OFFSET
0,1
COMMENTS
The right-hand side of Helmut Postl's identity F(2*n+5) + F(n)*F(n+3)^5 = F(n+5)*F(n+2)^5.
LINKS
Colin Barker, Table of n, a(n) for n = 0..750
Index entries for linear recurrences with constant coefficients, signature (13,104,-260,-260,104,13,-1).
FORMULA
From Colin Barker, Aug 31 2016: (Start)
a(n) = 13*a(n-1)+104*a(n-2)-260*a(n-3)-260*a(n-4)+104*a(n-5)+13*a(n-6)-a(n-7) for n>6 .
G.f.: (5+191*x-689*x^2-766*x^3+311*x^4+39*x^5-3*x^6) / ((1+x)*(1-18*x+x^2)*(1-3*x+x^2)*(1+7*x+x^2)).
(End)
MATHEMATICA
Table[(Fibonacci[n + 5] Fibonacci[n + 2]^5), {n, 0, 20}] (* Vincenzo Librandi, Sep 02 2016 *)
PROG
(PARI) Vec((5+191*x-689*x^2-766*x^3+311*x^4+39*x^5-3*x^6)/((1+x)*(1-18*x+x^2)*(1-3*x+x^2)*(1+7*x+x^2)) + O(x^20)) \\ Colin Barker, Aug 31 2016
(Magma) [Fibonacci(n+5)*Fibonacci(n+2)^5: n in [0..25]]; // Vincenzo Librandi, Sep 92 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Aug 31 2016
STATUS
approved