%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
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