A171180 a(n) = (4*n + 1)^(1/2)/(4*n + 1)*((1 - p)*q^n - (1 - q)*p^n), where p = (1 - (4*n + 1)^(1/2))/2 and q = (1 + (4*n + 1)^(1/2))/2.

1, 3, 7, 29, 96, 463, 1905, 10233, 49159, 287891, 1557744, 9814741, 58451849, 392539575, 2532516511, 17999936497, 124360077816, 930257069563, 6822980957481, 53470578301581, 413527226164711, 3382254701784223, 27432377661111360, 233410016529114601

Offset: 1

Author

Gary Detlefs, Dec 04 2009

Keywords

• nonn

Comments

If a sequence (s(n): n >= 0) is of the form s(0) = x, s(1) = x, and s(n) = s(n-1) + k*s(n-2) for n >= 2 (for some integer k >= 1 and some number x), then s(k) = a(k)*x. For example, if k = 6 and x = 3, then (s(n): n = 0..6) = (3, 3, 21, 39, 165, 399, 1389) and s(6) = 1389 = 463*3 = a(6)*x. [Edited by Petros Hadjicostas, Dec 26 2019]

Links

• A. G. Shannon and J. V. Leyendekkers, The Golden Ratio family and the Binet equation, Notes on Number Theory and Discrete Mathematics, 21(2) (2015), 35-42.

Crossrefs

Cf. A350467.

Programs

Mathematica

Table[Sum[Binomial[n - k, k]*n^k, {k, 0, n}], {n, 1, 25}] (* Vaclav Kotesovec, Jan 08 2024 *)
Table[Hypergeometric2F1[(1 - n)/2, -n/2, -n, -4*n], {n, 1, 25}] (* Vaclav Kotesovec, Jan 08 2024 *)

PARI

{a(n)=polcoeff(1/(1-x-n*x^2+x*O(x^n)), n)} \\ Paul D. Hanna, Dec 27 2012

Formula

a(n) = A193376(n,n). - Olivier Gérard, Jul 25 2011

a(n) = [x^n] 1/(1 - x - n*x^2). - Paul D. Hanna, Dec 27 2012

From Vaclav Kotesovec, Jan 08 2024: (Start)

a(n) = Sum_{k=0..n} binomial(n-k,k) * n^k.

a(n) ~ exp(sqrt(n)/2) * n^(n/2) / 2 * (1 + 23/(48*sqrt(n))). (End)