A215039 a(n) = Fibonacci(2*n)^3, n>=0.
0, 1, 27, 512, 9261, 166375, 2985984, 53582633, 961504803, 17253512704, 309601747125, 5555577996431, 99690802348032, 1788878864685457, 32100128763082731, 576013438873664000, 10336141770970357629, 185474538438612378103
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..790
- E. Kilic, Y. T. Ulutas, N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, Table 1, k=3.
- Index entries for linear recurrences with constant coefficients, signature (21,-56,21,-1).
Programs
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GAP
List([0..20], n-> Fibonacci(2*n)^3 ); # G. C. Greubel, Dec 22 2019
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Magma
[Fibonacci(2*n)^3: n in [0..20]]; // G. C. Greubel, Dec 22 2019
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Maple
with(combinat); seq( fibonacci(2*n)^3, n=0..20); # G. C. Greubel, Dec 22 2019
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Mathematica
Fibonacci[2*(Range[21]-1)]^3 (* G. C. Greubel, Dec 22 2019 *)
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PARI
vector(21, n, fibonacci(2*(n-1)) ); \\ G. C. Greubel, Dec 22 2019
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Sage
[fibonacci(2*n)^3 for n in (0..20)] # G. C. Greubel, Dec 22 2019
Formula
a(n) = F(2*n)^3, n>=0, with F=A000045.
O.g.f.: x*(1+6*x+x^2)/((1-3*x+x^2)*(1-18*x+x^2)) (from the bisection (even part) of A056570).
a(n) = (F(6*n) - 3*F(2*n))/5, n>=0.
a(n+2) - 18*a(n+1) + a(n) - 9*F(2*(n+1)) = 0, n>=0. From the F_n^3 recurrence (see a comment and references on A055870, use row n=4) together with the recurrence appearing in the solution of exercise 6.58, p. 315, on p. 556 of the second edition of the Graham-Knuth-Patashnik book (reference given on A007318), both with n -> 2*n. See also Koshy's book (reference given on A065563) p. 87, 1. and p. 89, 32. (with a - sign) and 33. - Wolfdieter Lang, Aug 11 2012
Comments