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%I #11 Jul 26 2022 13:46:49
%S 0,1,4,27,240,2625,34020,509355,8648640,164189025,3445942500,
%T 79222218075,1979900722800,53443570205025,1549547301802500,
%U 48028060502296875,1584712538529120000,55458748565165570625
%N Number of cycles with entries of opposite parities in all fixed-point-free involutions of {1,2,...,2n}.
%F a(n)=n^2*(2n-3)!!
%F a(n)=Sum(k*A161119(n,k), k=0..n)
%F E.g.f.: x(1-x)/(1-2x)^(3/2). [From _Paul Barry_, Sep 13 2010]
%F E.g.f.: x/2*U(0) where U(k)= 1 + (2*k+1)/(1 - x/(x + (k+1)/U(k+1))) ; (continued fraction, 3-step). - _Sergei N. Gladkovskii_, Sep 25 2012
%F D-finite with recurrence a(n) +(-2*n-1)*a(n-1) +a(n-2) +(-2*n+7)*a(n-3)=0. - _R. J. Mathar_, Jul 26 2022
%e a(2)=4 because in the 3 fixed-point-free involutions of {1,2,3,4}, namely (12)(34), (13)(24), (14)(23), we have a total of 4 cycles with entries of opposite parities.
%p seq(n^2*(product(2*j-1, j = 1 .. n-1)), n = 0 .. 18);
%Y Cf. A161119, A161121, A161122.
%K nonn
%O 0,3
%A _Emeric Deutsch_, Jun 02 2009
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