cp's OEIS Frontend

A274526 a(n) = ((1 + sqrt(11))^n - (1 - sqrt(11))^n)/sqrt(11).

%I A274526 #9 Feb 16 2025 08:33:36
%S A274526 0,2,4,28,96,472,1904,8528,36096,157472,675904,2926528,12612096,
%T A274526 54489472,235099904,1015094528,4381188096,18913321472,81638523904,
%U A274526 352410262528,1521205764096,6566514153472,28345085947904,122355313430528,528161486340096
%N A274526 a(n) = ((1 + sqrt(11))^n - (1 - sqrt(11))^n)/sqrt(11).
%C A274526 Number of zeros in substitution system {0 -> 11111, 1 -> 1001} at step n from initial string "1" (see example).
%H A274526 Ilya Gutkovskiy, <a href="/A274526/a274526.pdf">Illustration (substitution system {0 -> 11111, 1 -> 1001})</a>
%H A274526 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SubstitutionSystem.html">Substitution System</a>
%H A274526 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,10)
%F A274526 O.g.f.: 2*x/(1 - 2*x - 10*x^2).
%F A274526 E.g.f.: 2*exp(x)*sinh(sqrt(11)*x)/sqrt(11).
%F A274526 Dirichlet g.f.: (PolyLog(s,1+sqrt(11)) - PolyLog(s,1-sqrt(11)))/sqrt(11), where PolyLog(s,x) is the polylogarithm function.
%F A274526 a(n) = 2*a(n-1) + 10*a(n-2).
%F A274526 a(n) = 2*A083102(n-1), n>0.
%F A274526 Lim_{n->infinity} a(n+1)/a(n) = 1 + sqrt(11) = 1 + A010468.
%e A274526 Evolution from initial string "1": 1 -> 1001 -> 100111111111111001 -> 1001111111111110011001100110011001100110011001100110011001100111111111111001 -> ...
%e A274526 Therefore, number of zeros at step n:
%e A274526 a(0) = 0;
%e A274526 a(1) = 2;
%e A274526 a(2) = 4;
%e A274526 a(3) = 28, etc.
%t A274526 LinearRecurrence[{2, 10}, {0, 2}, 25]
%o A274526 (PARI) concat(0, Vec(2*x/(1-2*x-10*x^2) + O(x^99))) \\ _Altug Alkan_, Jun 27 2016
%Y A274526 Cf. A010468, A083102.
%K A274526 nonn,easy
%O A274526 0,2
%A A274526 _Ilya Gutkovskiy_, Jun 27 2016