A274709 A statistic on orbital systems over n sectors: the number of orbitals which rise to maximum height k over the central circle.
1, 1, 1, 1, 3, 3, 2, 3, 1, 10, 15, 5, 5, 9, 5, 1, 35, 63, 35, 7, 14, 28, 20, 7, 1, 126, 252, 180, 63, 9, 42, 90, 75, 35, 9, 1, 462, 990, 825, 385, 99, 11, 132, 297, 275, 154, 54, 11, 1, 1716, 3861, 3575, 2002, 702, 143, 13, 429, 1001, 1001, 637, 273, 77, 13, 1
Offset: 0
Examples
Triangle read by rows, n>=0. The length of row n is floor((n+2)/2). [ n] [k=0,1,2,...] [row sum] [ 0] [ 1] 1 [ 1] [ 1] 1 [ 2] [ 1, 1] 2 [ 3] [ 3, 3] 6 [ 4] [ 2, 3, 1] 6 [ 5] [ 10, 15, 5] 30 [ 6] [ 5, 9, 5, 1] 20 [ 7] [ 35, 63, 35, 7] 140 [ 8] [ 14, 28, 20, 7, 1] 70 [ 9] [126, 252, 180, 63, 9] 630 [10] [ 42, 90, 75, 35, 9, 1] 252 [11] [462, 990, 825, 385, 99, 11] 2772 [12] [132, 297, 275, 154, 54, 11, 1] 924 T(6, 2) = 5 because the five orbitals [-1, 1, 1, 1, -1, -1], [1, -1, 1, 1, -1, -1], [1, 1, -1, -1, -1, 1], [1, 1, -1, -1, 1, -1], [1, 1, -1, 1, -1, -1] raise to maximal height of 2 over the central circle.
Links
- Peter Luschny, The lost Catalan numbers
- Peter Luschny, Orbitals
Crossrefs
Programs
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Maple
S := proc(n,k) option remember; `if`(k>n or k<0, 0, `if`(n=k, 1, S(n-1,k-1)+ modp(n-k,2)*S(n-1,k)+S(n-1,k+1))) end: T := (n,k) -> S(n,2*k); seq(print(seq(T(n,k), k=0..iquo(n,2))), n=0..12);
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Sage
from itertools import accumulate # Brute force counting def unit_orbitals(n): sym_range = [i for i in range(-n+1, n, 2)] for c in Combinations(sym_range, n): P = Permutations([sgn(v) for v in c]) for p in P: yield p def max_orbitals(n): if n == 0: return [1] S = [0]*((n+2)//2) for u in unit_orbitals(n): L = list(accumulate(u)) S[max(L)] += 1 return S for n in (0..10): print(max_orbitals(n))
Comments