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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.

%I A115455 #24 Jul 23 2017 16:56:39
%S A115455 1,0,1,1,4,13,59,308,1871,12879,99144,843735,7865177,79698760,
%T A115455 872235089,10253148625,128839087676,1723418002261,24450430660739,
%U A115455 366702601116524,5796979684239647,96339860422218143,1679159568980521104,30628034488033962287
%N 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.
%H A115455 Alois P. Heinz, <a href="/A115455/b115455.txt">Table of n, a(n) for n = 0..200</a>
%H A115455 R. P. Stanley, <a href="http://www.ams.org/amsmtgs/colloq-10.pdf">Permutations</a>, Joint Mathematics Meeting, 2009.
%H A115455 R. P. Stanley, <a href="http://dx.doi.org/10.1016/j.jcta.2006.06.008">Alternating permutations and symmetric functions</a>, J. Comb. Theory A 114 (3) (2007) 436-460
%F A115455 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)).
%e A115455 a(3)=1 because there is one reverse alternating fixed-point-free involution on 1,...,6, viz., 351624.
%t A115455 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 *)
%Y A115455 Cf. A000085, A000111, A001147, A007779.
%K A115455 easy,nonn
%O A115455 0,5
%A A115455 _Richard Stanley_, Jan 22 2006