A110224 a(n) = Fibonacci(n)^3 + Fibonacci(n+1)^3.

1, 2, 9, 35, 152, 637, 2709, 11458, 48565, 205679, 871344, 3690953, 15635321, 66231970, 280563633, 1188485803, 5034507976, 21326515877, 90340574445, 382688808866, 1621095817661, 6867072066967, 29089384105824

Offset

0,2

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..172

Diego Marques and Alain Togbé, On the sum of powers of two consecutive Fibonacci numbers, Proc. Japan Acad. Ser. A Math. Sci., Volume 86, Number 10 (2010), 174-176.

Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Formula

G.f.: (1 - x - 3*x^2 - x^3)/(1 - 3*x - 6*x^2 + 3*x^3 + x^4) = (1 - x - 3*x^2 - x^3)/((1 + x - x^2)*(1 - 4x - x^2)).

a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4).

a(n) = (3*(-1)^n*Fibonacci(n-1) + 2*Fibonacci(3*n+2))/5.

Mathematica

Total/@Partition[Fibonacci[Range[0, 30]]^3, 2, 1] (* or *) LinearRecurrence [{3, 6, -3, -1}, {1, 2, 9, 35}, 30] (* Harvey P. Dale, May 29 2013 *)

Prog

(Magma) [Fibonacci(n)^3 + Fibonacci(n+1)^3: n in [0..30]]; // Vincenzo Librandi, Jun 05 2011
(PARI) a(n)=fibonacci(n)^3+fibonacci(n+1)^3 \\ Charles R Greathouse IV, Jun 05 2011
(Sage) [sum(fibonacci(n+k)^3 for k in (0..1)) for n in (0..30)] # G. C. Greubel, Mar 18 2019

Crossrefs

Cf. A056570.

Sequence in context: A150945 A150946 A150947 * A363549 A150948 A150949

Adjacent sequences: A110221 A110222 A110223 * A110225 A110226 A110227

Keyword

easy,nonn

Author

Paul Barry, Jul 16 2005

Status

approved