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A274878
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A statistic on orbital systems over n sectors: the number of orbitals with span k.
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10
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1, 1, 0, 2, 0, 6, 0, 2, 4, 0, 10, 20, 0, 2, 12, 6, 0, 14, 84, 42, 0, 2, 28, 32, 8, 0, 18, 252, 288, 72, 0, 2, 60, 120, 60, 10, 0, 22, 660, 1320, 660, 110, 0, 2, 124, 390, 300, 96, 12, 0, 26, 1612, 5070, 3900, 1248, 156, 0, 2, 252, 1176, 1260, 588, 140, 14
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OFFSET
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0,4
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COMMENTS
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The definition of an orbital system is given in A232500 (see also the illustration there). The number of orbitals over n sectors is counted by the swinging factorial A056040.
The 'span' of an orbital w is the difference between the highest and the lowest level of the orbital system touched by w.
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LINKS
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EXAMPLE
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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] [0, 2] 2
[ 3] [0, 6] 6
[ 4] [0, 2, 4] 6
[ 5] [0, 10, 20] 30
[ 6] [0, 2, 12, 6] 20
[ 7] [0, 14, 84, 42] 140
[ 8] [0, 2, 28, 32, 8] 70
[ 9] [0, 18, 252, 288, 72] 630
[10] [0, 2, 60, 120, 60, 10] 252
T(6, 3) = 6 because the span of the following six orbitals is 3:
[-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], [1, 1, 1, -1, -1, -1].
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PROG
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(Sage) # uses[unit_orbitals from A274709]
from itertools import accumulate
# Brute force counting.
def orbital_span(n):
if n == 0: return [1]
S = [0]*((n+2)//2)
for u in unit_orbitals(n):
L = list(accumulate(u))
S[max(L) - min(L)] += 1
return S
for n in (0..10): print(orbital_span(n))
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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