login
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

%I #16 Dec 30 2023 17:49:20

%S 1,1,0,0,2,2,0,0,12,12,0,0,120,120,0,0,1680,1680,0,0,30240,30240,0,0,

%T 665280,665280,0,0,17297280,17297280,0,0,518918400,518918400,0,0,

%U 17643225600,17643225600,0,0,670442572800,670442572800,0,0,28158588057600,28158588057600,0,0,1295295050649600

%N 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!.

%C Number of solutions to the rook problem on an n X n board having a certain symmetry group (see Robinson for details).

%C A037224 is an essentially identical sequence.

%D R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).

%F For asymptotics see the Robinson paper.

%F 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

%p 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;

%p For Maple program see A000903.

%t Table[If[MemberQ[{2,3},Mod[n,4]],0,((2Floor[n/4])!/Floor[n/4]!)],{n,0,50}] (* _Harvey P. Dale_, Dec 30 2023 *)

%Y If the duplicates and zeros are omitted we get A001813.

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

%K nonn

%O 0,5

%A _N. J. A. Sloane_, Sep 23 2006