OFFSET
0,3
COMMENTS
Bisection (even part) of A056570. From this follows the o.g.f., and its partial fraction decomposition leads to the explicit formula given below. The recurrences for F(2*n) and F(6*n)/8 are used in this computation. They follow from the fact that F(2*n) = S(n-1,3), and F(6*n)/8 = S(n-1,18), with Chebyshev's S(n,x) = U(n,x/2) polynomial of the second kind (see A001906 and A049660, respectively).
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).
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
MAPLE
with(combinat); seq( fibonacci(2*n)^3, n=0..20); # G. C. Greubel, Dec 22 2019
MATHEMATICA
Fibonacci[2*(Range[21]-1)]^3 (* G. C. Greubel, Dec 22 2019 *)
PROG
(PARI) vector(21, n, fibonacci(2*(n-1)) ); \\ G. C. Greubel, Dec 22 2019
(Magma) [Fibonacci(2*n)^3: n in [0..20]]; // G. C. Greubel, Dec 22 2019
(Sage) [fibonacci(2*n)^3 for n in (0..20)] # G. C. Greubel, Dec 22 2019
(GAP) List([0..20], n-> Fibonacci(2*n)^3 ); # G. C. Greubel, Dec 22 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Aug 10 2012
STATUS
approved