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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!. 3
 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). A037224 is an essentially identical sequence. REFERENCES R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976). LINKS Table of n, a(n) for n=0..48. FORMULA For asymptotics see the Robinson paper. a(n) = (1/2 + (-1)^(n/2 - 1/4 + (-1)^n/4)/2) * ((n/2 - 3/4 + (-1)^n/4 + (-1)^(n/2 - 1/4 + (-1)^n/4)/2)! / ((n/4 - 3/8 + (-1)^n/8 + (-1)^(n/2 - 1/4 + (-1)^n/4)/4)!)). - Wesley Ivan Hurt, Mar 30 2015 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. MATHEMATICA Table[If[MemberQ[{2, 3}, Mod[n, 4]], 0, ((2Floor[n/4])!/Floor[n/4]!)], {n, 0, 50}] (* Harvey P. Dale, Dec 30 2023 *) CROSSREFS If the duplicates and zeros are omitted we get A001813. Cf. A000898, A000899, A000900, A000901, A000902, A000903, A037224. Sequence in context: A182107 A337999 A037224 * A352661 A283494 A353254 Adjacent sequences: A122667 A122668 A122669 * A122671 A122672 A122673 KEYWORD nonn AUTHOR N. J. A. Sloane, Sep 23 2006 STATUS approved

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Last modified September 8 05:26 EDT 2024. Contains 375751 sequences. (Running on oeis4.)