n + 1) 384 (n + 3) a(n) - ----------------------- + ---------------- n + 8 n + 8 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7564, a[6] = 44982, a[7] = 267510, a[8] = 1590960 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 1811 30474361\ 4.266 6 (1/n) |1 + ----- + --------| | 136 n 2| \ 36992 n / Note that everything is rigorous except the constant in front, 4.266 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+6*t^2-t+12*t^4-2*t^3+2*t*(6*t-1)*(8*t^3-2*t^2+6*t-1)*P+(6*t-1)*(4*t^2-t+1)*( 8*t^3-2*t^2+6*t-1)*P^2, [-384*(n+3)/(n+8)+16*(53+16*n)/(n+8)*N-4*(123+62*n)/(n+ 8)*N^2+4*(17+21*n)/(n+8)*N^3+2*(515+56*n)/(n+8)*N^4-2*(319+40*n)/(n+8)*N^5+2*( 189+26*n)/(n+8)*N^6-(98+13*n)/(n+8)*N^7+N^8, [6, 36, 214, 1272, 7564, 44982, 267510, 1590960]], 4.266*6^n*(1/n)^(1/2)*(1+1811/136/n+30474361/36992/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7564, 44982, 267510, 1590960, 9462304, 56279824, 334754740, 1991218422, 11844840490, 70462437914, 419183549742, 2493841783056, 14837189545148, 88277993824228, 525256344529584, 3125418503868136, 18597863189912640, 110671524361870956, 658607517979402308, 3919542687612002598, 23327167582804676058, 138837405019825914486, 826359082608264642994, 4918684091984746497586, 29278365853496908110682, 174285991447733509340302, 1037518480568323458299822, 6176563047215064400299456, 36771856692794288245706172, 218928270022255688105292880, 1303483988146515002030433268, 7761168048728232251269689180, 46213191845093350822617773756, 275183440000330223507481674312, 1638687104909021652440222956736, 9758591664940169274223117265304, 58115982150651571900089785812320, 346115700201681931796498777955984, 2061409416962276939039306251901872, 12277908816166423734366705287412252, 73131031238216130129010557201408484, 435608243706209161655584765074557708, 2594821770283355790905159092446544980, 15457383340939303467946023312412958086, 92083405830508669102507402359742482818, 548584675492787626673637920117515577510] Proposition Number , 68, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [3, 4, 5] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 5 3 2 -1 + (6 t - 1) (4 t - 4 t + 6 t - 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 11) a(n + 5) 36 (n + 5) a(n + 4) 2 (2 n + 9) a(n + 3) a(n + 6) = --------------------- - ------------------- - -------------------- n + 6 n + 6 n + 6 24 (n + 4) a(n + 2) 2 (2 n + 7) a(n + 1) 24 (n + 3) a(n) + ------------------- + -------------------- - --------------- n + 6 n + 6 n + 6 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272, a[5] = 7562, a[6] = 44958 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 521 2502529\ 4.205 6 (1/n) |1 + ---- + -------| | 40 n 2| \ 3200 n / Note that everything is rigorous except the constant in front, 4.205 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1+(6*t-1)*(4*t^5-4*t^3+6*t-1)*P^2, [24*(n+3)/(n+6)-2*(2*n+7)/(n+6)*N-24*(n+4)/ (n+6)*N^2+2*(2*n+9)/(n+6)*N^3+36*(n+5)/(n+6)*N^4-6*(2*n+11)/(n+6)*N^5+N^6, [6, 36, 214, 1272, 7562, 44958]], 4.205*6^n*(1/n)^(1/2)*(1+521/40/n+2502529/3200/n^ 2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7562, 44958, 267300, 1589316, 9450196, 56194086, 334163976, 1987228726, 11818307256, 70288111376, 418049425304, 2486523268404, 14790285061524, 87979129336868, 523361595527232, 3113458587734110, 18522660336804468, 110200263338247084, 655663298780619876, 3901198605504457830, 23213154398977342440, 138130359633331334748, 821983264478393999764, 4891652813438850746832, 29111665836085553119332, 173259571618490266150768, 1031207646797687254066248, 