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A215040
a(n) = F(2*n+1)^3, n>=0, with F = A000045 (Fibonacci).
1
1, 8, 125, 2197, 39304, 704969, 12649337, 226981000, 4073003173, 73087061741, 1311494070536, 23533806109393, 422297015640625, 7577812474746632, 135978327528030989, 2440032083025183109, 43784599166913148552, 785682752921379769625, 14098504953417839657513
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
Kenny B. Davenport, proposer, Problem B-1353, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 259.
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
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
KEYWORD
nonn,easy,changed
AUTHOR
Wolfdieter Lang, Aug 10 2012
STATUS
approved