A115455 a(n) = number of reverse alternating fixed-point-free involutions w on 1,2,...,2n, i.e., w(1) < w(2) > w(3) < w(4) > ... < w(2n), w^2=1 and w(i) != i for all i.
1, 0, 1, 1, 4, 13, 59, 308, 1871, 12879, 99144, 843735, 7865177, 79698760, 872235089, 10253148625, 128839087676, 1723418002261, 24450430660739, 366702601116524, 5796979684239647, 96339860422218143, 1679159568980521104, 30628034488033962287
Offset: 0
Examples
a(3)=1 because there is one reverse alternating fixed-point-free involution on 1,...,6, viz., 351624.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..200
- R. P. Stanley, Permutations, Joint Mathematics Meeting, 2009.
- R. P. Stanley, Alternating permutations and symmetric functions, J. Comb. Theory A 114 (3) (2007) 436-460
Programs
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Mathematica
Table[SeriesCoefficient[(1-x^2)^(-1/4)*(1+x)^(-1/2)*Sum[(-1)^k*EulerE[2*k]*(1/4*Log[(1+x)/(1-x)])^k/k!,{k,0,n}],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Apr 29 2014 *)
Formula
G.f.: (1-x^2)^{-1/4} (1+x)^{-1/2} Sum_{k>=0} E_{2k} v^k/k!, where E_{2k} is an Euler number and v = (1/4)*log((1+x)/(1-x)).