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A267060
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a(n) = number of different ways to seat a set of n married male-female couples at a round table so that men and women alternate and every man is separated by at least d = 2 men from his wife.
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2
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0, 0, 0, 0, 24, 240, 22320, 1330560, 112210560, 11183235840, 1340192044800, 189443216793600, 31267307962598400, 5964702729085900800, 1303453560329957836800, 323680816052170536960000, 90679832709074132299776000, 28473630606612014817337344000
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OFFSET
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1,5
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COMMENTS
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We assume that the chairs are uniform and indistinguishable.
First we arrange the females in alternating seats by circular permutation, there are (n-1)! ways. Secondly, we evaluate the number F_{n}, ways of arranging males in the remaining seats as mentioned in the definition above.
By the principle of inclusion-exclusion and theory of rook polynomial Vl, we obtain that a_{n} = (n-1)!*F_{n}, F_{n} = sum(-1)^{k}*r_{k}(B3)*(n-k)! where r_{k}(B3) is the number of ways of putting k non-taking rooks on positions 1's of B3, and the rook polynomials are R_{B3}(x) = sum r_{k}(B3)*x^{k}.
Also F_{n} = per(B3), here per(B3) denotes the permanent of matrix (board) B3, but it is very difficult problem to evaluate the value, per(B3).
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REFERENCES
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G. Polya, Aufgabe 424, Arch. Math. Phys. (3) 20 (1913) 271.
John Riordan. The enumeration of permutations with three-ply staircase restrictions.
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LINKS
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FORMULA
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EXAMPLE
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For d=1, the sequence a_{n} is the classical menage sequence A094047.
For d=2 (the current sequence), the F(n)s are 0, 0, 0, 0, 1, 2, 31, 264, 2783, 30818, 369321, ... which is A004307(n) then the sequence a_{n} is 0, 0, 0, 0, 24, 240, 22320, 1330560, 112210560, 11183235840, 1340192044800,...
For d=3, the F(n)s are 0, 0, 0, 0, 0, 0, 1, 2, 78, 888, 13909, ... which is A184965, and a(n) = (n-1)!*A184965(n).
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MATHEMATICA
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b[n_, n0_] := Permanent[Table[If[(0 <= j - i && j - i < n - n0) || j - i < -n0, 1, 0], {i, 1, n}, {j, 1, n}]];
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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