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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!.

%I A122670 #16 Dec 30 2023 17:49:20
%S A122670 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 A122670 665280,665280,0,0,17297280,17297280,0,0,518918400,518918400,0,0,
%U A122670 17643225600,17643225600,0,0,670442572800,670442572800,0,0,28158588057600,28158588057600,0,0,1295295050649600
%N 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!.
%C A122670 Number of solutions to the rook problem on an n X n board having a certain symmetry group (see Robinson for details).
%C A122670 A037224 is an essentially identical sequence.
%D A122670 R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
%F A122670 For asymptotics see the Robinson paper.
%F A122670 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 A122670 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 A122670 For Maple program see A000903.
%t A122670 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 A122670 If the duplicates and zeros are omitted we get A001813.
%Y A122670 Cf. A000898, A000899, A000900, A000901, A000902, A000903, A037224.
%K A122670 nonn
%O A122670 0,5
%A A122670 _N. J. A. Sloane_, Sep 23 2006