Rational generating functions for the Certain Stanley-Stern Sums By Shalosh B. Ekhad Theorem Number, 1 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- 2 \ j -t + 1 ) Z[j] t = ---------------- / 3 2 ----- -t - t - t + 1 j = 0 Let n - 1 --------' ' | | F[n](x) = | | | | | | i = 0 Z[i + 2] (Z[i + 1] + Z[i + 2]) (Z[i] + Z[i + 1] + Z[i + 2]) (1 + x + x + x ) Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ H(n) = ) a(n, k) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 1 and v is a certain vector of length, 1 that are too big to display. At any rate we can use them to find the first, 31, terms starting at n=0, for the sake of the OEIS. [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736, 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976] ----------------------------- This took, 0.005, seconds. Theorem Number, 2 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- 2 \ j -t + 1 ) Z[j] t = ---------------- / 3 2 ----- -t - t - t + 1 j = 0 Let n - 1 --------' ' | | F[n](x) = | | | | | | i = 0 Z[i + 2] (Z[i + 1] + Z[i + 2]) (Z[i] + Z[i + 1] + Z[i + 2]) (1 + x + x + x ) Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ H(n) = ) a(n, k) a(n, k + 1) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 118 and v is a certain vector of length, 118 that are too big to display. At any rate we can use them to find the first, 31, terms starting at n=0, for the sake of the OEIS. [0, 3, 31, 278, 2414, 21245, 185413, 1615519, 14087271, 122769986, 1069826009, 9322756293, 81237467781, 707887924086, 6168380649380, 53749697447416, 468360696808845, 4081167732206704, 35562166836696015, 309878846990028616, 2700197777043736460, 23528767732758920650, 205023093462987604684, 1786513782536621477454, 15567180235679878542486, 135648043762700109300031, 1181999015903573748133834, 10299607951495440493452486, 89747895271658216825993108, 782037990238211875803404797, 6814459726034900678164714108] ----------------------------- This took, 3.438, seconds. Theorem Number, 3 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- 2 \ j -t + 1 ) Z[j] t = ---------------- / 3 2 ----- -t - t - t + 1 j = 0 Let n - 1 --------' ' | | F[n](x) = | | | | | | i = 0 Z[i + 2] (Z[i + 1] + Z[i + 2]) (Z[i] + Z[i + 1] + Z[i + 2]) (1 + x + x + x ) Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ H(n) = ) a(n, k) a(n, k + 1) a(n, k + 2) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 5004 and v is a certain vector of length, 5004 that are too big to display. At any rate we can use them to find the first, 31, terms starting at n=0, for the sake of the OEIS. [0, 2, 61, 1296, 24921, 489582, 9350549, 177963168, 3391064002, 64487799588, 1226118640377, 23312855235002, 443191812789105, 8425194093965061, 160163679385413330, 3044682965559992551, 57878731473576716586, 1100259164856451370190, 20915599142610140712072, 397599157060411258571016, 7558235994899249030612185, 143679680114476726692933873, 2731305194823117973566190957, 51921243751545534807869017784, 987006314293203617433741663451, 18762675639030758203445900540635, 356672484945172704455067474098158, 6780230251423357902360833593912420, 128890015722881276392989844434764090, 2450158110066022675674359852680097245, 46576724557545302943309527086461340930] ----------------------------- This took, 2033.657, seconds. ----------------------------------------- This concludes this article that took, 2132.332, seconds to produce.