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A277091 a(n) = ((1 + sqrt(15))^n - (1 - sqrt(15))^n)/sqrt(15).

%I A277091 #10 Feb 16 2025 08:33:36
%S A277091 0,2,4,36,128,760,3312,17264,80896,403488,1939520,9527872,46209024,
%T A277091 225808256,1098542848,5358401280,26096402432,127210422784,
%U A277091 619770479616,3020486878208,14717760471040,71722337236992,349493321068544,1703099363454976,8299105221869568,40441601532108800
%N A277091 a(n) = ((1 + sqrt(15))^n - (1 - sqrt(15))^n)/sqrt(15).
%C A277091 Number of zeros in substitution system {0 -> 1111111, 1 -> 1001} at step n from initial string "1" (see example).
%H A277091 Ilya Gutkovskiy, <a href="/A277091/a277091.pdf">Illustration (substitution system {0 -> 1111111, 1 -> 1001}) and similar sequences</a>
%H A277091 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SubstitutionSystem.html">Substitution System</a>
%H A277091 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,14).
%F A277091 O.g.f.: 2*x/(1 - 2*x - 14*x^2).
%F A277091 E.g.f.: 2*sinh(sqrt(15)*x)*exp(x)/sqrt(15).
%F A277091 a(n) = 2*a(n-1) + 14*a(n-2).
%F A277091 Lim_{n->infinity} a(n+1)/a(n) = 1 + sqrt(15) = 1 + A010472.
%e A277091 Evolution from initial string "1": 1 -> 1001 -> 1001111111111111111001 -> ...
%e A277091 Therefore, number of zeros at step n:
%e A277091 a(0) = 0;
%e A277091 a(1) = 2;
%e A277091 a(2) = 4, etc.
%t A277091 LinearRecurrence[{2, 14}, {0, 2}, 26]
%o A277091 (PARI) concat(0, Vec(2*x/(1-2*x-14*x^2) + O(x^99))) \\ _Altug Alkan_, Oct 01 2016
%Y A277091 Cf. A010472, A103435, A274520, A274526.
%K A277091 nonn,easy
%O A277091 0,2
%A A277091 _Ilya Gutkovskiy_, Sep 29 2016