OFFSET
0,2
COMMENTS
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.
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
KEYWORD
nonn,easy,changed
AUTHOR
Wolfdieter Lang, Aug 10 2012
STATUS
approved