6137813457870869748529148, 36534223227564741192094536, 217472658430937710414174884, 1294577337077895162316177296, 7706724699788043479343760612, 45880712769880308338900727384, 273154832326134868125779903736, 1626319971156341349181021068352, 9683256394543314610491985962798, 57657412536593910625062421487180, 343326320640053256691648058786504, 2044453480460431778198363148838404, 12174902716459651798539044567315652, 72505649567614945427959319917982692, 431813508858897006790224811251688428, 2571808203855980343902893285555965280, 15317886647816722444541262208671130286, 91238259470573690493118524002219522676, 543466704651662865564987092697981902116] Proposition Number , 69, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [4, 2, 5] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] Proposition Number , 70, : Let a(n) be the number of n-lettered words in the alphabet, {1, 2, 3, 4, 5, 6} With as many occurrences of the substring (consecutive subword), [1, 2, 3] as those of, [4, 5, 6] Let P(t), be the ordinary generating function of a(n) infinity ----- \ n P(t) = ) a(n) t / ----- n = 0 Then P=P(t) satisfies the quadratic equation 3 2 -1 - (6 t - 1) (4 t - 6 t + 1) P = 0 Not only that, a(n) safisfies the following linear recurrence 6 (2 n + 7) a(n + 3) 36 (n + 3) a(n + 2) 2 (2 n + 5) a(n + 1) a(n + 4) = -------------------- - ------------------- - -------------------- n + 4 n + 4 n + 4 24 (n + 2) a(n) + --------------- n + 4 subject to the initial conditions a[1] = 6, a[2] = 36, a[3] = 214, a[4] = 1272 Finally, the asymptotics of a(n) to order, 2, is : n 1/2 / 101 94105 \ 4.146 6 (1/n) |1 + --- + ------| | 8 n 2| \ 128 n / Note that everything is rigorous except the constant in front, 4.146 that is a mere non-rigorous estimate Sketch of proof: The above has been proved completely rigorously internally, and I will spare you the details. If you insist you can get them yourself by tracing the code, or else donate 50 dollars to the Society for the Liberation of Computerkind. QED. For the sake of people who want to use these formulas here are the quardaric equation, recurrence operator, and asymptotics in Maple input format -1-(6*t-1)*(4*t^3-6*t+1)*P^2, [-24*(n+2)/(n+4)+2*(2*n+5)/(n+4)*N+36*(n+3)/(n+4) *N^2-6*(2*n+7)/(n+4)*N^3+N^4, [6, 36, 214, 1272]], 4.146*6^n*(1/n)^(1/2)*(1+101 /8/n+94105/128/n^2) For the sake of Sloane, here are the first 50 terms of the sequece [1, 6, 36, 214, 1272, 7560, 44934, 267084, 1587600, 9437452, 56103360, 333536544, 1982980294, 11789999220, 70101859752, 416836430388, 2478689575632, 14740049205072, 87658901250972, 521330799270168, 3100637441452608, 18442033488702744, 109694994609429504, 652506692363505792, 3881532719375197254, 23090940411468661188, 137372576135197716216, 817294320885186675556, 4862693522213352420432, 28933119143326148361840, 172160498604288605914804, 1024452026648138834664744, 6096345263472444342028704, 36279998056790655936162216, 215915927802600453201272448, 1285055206530424090705339200, 7648539637606680435299340316, 45525512168325170155102352136, 270988394208479818534217290320, 1613117578354021997894354117128, 9602864148666899453324750301216, 57168253086022780616700687498912, 340352056289375957527227956228472, 2026380983486905729839789629753904, 12065158506396435697007580498151680, 71839634310285823980500414613287472, 427773902379016214248656872598484992, 2547319909014030424051349818294067712, 15169513923376050186451359577976457798, 90339721768951142132586090821637256164, 538027752862270613488088610099729283608] It took, 1314.681, seconds to generate this book.