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A161120
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Number of cycles with entries of opposite parities in all fixed-point-free involutions of {1,2,...,2n}.
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4
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0, 1, 4, 27, 240, 2625, 34020, 509355, 8648640, 164189025, 3445942500, 79222218075, 1979900722800, 53443570205025, 1549547301802500, 48028060502296875, 1584712538529120000, 55458748565165570625
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OFFSET
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0,3
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LINKS
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FORMULA
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a(n)=n^2*(2n-3)!!
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
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EXAMPLE
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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.
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MAPLE
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seq(n^2*(product(2*j-1, j = 1 .. n-1)), n = 0 .. 18);
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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