Theorem: Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 7, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 7, 12, 21, 37, 65, 114, 200, 351, 616, 1081, 1897, 3329, 5842, 10252, 17991, 31572, 55405, 97229, 170625] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficients a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) with initial conditions, a(1) = 4, a(2) = 7, a(3) = 12, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) with initial conditions, b(1) = 4, b(2) = 7, b(3) = 12, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 2) b(n) - b(n + 1) b(n + 2) b(n) - b(n + 1) -1/2 <= - -------------------------, - ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (100 + 12 69 ) 2 (100 + 12 69 ) [------------------- + --------------------- + 2/3, - ------------------- 6 1/2 1/3 12 3 (100 + 12 69 ) 1 - ----------------------- + 2/3 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (100 + 12 69 ) / 1/2 1/3 (100 + 12 69 ) 1 - ------------------- - ----------------------- + 2/3 12 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (100 + 12 69 ) / In floating-point [1.754877667, 0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] The largest root is, 1.754877667 and the remaining roots are [0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] whose absolute values are [0.7548776666, 0.7548776666] so the largest absolute value is, 0.7548776666 that is less than 1, It follows that the sequence 2 b(n + 2) b(n) - b(n + 1) c(n) = - ------------------------- b(n) satisfies the inequality n | c(n) | <= A 0.7548776666 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, -0.4285714286, -0.2500000000, 0.1904761905, 0.1891891892, -0.06153846154, -0.1228070175, 0.005000000000, 0.07122507122, 0.01461038961, -0.03700277521, -0.01739588824, 0.01682186843, 0.01403628894, -0.006145142411, -0.009504752376, 0.001171924490, 0.005703456367, 0.0007302348065, -0.003071062271, -0.001168903168, 0.001463490759, 0.001024822391, -0.0005827491423, -0.0007268299193, 0.0001539116948, 0.0004519041666, 0.00002306671899, -0.0002518590338, -0.00007488062000, 0.0001251645128, 0.00007335061177, -0.00005334390925, -0.00005487391748, -6 0.00001694668605, 0.00003542338034, -0.9738428538 10 , -0.00002042437999, -5 -5 -5 -0.4451536794 10 , 0.00001054746355, 0.5122083903 10 , -0.4754832539 10 , -5 -5 -5 -0.4084285431 10 , 0.1708345581 10 , 0.2746144054 10 , -6 -5 -6 -0.3003429039 10 , -0.1638484280 10 , -0.2304816029 10 , -6 -6 -6 0.8771781708 10 , 0.3463536640 10 , -0.4149524457 10 , -6 -6 -6 -0.2990803846 10 , 0.1631453405 10 , 0.2104186200 10 , -7 -6 -8 -0.4138848514 10 , -0.1300502497 10 , -0.8293394300 10 , -7 0.7207497597 10 ] The largest is 0.2500000000 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED.