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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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3
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0, 1, 4, 27, 240, 2625, 34020, 509355, 8648640, 164189025, 3445942500, 79222218075, 1979900722800, 53443570205025, 1549547301802500, 48028060502296875, 1584712538529120000, 55458748565165570625
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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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 (pbarry(AT)wit.ie), Sep 13 2010]
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EXAMPLE
| 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
| seq(n^2*(product(2*j-1, j = 1 .. n-1)), n = 0 .. 18);
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CROSSREFS
| A161119, A161121, A161122
Sequence in context: A160883 A185655 A181146 * A183430 A121063 A051863
Adjacent sequences: A161117 A161118 A161119 * A161121 A161122 A161123
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KEYWORD
| nonn
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 02 2009
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