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%I #10 Feb 16 2024 16:56:48
%S 1,2,18,360,12600,680400,52390800,5448643200,735566832000,
%T 125046361440000,26134689540960000,6585941764321920000,
%U 1969196587532254080000,689218805636288928000000
%N a(n)=(n+1)(2n)!/2^n.
%C Define T(n,k)=((1+(-1)^n)/2)*C(k-1+n/2, n/2)*n!/2^(n/2). Then T(n,k) has e.g.f. 1/sum{j=0..k, C(k,j)*(-1)^j*x^(2j)/2^j}. T(n,1) is A000680 with interpolated zeros. T(n,3) is A132912.
%H Daniele Marchei, Emanuela Merelli, and Andrew Francis, <a href="https://arxiv.org/abs/2402.07874">Factorizing the Brauer monoid in polynomial time</a>, arXiv:2402.07874 [math.RA], 2024. See p. 24.
%F E.g.f.: 1/(1-x^2+x^4/4) (with interpolated zeros);
%F a(n)-(n+1)*(2*n-1)*a(n-1)=0. - _R. J. Mathar_, Nov 05 2012
%t Table[(n+1) (2n)!/2^n,{n,0,20}] (* _Harvey P. Dale_, Jun 02 2020 *)
%K easy,nonn
%O 0,2
%A _Paul Barry_, Sep 04 2007
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