A098531 Sum of fifth powers of first n Fibonacci numbers.

0, 1, 2, 34, 277, 3402, 36170, 407463, 4491564, 49926988, 553211363, 6137270812, 68054635036, 754774491429, 8370420537086, 92830050637086, 1029498223070793, 11417322172518550, 126619992693837974, 1404237451180502875, 15573231068749231000

Offset: 0

Author

Benoit Cloitre, Sep 12 2004

Keywords

• nonn
• easy

Comments

Prime p divides a((p-1)/2) for p = {29,89,101,181,229,...} = A047650[n]. Primes for which golden mean tau is a quadratic residue or Primes of the form x^2 + 20y^2. - Alexander Adamchuk, Aug 07 2006

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Index entries for linear recurrences with constant coefficients, signature (9,32,-100,20,48,-7,-1).

Crossrefs

Cf. A000071, A001654, A005968, A005969, A047650, A056572, A098532, A098533, A119286, A128697.

Programs

Magma

[(&+[Fibonacci(k)^5:k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 17 2018

Mathematica

Accumulate[Fibonacci[Range[0,20]]^5]  (* Harvey P. Dale, Jan 14 2011 *)
CoefficientList[Series[x*(1-7*x-16*x^2+7*x^3+x^4)/((1-x)*(1+4*x-x^2)*(1-x-x^2)*(1-11*x-x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 13 2012 *)

PARI

a(n)=sum(i=0,n,fibonacci(i)^5)

Formula

a(n) = -7/22 + 2*F(n+2)/5 + (F(5*(n+1)) + F(5*n))/(5*55) - (-1)^n*(F(3*(n+1)) - F(3*n))/(2*10), where F=A000045. One may use F(5*(n+1)) + F(5*n) = F(5*n+1) + 4*F(5*n+2) (due to the Binet-de Moivre formula).

G.f.: x*(1-7*x-16*x^2+7*x^3+x^4)/((1-x)*(1+4*x-x^2)*(1-x-x^2)*(1-11*x-x^2)). - Bruno Berselli, Oct 12 2012

Extensions

Formula corrected, with the author's consent, by Wolfdieter Lang, Oct 12 2012