A274706 Irregular triangle read by rows. T(n,k) (n >= 0) is a statistic on orbital systems over n sectors: the number of orbitals which have an integral whose absolute value is k.
1, 1, 0, 2, 0, 4, 2, 2, 0, 2, 0, 2, 6, 4, 6, 4, 4, 4, 2, 0, 6, 0, 6, 0, 4, 0, 2, 0, 2, 6, 24, 16, 20, 14, 16, 12, 8, 6, 8, 4, 4, 2, 8, 0, 14, 0, 14, 0, 10, 0, 10, 0, 6, 0, 4, 0, 2, 0, 2, 36, 52, 68, 48, 64, 48, 48, 40, 44, 32, 36, 24, 22, 16, 16, 8, 10, 8, 4, 4, 2
Offset: 0
Examples
The length of row n is 1+floor(n^2//4). The triangle begins: [n] [k=0,1,2,...] [row sum] [0] [1] 1 [1] [1] 1 [2] [0, 2] 2 [3] [0, 4, 2] 6 [4] [2, 0, 2, 0, 2] 6 [5] [6, 4, 6, 4, 4, 4, 2] 30 [6] [0, 6, 0, 6, 0, 4, 0, 2, 0, 2] 20 [7] [6, 24, 16, 20, 14, 16, 12, 8, 6, 8, 4, 4, 2] 140 [8] [8, 0, 14, 0, 14, 0, 10, 0, 10, 0, 6, 0, 4, 0, 2, 0, 2] 70 T(5, 4) = 4 because the integral of four orbitals have the absolute value 4: Integral([-1, -1, 1, 1, 0]) = -4, Integral([0, -1, -1, 1, 1]) = -4, Integral([0, 1, 1, -1, -1]) = 4, Integral([1, 1, -1, -1, 0]) = 4.
Links
- Peter Luschny, Orbitals
Crossrefs
Programs
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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 orbital_integral(n): if n == 0: return [1] S = [0]*(1+floor(n^2//4)) for u in unit_orbitals(n): L = list(accumulate(accumulate(u))) S[abs(L[-1])] += 1 return S for n in (0..8): print(orbital_integral(n))
Comments