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a(n) = Sum_{k=0..n} binomial(n,k) * 10^(n-k) * Fibonacci(n+k).
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%I #42 Nov 02 2024 06:49:35

%S 0,11,143,3058,55341,1052755,19717984,371084087,6973353387,

%T 131101759514,2464418392865,46327530894271,870879506447808,

%U 16371134451297043,307750614069672631,5785211638097121890,108752568228856901349,2044371455527726003547,38430858858805840293152

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

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

%C 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.

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

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

%C Notable values of m include:

%C m = 1: Fibonacci(3n),

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

%C m = -1: the zero sequence,

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

%C 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).

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

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

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

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (13,109).

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

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

%F G.f.: 11*x*/(1 - 13*x - 109*x^2). - Corrected by _Georg Fischer_, May 10 2019

%t Table[Sum[Binomial[n, k]*10^(n - k)*Fibonacci[n + k], {k, 0, n}], {n, 0, 25}]

%t FullSimplify[Table[((13 + 11 Sqrt[5])^n - (13 - 11 Sqrt[5])^n)/(2^n Sqrt[5]), {n, 0, 25}]]

%t LinearRecurrence[{13,109},{0,11},30] (* _Harvey P. Dale_, Jul 31 2018 *)

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

%K nonn,easy,changed

%O 0,2

%A _John Molokach_, Jul 27 2013