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%I #21 Apr 12 2017 23:17:55
%S 0,1,244,33012,4117113,507401488,62424765712,7678070811369,
%T 944346243245076,116147016764564500,14285140634333292625,
%U 1756956185432949082176,216091326285380812359744,26577476188001703626949937
%N Partial sum of fifth power of the even-indexed Fibonacci numbers.
%C 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.
%C 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).
%H G. C. Greubel, <a href="/A217471/b217471.txt">Table of n, a(n) for n = 0..475</a>
%H R. S. Melham, <a href="http://www.fq.math.ca/Papers1/46_47-4/Melham.pdf">Some conjectures concerning sums of odd powers of Fibonacci and Lucas numbers</a>, The Fibonacci Quart. 46/47 (2008/2009), no. 4, 312-315.
%H K. Ozeki, <a href="http://www.fq.math.ca/Papers1/46_47-2/Ozeki.pdf">On Melham's sum</a>, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
%H H. Prodinger, <a href="http://www.fq.math.ca/Papers1/46_47-3/Prodinger.pdf">On a sum of Melham and its variants</a>, The Fibonacci Quart. 46/47 2008/2009), no. 3, 207-215.
%F a(n) = Sum_{k=0..n} F(2*k)^5, n>=0.
%F 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)).
%F a(n) = 2*(F(2*(n+1)) - F(2*n))/5 - F(3*(2*n+1))/20 +
%F (F(10*(n+1)) - F(10*n))/F(10)^2 - 7/22 (from the partial fraction decomposition of the o.g.f.).
%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).
%F 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).
%e a(2) = 244 = 2*(8-3)/5 - 610/20 + (832040-6765)/55^2 - 7/22.
%e a(2) = 244 = (1/11)*5^5 - (15/44)*5^3 + (25/44)*5 - 7/22.
%e a(2) = 244 = (5-1)^2*(4*5^3 + 8*5^2 - 3*5 - 14)/44
%e = (4*5^3 + 8*5^2 - 3*5 - 14)*(4/11).
%t Table[Sum[Fibonacci[2*k]^5, {k, 0, n}], {n, 0, 50}] (* _G. C. Greubel_, Apr 12 2017 *)
%o (PARI) a(n) = sum(k=1, n, fibonacci(2*k)^5); \\ _Michel Marcus_, Feb 29 2016
%Y Cf. A163198 (third powers).
%K nonn,easy
%O 0,3
%A _Wolfdieter Lang_, Oct 11 2012
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