OFFSET
0,4
COMMENTS
a(0) = 0 assuming 0^0 = 1, or using the limit for n -> 0 (assuming n is a real variable); the same value for a(0) arises from other formulae for this sequence.
LINKS
Eric Weisstein's World of Mathematics, Fibonacci Polynomial
Wikipedia, Fibonacci polynomials
FORMULA
a(n) = 2^(1 - n) * Sum_{k=0..floor(n/2)} binomial(n, 2*k + 1)*(4*n^2 + 1)^k.
a(n) = 2 * (i*n)^n * sinh(n*arctanh(sqrt(4*n^2 + 1)))/sqrt(4*n^2 + 1), assuming 0^0 = 1 for n = 0.
For n > 0, a(n) = n^(n - 1) * F_n(1/n), where F_n(x) is the Fibonacci polynomial.
For n > 0, a(n) = sqrt(Pi/4)*i*(-i*n)^n*LegendreP((n - 1)/2, -1/2, -1/(2*n^2) - 1) / (4*n^2 + 1)^(1/4). - Peter Luschny, Oct 15 2018
MATHEMATICA
a[0] = Limit[n^(n - 1) Fibonacci[n, 1/n], n -> 0]; (* a[0] = 0 *)
a[n_] := a[n] = n^(n - 1) Fibonacci[n, 1/n];
Table[a[n], {n, 0, 19}]
PROG
(PARI) for(n=0, 20, print1( 2^(1-n)*sum(k=0, floor(n/2), binomial(n, 2*k+1)*(4*n^2+1)^k) , ", ")) \\ G. C. Greubel, Oct 15 2018
(Magma) [2^(1-n)*(&+[Binomial(n, 2*k+1)*(4*n^2+1)^k: k in [0..Floor(n/2)]]): n in [0..20]]; // G. C. Greubel, Oct 15 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Reshetnikov, Oct 15 2018
STATUS
approved