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, 785682752921379769625, 14098504953417839657513

Offset: 0

Author

Wolfdieter Lang, Aug 10 2012

Keywords

• nonn
• easy

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
• Kenny B. Davenport, proposer, Problem B-1353, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 259.

Crossrefs

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

Programs

Mathematica

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

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.

a(n) = 1 + (4*Sum_{k=1..n} F(6*k) + 3*Sum_{k=1..n} F(2*k)) / 5 (Davenport, 2024). - Amiram Eldar, Aug 29 2024