A161120 Number of cycles with entries of opposite parities in all fixed-point-free involutions of {1,2,...,2n}.

0, 1, 4, 27, 240, 2625, 34020, 509355, 8648640, 164189025, 3445942500, 79222218075, 1979900722800, 53443570205025, 1549547301802500, 48028060502296875, 1584712538529120000, 55458748565165570625

Offset: 0

Author

Emeric Deutsch, Jun 02 2009

Keywords

• nonn

Examples

			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.
		

Crossrefs

Cf. A161119, A161121, A161122.

Programs

Maple

seq(n^2*(product(2*j-1, j = 1 .. n-1)), n = 0 .. 18);

Formula

a(n)=n^2*(2n-3)!!

a(n)=Sum(k*A161119(n,k), k=0..n)

E.g.f.: x(1-x)/(1-2x)^(3/2). [From Paul Barry, Sep 13 2010]

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

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