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A247496
a(n) = n!*[x^n](exp(n*x)*BesselI_{1}(2*x)/x), n>=0, main diagonal of A247495.
7
1, 1, 5, 36, 354, 4425, 67181, 1200745, 24699662, 574795035, 14930563042, 428235433978, 13442267711940, 458373150076335, 16872717817840509, 666835739823870900, 28163028244810505622, 1265837029802096365275, 60330098878933736719190, 3039079334694016053006276
OFFSET
0,3
COMMENTS
Also coefficient of x^n in the expansion of 1/(n+1) * (1 + n*x + x^2)^(n+1). - Seiichi Manyama, May 06 2019
LINKS
FORMULA
a(n) = Sum_{j=0..floor(n/2)} ((j+1)*n^(n-2*j)*n!)/((j+1)!^2*(n-2*j)!).
a(n) ~ BesselI(1,2) * n^n. - Vaclav Kotesovec, Dec 12 2014
From Ilya Gutkovskiy, Sep 21 2017: (Start)
a(n) = [x^n] (1 - n*x - sqrt(1 - 2*n*x + (n^2 - 4)*x^2))/(2*x^2).
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)
MATHEMATICA
Flatten[{1, Table[n^n*HypergeometricPFQ[{1/2-n/2, -n/2}, {2}, 4/n^2], {n, 1, 20}]}] (* Vaclav Kotesovec, Dec 12 2014 *)
PROG
(Sage)
a = lambda n: 1 if n==0 else n^n*hypergeometric([1/2-n/2, -n/2], [2], 4/n^2).simplify()
[a(n) for n in range(20)]
(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
(PARI) {a(n) = polcoef((1+n*x+x^2)^(n+1)/(n+1), n)} \\ Seiichi Manyama, May 06 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Peter Luschny, Dec 12 2014
STATUS
approved