OFFSET
0,1
COMMENTS
The right-hand side of Helmut Postl's identity F(2n+6) + F(n)*F(n+4)^3 = F(n+6)*F(n+2)^3, n even.
LINKS
Colin Barker, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (55,-385,385,-55,1).
FORMULA
From Colin Barker, Aug 31 2016: (Start)
a(n) = 55*a(n-1)-385*a(n-2)+385*a(n-3)-55*a(n-4)+a(n-5) for n>4.
G.f.: (8+127*x+55*x^2-x^3) / ((1-x)*(1-47*x+x^2)*(1-7*x+x^2)).
(End)
MATHEMATICA
Table[(Fibonacci[2 n + 6] Fibonacci[2 n + 2]^3), {n, 0, 20}] (* Vincenzo Librandi, Sep 02 2016 *)
PROG
(PARI) Vec((8+127*x+55*x^2-x^3)/((1-x)*(1-47*x+x^2)*(1-7*x+x^2)) + O(x^20)) \\ Colin Barker, Aug 31 2016
(Magma) [Fibonacci(2*n+6)*Fibonacci(2*n+2)^3: n in [0..25]]; // Vincenzo Librandi, Sep 02 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Aug 31 2016
STATUS
approved