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

Author

Ilya Gutkovskiy, Sep 29 2016

Keywords

• nonn
• easy

Comments

Number of zeros in substitution system {0 -> 1111111, 1 -> 1001} at step n from initial string "1" (see example).

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).

Crossrefs

Cf. A010472, A103435, A274520, A274526.

Programs

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.