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%I #28 Mar 19 2020 14:20:26
%S 1,1,5,36,354,4425,67181,1200745,24699662,574795035,14930563042,
%T 428235433978,13442267711940,458373150076335,16872717817840509,
%U 666835739823870900,28163028244810505622,1265837029802096365275,60330098878933736719190,3039079334694016053006276
%N a(n) = n!*[x^n](exp(n*x)*BesselI_{1}(2*x)/x), n>=0, main diagonal of A247495.
%C Also coefficient of x^n in the expansion of 1/(n+1) * (1 + n*x + x^2)^(n+1). - _Seiichi Manyama_, May 06 2019
%H Seiichi Manyama, <a href="/A247496/b247496.txt">Table of n, a(n) for n = 0..386</a>
%F a(n) = Sum_{j=0..floor(n/2)} ((j+1)*n^(n-2*j)*n!)/((j+1)!^2*(n-2*j)!).
%F a(n) ~ BesselI(1,2) * n^n. - _Vaclav Kotesovec_, Dec 12 2014
%F From _Ilya Gutkovskiy_, Sep 21 2017: (Start)
%F a(n) = [x^n] (1 - n*x - sqrt(1 - 2*n*x + (n^2 - 4)*x^2))/(2*x^2).
%F a(n) = [x^n] 1/(1 - n*x - x^2/(1 - n*x - x^2/(1 - n*x - x^2/(1 - n*x - x^2/(1 - ...))))), a continued fraction. (End)
%t Flatten[{1,Table[n^n*HypergeometricPFQ[{1/2-n/2, -n/2}, {2}, 4/n^2],{n,1,20}]}] (* _Vaclav Kotesovec_, Dec 12 2014 *)
%o (Sage)
%o a = lambda n: 1 if n==0 else n^n*hypergeometric([1/2-n/2, -n/2], [2], 4/n^2).simplify()
%o [a(n) for n in range(20)]
%o (PARI) {a(n) = sum(k=0, n\2, n^(n-2*k)*binomial(n, 2*k)*binomial(2*k, k)/(k+1))} \\ _Seiichi Manyama_, May 05 2019
%o (PARI) {a(n) = polcoef((1+n*x+x^2)^(n+1)/(n+1), n)} \\ _Seiichi Manyama_, May 06 2019
%Y Cf. A001006, A186925, A247495, A292716.
%K nonn
%O 0,3
%A _Peter Luschny_, Dec 12 2014
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