%I #13 May 23 2014 13:53:48

%S 1,0,6,6,36,88,376,1096,4476,14200,57284,190206,764812,2615268,

%T 10499504,36677626,147110276,522288944

%N Number of balanced orbitals over an odd number of sectors.

%C See A241810 and A232500 for the combinatorial definitions.

%F a(n) = A241810(2*n+1).

%t np[z_]:=Module[{i,j},For[i=Length[z],i>1&&z[[i-1]]>=z[[i]],i--]; For[j=Length[z],z[[j]]<=z[[i-1]],j--]; Join[Take[z,i-2],{z[[j]]}, Reverse[Drop[ReplacePart[z,z[[i-1]],j],i-1]]]]; o=Table[1,{16}];

%t Print[1]; Do[p=Join[-Take[o,n],{0},Take[o,n]]; c=0; Do[If[Accumulate[Accumulate[p]][[-1]]==0,c++]; p=np[p],{(2*n+1)!/(2*n!^2)}]; Print[2*c],{n,16}]

%t (* _Hans Havermann_, May 10 2014 *)

%o (Sage)

%o def A242087(n):

%o if n == 0: return 1

%o A = 0; T = [0]

%o for i in (1..n):

%o T.append(-1); T.append(1)

%o for p in Permutations(T):

%o P = 0; S = 0

%o for k in (0..2*n):

%o P += p[k]; S += P

%o if S == 0: A += 1

%o return A

%o [A242087(n) for n in (0..10)]

%K nonn,more

%O 0,3

%A _Peter Luschny_, May 04 2014

%E More terms from _Hans Havermann_, May 10 2014

%E a(17) from _Hans Havermann_, May 23 2014