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Search: id:A161120
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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
(list; graph; listen)
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OFFSET
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0,3
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FORMULA
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a(n)=n^2*(2n-3)!!
a(n)=Sum(k*A161119(n,k), k=0..n)
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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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A161119, A161121, A161122
Sequence in context: A091121 A026005 A059391 this_sequence A121063 A051863 A000699
Adjacent sequences: A161117 A161118 A161119 this_sequence A161121 A161122 A161123
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KEYWORD
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nonn
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 02 2009
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