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A217471
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Partial sum of fifth power of the even-indexed Fibonacci numbers.
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3
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0, 1, 244, 33012, 4117113, 507401488, 62424765712, 7678070811369, 944346243245076, 116147016764564500, 14285140634333292625, 1756956185432949082176, 216091326285380812359744, 26577476188001703626949937
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OFFSET
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0,3
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COMMENTS
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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).
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LINKS
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K. Ozeki, On Melham's sum, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
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FORMULA
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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).
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EXAMPLE
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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).
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MATHEMATICA
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Table[Sum[Fibonacci[2*k]^5, {k, 0, n}], {n, 0, 50}] (* G. C. Greubel, Apr 12 2017 *)
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PROG
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(PARI) a(n) = sum(k=1, n, fibonacci(2*k)^5); \\ Michel Marcus, Feb 29 2016
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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