A320565 a(n) = ((1 + sqrt(4*n^2 + 1))^n - (1 - sqrt(4*n^2 + 1))^n)/(2^n * sqrt(4*n^2 + 1)).

0, 1, 1, 10, 33, 701, 4033, 132301, 1089921, 48460114, 520210801, 29215223489, 386721507745, 26250621340841, 413242502386337, 32899021525375426, 600383148312628737, 54846079150716441949, 1138470675779123657425, 117372939125452004885621

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

Table of n, a(n) for n=0..19.

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

Cf. A084844, A320519, A320534.

Sequence in context: A264251 A140866 A219848 * A177221 A045087 A324746

Adjacent sequences: A320562 A320563 A320564 * A320566 A320567 A320568

Keyword

nonn

Author

Vladimir Reshetnikov, Oct 15 2018

Status

approved