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!.

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

Offset: 0

Author

N. J. A. Sloane, Sep 23 2006

Keywords

• nonn

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).

Crossrefs

If the duplicates and zeros are omitted we get A001813.

Cf. A000898, A000899, A000900, A000901, A000902, A000903, A037224.

Programs

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 *)

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