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

0,5

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