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A241810 Number of balanced orbitals over n sectors. 3

%I #13 May 23 2014 13:54:00

%S 1,1,0,0,2,6,0,6,8,36,0,88,58,376,0,1096,526,4476,0,14200,5448,57284,

%T 0,190206,61108,764812,0,2615268,723354,10499504,0,36677626,8908546,

%U 147110276,0,522288944,113093022

%N Number of balanced orbitals over n sectors.

%C For the combinatorial definitions see A232500. An orbital is balanced if its integral is 0. The integral of an orbital w over n sectors is sum(1<=k<=n, sum(1<=i<=k, w(i))) where w(i) are the jumps of the orbital represented by -1, 0, 1.

%F a(2*n) = A204459(2, n).

%F a(2*n+1) = A242087(n).

%F a(4*n) = A063074(n) = A029895(2*n) = A067059(2*n, 2*n).

%F a(4*n+2) = 0 for all n (proved by H. Havermann).

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

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

%o (Sage)

%o def A241810(n):

%o if n == 0: return 1

%o A = 0

%o T = [0] if is_odd(n) else []

%o for i in (1..n//2):

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

%o for p in Permutations(T):

%o P = 0; S = 0

%o for k in (0..n-1):

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

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

%o return A

%o [A241810(n) for n in (0..32)]

%Y Cf. A232500, A242087.

%K nonn,more

%O 0,5

%A _Peter Luschny_, Apr 29 2014

%E More terms from _Hans Havermann_, May 10 2014

%E a(35), a(36) from _Hans Havermann_, May 23 2014

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