A225799 a(n) = Sum_{k=0..n} binomial(n,k) * 10^(n-k) * Fibonacci(n+k).

0, 11, 143, 3058, 55341, 1052755, 19717984, 371084087, 6973353387, 131101759514, 2464418392865, 46327530894271, 870879506447808, 16371134451297043, 307750614069672631, 5785211638097121890, 108752568228856901349, 2044371455527726003547, 38430858858805840293152

Offset: 0

Author

John Molokach, Jul 27 2013

Keywords

• nonn
• easy

Comments

This sequence is part of a family of Fibonacci-like sequences, where:

Sum_{k=0..n} binomial(n,k)*m^(n-k)*Fibonacci(n+k) produces a sequence whose terms are divisible by (m+1); m>=1.

A recurrence relation for a(n) (m not equal to zero) is:

a(n) = (m+3)*a(n-1) + (m^2+m-1)*a(n-2); a(0)=0, a(1)=m+1.

Notable values of m include:

m = 1: Fibonacci(3n),

m = 0: Fibonacci(2n) (using recurrence relation only - the sum above is undefined for m=0),

m = -1: the zero sequence,

m = -2: (-1)*Fibonacci(n), or A152163(n+2).

For any value of m, the sequence gives a(n*k) divisible by a(n); n>=1, k>=1, m not equal to -1 (zero is not divisible by zero).

Equivalent sequences are given by: Sum_{k=0..n} binomial(n,k) * (m+1)^k * Fibonacci(k).

When these sequences are divided by m+1, we obtain the family of sequences A057088, A015553, A087567, A087579, A087584, A087603, and so on.

Another interesting value of m, m = -3, gives a(2n-1)= -2 * 5^(n-1); a(2n)=0.

Links

• Index entries for linear recurrences with constant coefficients, signature (13,109).

Crossrefs

Cf. A000045, A027941, A152163, A014445, A057088, A015553, A087567, A087579, A087584, A087603.

Programs

Mathematica

Table[Sum[Binomial[n, k]*10^(n - k)*Fibonacci[n + k], {k, 0, n}], {n, 0, 25}]
FullSimplify[Table[((13 + 11 Sqrt[5])^n - (13 - 11 Sqrt[5])^n)/(2^n Sqrt[5]), {n, 0, 25}]]
LinearRecurrence[{13,109},{0,11},30] (* Harvey P. Dale, Jul 31 2018 *)

Formula

a(n) = ((13 + 11*sqrt(5))^n - (13 - 11*sqrt(5))^n)/(2^n*sqrt(5)).

a(n) = 13*a(n-1) + 109*a(n-2); a(0)=0, a(1)=11.

G.f.: 11*x*/(1 - 13*x - 109*x^2). - Corrected by Georg Fischer, May 10 2019