OFFSET
0,2
COMMENTS
More generally for any complex number z, sequence a(n)=Sum_{k=0..n} z^k*F(k) satisfies the recurrence : a(0)=0, a(1)=z, a(2)=z(z+1), for n>2 a(n)=(z+1)*a(n-1)+z*(z-1)*a(n-2)-z^2*a(n-3)
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,12,-16).
FORMULA
a(0)=0, a(1)=4, a(2)=20, a(n)=5a(n-1)+12a(n-2)-16a(n-3).
O.g.f.: 4*x/((x-1)*(16*x^2+4*x-1)). - R. J. Mathar, Dec 05 2007
PROG
(PARI) a(n)=if(n<0, 0, sum(k=0, n, fibonacci(k)*4^k))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Benoit Cloitre, May 29 2003
STATUS
approved