A277091 a(n) = ((1 + sqrt(15))^n - (1 - sqrt(15))^n)/sqrt(15).
0, 2, 4, 36, 128, 760, 3312, 17264, 80896, 403488, 1939520, 9527872, 46209024, 225808256, 1098542848, 5358401280, 26096402432, 127210422784, 619770479616, 3020486878208, 14717760471040, 71722337236992, 349493321068544, 1703099363454976, 8299105221869568, 40441601532108800
Offset: 0
Examples
Evolution from initial string "1": 1 -> 1001 -> 1001111111111111111001 -> ... Therefore, number of zeros at step n: a(0) = 0; a(1) = 2; a(2) = 4, etc.
Links
- Ilya Gutkovskiy, Illustration (substitution system {0 -> 1111111, 1 -> 1001}) and similar sequences
- Eric Weisstein's World of Mathematics, Substitution System
- Index entries for linear recurrences with constant coefficients, signature (2,14).
Programs
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Mathematica
LinearRecurrence[{2, 14}, {0, 2}, 26] -
PARI
concat(0, Vec(2*x/(1-2*x-14*x^2) + O(x^99))) \\ Altug Alkan, Oct 01 2016
Formula
O.g.f.: 2*x/(1 - 2*x - 14*x^2).
E.g.f.: 2*sinh(sqrt(15)*x)*exp(x)/sqrt(15).
a(n) = 2*a(n-1) + 14*a(n-2).
Lim_{n->infinity} a(n+1)/a(n) = 1 + sqrt(15) = 1 + A010472.
Comments