A215040 a(n) = F(2*n+1)^3, n>=0, with F = A000045 (Fibonacci).

1, 8, 125, 2197, 39304, 704969, 12649337, 226981000, 4073003173, 73087061741, 1311494070536, 23533806109393, 422297015640625, 7577812474746632, 135978327528030989, 2440032083025183109, 43784599166913148552

Offset

0,2

Comments

Bisection (odd part) of A056570. From this follows the o.g.f., and its partial fraction decomposition leads to the explicit formula given below. For the computation see A215039 for a comment on Chebyshev's S-polynomials.

Links

Harvey P. Dale, Table of n, a(n) for n = 0..797

Formula

a(n) = F(2*n+1)^3, n>=0, with F=A000045.

O.g.f.: (1-x)*(1-12*x+x^2)/((1-3*x+x^2)*(1-18*x+x^2)) (from the bisection (odd part) of A056570.

a(n) = (12*F(2*n+1) + F(6*(n+1)) - F(6*n))/20, n>=0.

Mathematica

Fibonacci[2*Range[0, 20]+1]^3 (* Harvey P. Dale, Jan 24 2013 *)

Crossrefs

Cf. A000045, A056570, A163200 (partial sums).

Sequence in context: A061103 A264143 A110272 * A033536 A355762 A215793

Adjacent sequences: A215037 A215038 A215039 * A215041 A215042 A215043

Keyword

nonn,easy

Author

Wolfdieter Lang, Aug 10 2012

Status

approved