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A122670
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If n mod 4 = 2 or n mod 4 = 3 then a(n) = 0 else let m=floor(n/4), then a(n) = (2*m)!/m!.
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2
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1, 1, 0, 0, 2, 2, 0, 0, 12, 12, 0, 0, 120, 120, 0, 0, 1680, 1680, 0, 0, 30240, 30240, 0, 0, 665280, 665280, 0, 0, 17297280, 17297280, 0, 0, 518918400, 518918400, 0, 0, 17643225600, 17643225600, 0, 0, 670442572800, 670442572800, 0, 0, 28158588057600, 28158588057600, 0, 0, 1295295050649600
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OFFSET
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0,5
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COMMENTS
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Number of solutions to the rook problem on an n X n board having a certain symmetry group (see Robinson for details).
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REFERENCES
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R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
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LINKS
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Table of n, a(n) for n=0..48.
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FORMULA
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For asymptotics see the Robinson paper.
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MAPLE
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R:=proc(n) local m; if n mod 4 = 2 or n mod 4 = 3 then RETURN(0); fi; m:=floor(n/4); (2*m)!/m!; end;
For Maple program see A000903.
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CROSSREFS
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If the duplicates and zeros are omitted we get A001813.
Cf. A000898, A000899, A000900, A000901, A000902, A000903.
Sequence in context: A063865 A182107 A037224 * A190389 A176127 A087637
Adjacent sequences: A122667 A122668 A122669 * A122671 A122672 A122673
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane, Sep 23 2006
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STATUS
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approved
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