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%I #31 Jan 08 2024 08:31:59
%S 1,3,7,29,96,463,1905,10233,49159,287891,1557744,9814741,58451849,
%T 392539575,2532516511,17999936497,124360077816,930257069563,
%U 6822980957481,53470578301581,413527226164711,3382254701784223,27432377661111360,233410016529114601
%N 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.
%C 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]
%H A. G. Shannon and J. V. Leyendekkers, <a href="http://nntdm.net/volume-21-2015/number-2/35-42/">The Golden Ratio family and the Binet equation</a>, Notes on Number Theory and Discrete Mathematics, 21(2) (2015), 35-42.
%F a(n) = A193376(n,n). - _Olivier Gérard_, Jul 25 2011
%F a(n) = [x^n] 1/(1 - x - n*x^2). - _Paul D. Hanna_, Dec 27 2012
%F From _Vaclav Kotesovec_, Jan 08 2024: (Start)
%F a(n) = Sum_{k=0..n} binomial(n-k,k) * n^k.
%F a(n) ~ exp(sqrt(n)/2) * n^(n/2) / 2 * (1 + 23/(48*sqrt(n))). (End)
%t Table[Sum[Binomial[n - k, k]*n^k, {k, 0, n}], {n, 1, 25}] (* _Vaclav Kotesovec_, Jan 08 2024 *)
%t Table[Hypergeometric2F1[(1 - n)/2, -n/2, -n, -4*n], {n, 1, 25}] (* _Vaclav Kotesovec_, Jan 08 2024 *)
%o (PARI) {a(n)=polcoeff(1/(1-x-n*x^2+x*O(x^n)), n)} \\ _Paul D. Hanna_, Dec 27 2012
%Y Cf. A350467.
%K nonn
%O 1,2
%A _Gary Detlefs_, Dec 04 2009
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