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 A122670 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!. 2
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS Number of solutions to the rook problem on an n X n board having a certain symmetry group (see Robinson for details). REFERENCES R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976). LINKS FORMULA For asymptotics see the Robinson paper. MAPLE 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. CROSSREFS 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 KEYWORD nonn AUTHOR N. J. A. Sloane, Sep 23 2006 STATUS approved

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