A241810 Number of balanced orbitals over n sectors.

1, 1, 0, 0, 2, 6, 0, 6, 8, 36, 0, 88, 58, 376, 0, 1096, 526, 4476, 0, 14200, 5448, 57284, 0, 190206, 61108, 764812, 0, 2615268, 723354, 10499504, 0, 36677626, 8908546, 147110276, 0, 522288944, 113093022

Offset

0,5

Comments

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.

Links

Table of n, a(n) for n=0..36.

Formula

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

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

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

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

Mathematica

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}];
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]
(* Hans Havermann, May 10 2014 *)

Prog

(Sage)
def A241810(n):
    if n == 0: return 1
    A = 0
    T = [0] if is_odd(n) else []
    for i in (1..n//2):
        T.append(-1); T.append(1)
    for p in Permutations(T):
        P = 0; S = 0
        for k in (0..n-1):
            P += p[k]; S += P
        if S == 0: A += 1
    return A
[A241810(n) for n in (0..32)]

Crossrefs

Cf. A232500, A242087.

Sequence in context: A327280 A350256 A345208 * A156991 A229586 A294789

Adjacent sequences: A241807 A241808 A241809 * A241811 A241812 A241813

Keyword

nonn,more

Author

Peter Luschny, Apr 29 2014

Extensions

More terms from Hans Havermann, May 10 2014

a(35), a(36) from Hans Havermann, May 23 2014

Status

approved