OFFSET
1,2
COMMENTS
For a Lucas sequence U(k,1), the sum of the cubes of the first n terms is divisible by the sum of the first n terms. This sequence corresponds to the case of k=3.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..200
FORMULA
a(n) = 11*a(n-1) - 33*a(n-2) + 33*a(n-3) - 11*a(n-4) + a(n-5). - Vaclav Kotesovec, May 23 2013
G.f.: x*(1-4*x+x^2)/((1-x)*(1-7*x+x^2)*(1-3*x+x^2)). [Bruno Berselli, Jun 07 2013]
MATHEMATICA
Table[Fibonacci[2*n+1]/4 + LucasL[4*n+2]/20 - 2/5, {n, 1, 20}] (* Vaclav Kotesovec, May 23 2013 *)
With[{f=Fibonacci[Range[2, 50, 2]]}, Accumulate[f^3]/Accumulate[f]] (* Harvey P. Dale, Feb 17 2020 *)
PROG
(PARI) Vec(x*(1-4*x+x^2)/((1-x)*(1-7*x+x^2)*(1-3*x+x^2)) + O(x^100)) \\ Altug Alkan, Dec 09 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Max Alekseyev, Nov 09 2012
STATUS
approved