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%I #11 Jan 24 2013 17:02:12
%S 1,8,125,2197,39304,704969,12649337,226981000,4073003173,73087061741,
%T 1311494070536,23533806109393,422297015640625,7577812474746632,
%U 135978327528030989,2440032083025183109,43784599166913148552
%N a(n) = F(2*n+1)^3, n>=0, with F = A000045 (Fibonacci).
%C 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.
%H Harvey P. Dale, <a href="/A215040/b215040.txt">Table of n, a(n) for n = 0..797</a>
%F a(n) = F(2*n+1)^3, n>=0, with F=A000045.
%F 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.
%F a(n) = (12*F(2*n+1) + F(6*(n+1)) - F(6*n))/20, n>=0.
%t Fibonacci[2*Range[0,20]+1]^3 (* _Harvey P. Dale_, Jan 24 2013 *)
%Y Cf. A000045, A056570, A163200 (partial sums).
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Aug 10 2012
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