A217471 Partial sum of fifth power of the even-indexed Fibonacci numbers.

0, 1, 244, 33012, 4117113, 507401488, 62424765712, 7678070811369, 944346243245076, 116147016764564500, 14285140634333292625, 1756956185432949082176, 216091326285380812359744, 26577476188001703626949937

Offset

0,3

Comments

For the o.g.f. for general powers of Fibonacci numbers F=A000045 see A056588 (row polynomials as numerators) and A055870 (row polynomials as denominator). The even part of the bisection leads to the o.g.f. for powers of F(2*n), and the partial sums of these powers are then given by dividing this o.g.f. by (1-x). For the o.g.f.s for F(n)^5 and F(2*n)^5 see A056572 and A215044, respectively.

The tables of the coefficient of the polynomials which appear in Ozeki's formula and in Melham's conjecture are found in A217472 and A217475, respectively (see References).

Links

G. C. Greubel, Table of n, a(n) for n = 0..475

R. S. Melham, Some conjectures concerning sums of odd powers of Fibonacci and Lucas numbers, The Fibonacci Quart. 46/47 (2008/2009), no. 4, 312-315.

K. Ozeki, On Melham's sum, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.

H. Prodinger, On a sum of Melham and its variants, The Fibonacci Quart. 46/47 2008/2009), no. 3, 207-215.

Formula

a(n) = Sum_{k=0..n} F(2*k)^5, n>=0.

O.g.f.: x*(1+99*x+416*x^2+99*x^3+x^4)/((1-3*x+x^2)*(1-18*x+x^2)*(1-123*x+x^2)*(1-x)).

a(n) = 2*(F(2*(n+1)) - F(2*n))/5 - F(3*(2*n+1))/20 +

(F(10*(n+1)) - F(10*n))/F(10)^2 - 7/22 (from the partial fraction decomposition of the o.g.f.).

a(n) = (1/11)*F(2*n+1)^5 - (15/44)*F(2*n+1)^3 + (25/44)*F(2*n+1) - 7/22 (from Ozeki reference, Theorem 2, p. 109 --- with a misprint -- and from Prodinger reference, p. 207).

a(n) =(F(2*n+1)-1)^2*(4*F(2*n+1)^3 + 8*F(2*n+1)^2 - 3*F(2*n+1) - 14)/44 (an example for Melham's conjecture, see the reference, eq. (2.7) for m=2).

Example

a(2) = 244 = 2*(8-3)/5 - 610/20 + (832040-6765)/55^2 - 7/22.
a(2) = 244 = (1/11)*5^5 - (15/44)*5^3 + (25/44)*5 - 7/22.
a(2) = 244 = (5-1)^2*(4*5^3 + 8*5^2 - 3*5 - 14)/44
           = (4*5^3 + 8*5^2 - 3*5 - 14)*(4/11).

Mathematica

Table[Sum[Fibonacci[2*k]^5, {k, 0, n}], {n, 0, 50}] (* G. C. Greubel, Apr 12 2017 *)

Prog

(PARI) a(n) = sum(k=1, n, fibonacci(2*k)^5); \\ Michel Marcus, Feb 29 2016

Crossrefs

Cf. A163198 (third powers).

Sequence in context: A194134 A085440 A218246 * A146552 A263948 A306981

Adjacent sequences: A217468 A217469 A217470 * A217472 A217473 A217474

Keyword

nonn,easy

Author

Wolfdieter Lang, Oct 11 2012

Status

approved