OFFSET
0,2
COMMENTS
More generally for any complex number z, the 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
MATHEMATICA
LinearRecurrence[{5, 12, -16}, {0, 4, 20}, 21] (* Amiram Eldar, Apr 29 2025 *)
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
