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

0, 2, 4, 28, 96, 472, 1904, 8528, 36096, 157472, 675904, 2926528, 12612096, 54489472, 235099904, 1015094528, 4381188096, 18913321472, 81638523904, 352410262528, 1521205764096, 6566514153472, 28345085947904, 122355313430528, 528161486340096

Offset

0,2

Comments

Number of zeros in substitution system {0 -> 11111, 1 -> 1001} at step n from initial string "1" (see example).

Links

Table of n, a(n) for n=0..24.

Ilya Gutkovskiy, Illustration (substitution system {0 -> 11111, 1 -> 1001})

Eric Weisstein's World of Mathematics, Substitution System

Index entries for linear recurrences with constant coefficients, signature (2,10)

Formula

O.g.f.: 2*x/(1 - 2*x - 10*x^2).

E.g.f.: 2*exp(x)*sinh(sqrt(11)*x)/sqrt(11).

Dirichlet g.f.: (PolyLog(s,1+sqrt(11)) - PolyLog(s,1-sqrt(11)))/sqrt(11), where PolyLog(s,x) is the polylogarithm function.

a(n) = 2*a(n-1) + 10*a(n-2).

a(n) = 2*A083102(n-1), n>0.

Lim_{n->infinity} a(n+1)/a(n) = 1 + sqrt(11) = 1 + A010468.

Example

Evolution from initial string "1": 1 -> 1001 -> 100111111111111001 -> 1001111111111110011001100110011001100110011001100110011001100111111111111001 -> ...
Therefore, number of zeros at step n:
a(0) = 0;
a(1) = 2;
a(2) = 4;
a(3) = 28, etc.

Mathematica

LinearRecurrence[{2, 10}, {0, 2}, 25]

Prog

(PARI) concat(0, Vec(2*x/(1-2*x-10*x^2) + O(x^99))) \\ Altug Alkan, Jun 27 2016

Crossrefs

Cf. A010468, A083102.

Sequence in context: A156449 A259375 A192374 * A264667 A356700 A357158

Adjacent sequences: A274523 A274524 A274525 * A274527 A274528 A274529

Keyword

nonn,easy

Author

Ilya Gutkovskiy, Jun 27 2016

Status

approved