A274520 a(n) = ((1 + sqrt(7))^n - (1 - sqrt(7))^n)/sqrt(7).

0, 2, 4, 20, 64, 248, 880, 3248, 11776, 43040, 156736, 571712, 2083840, 7597952, 27698944, 100985600, 368164864, 1342243328, 4893475840, 17840411648, 65041678336, 237125826560, 864501723136, 3151758405632, 11490527150080, 41891604733952, 152726372368384

Offset

0,2

Comments

Number of zeros in substitution system {0 -> 111, 1 -> 1001} at step n from initial string "1" (see example).

Links

Table of n, a(n) for n=0..26.

Ilya Gutkovskiy, Illustration (substitution system {0 -> 111, 1 -> 1001})

Eric Weisstein's World of Mathematics, Substitution System

Index entries for linear recurrences with constant coefficients, signature (2,6)

Formula

O.g.f.: 2*x/(1 - 2*x - 6*x^2).

E.g.f.: 2*exp(x)*sinh(sqrt(7)*x)/sqrt(7).

Dirichlet g.f.: (PolyLog(s,1+sqrt(7)) - PolyLog(s,1-sqrt(7)))/sqrt(7), where PolyLog(s,x) is the polylogarithm function.

a(n) = 2*a(n-1) + 6*a(n-2).

a(n) = 2*A083099(n).

Lim_{n->infinity} a(n+1)/a(n) = 1 + sqrt(7) = 1 + A010465.

Example

Evolution from initial string "1": 1 -> 1001 -> 10011111111001 -> 1001111111100110011001100110011001100110011111111001 -> ...
Therefore, number of zeros at step n:
a(0) = 0;
a(1) = 2;
a(2) = 4;
a(3) = 20, etc.

Mathematica

LinearRecurrence[{2, 6}, {0, 2}, 27]

Prog

(PARI) a(n)=([0, 1; 6, 2]^n*[0; 2])[1, 1] \\ Charles R Greathouse IV, Jul 26 2016

Crossrefs

Cf. A010465, A083099.

Sequence in context: A137697 A192380 A009336 * A337616 A238229 A192377

Adjacent sequences: A274517 A274518 A274519 * A274521 A274522 A274523

Keyword

nonn,easy

Author

Ilya Gutkovskiy, Jun 26 2016

Status

approved