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A275930
a(n) = F(n+5)*F(n+2)^5, where F = Fibonacci (A000045).
1
5, 256, 3159, 65625, 1114112, 20421115, 363484989, 6542701056, 117265259375, 2105190412273, 37769592176640, 677792498259891, 12162186734914229, 218243684178400000, 3916209628945328967, 70273629018014076105, 1261008431526362415104, 22627882807257322061611, 406040850098667041878125
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
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
Cf. A000045.
Sequence in context: A308652 A002770 A069071 * A181397 A283039 A055386
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Aug 31 2016
STATUS
approved