

A161120


Number of cycles with entries of opposite parities in all fixedpointfree involutions of {1,2,...,2n}.


4



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

0,3


LINKS



FORMULA

a(n)=n^2*(2n3)!!
E.g.f.: x(1x)/(12x)^(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, 3step).  Sergei N. Gladkovskii, Sep 25 2012
Dfinite with recurrence a(n) +(2*n1)*a(n1) +a(n2) +(2*n+7)*a(n3)=0.  R. J. Mathar, Jul 26 2022


EXAMPLE

a(2)=4 because in the 3 fixedpointfree 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.


MAPLE

seq(n^2*(product(2*j1, j = 1 .. n1)), n = 0 .. 18);


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



