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A274881
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A statistic on orbital systems over n sectors: the number of orbitals which have an ascent of length k.
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10
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1, 1, 0, 2, 0, 6, 0, 3, 3, 0, 18, 12, 0, 4, 12, 4, 0, 40, 80, 20, 0, 5, 40, 20, 5, 0, 75, 375, 150, 30, 0, 6, 120, 90, 30, 6, 0, 126, 1470, 882, 252, 42, 0, 7, 350, 371, 147, 42, 7, 0, 196, 5292, 4508, 1568, 392, 56, 0, 8, 1008, 1456, 672, 224, 56, 8
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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 ascent of an orbital is its longest up-run.
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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, 3, 3] 6
[ 5] [0, 18, 12] 30
[ 6] [0, 4, 12, 4] 20
[ 7] [0, 40, 80, 20] 140
[ 8] [0, 5, 40, 20, 5] 70
[ 9] [0, 75, 375, 150, 30] 630
[10] [0, 6, 120, 90, 30, 6] 252
[11] [0, 126, 1470, 882, 252, 42] 2772
[12] [0, 7, 350, 371, 147, 42, 7] 924
T(6,3) = 4 because four orbitals over six sectors have a maximal up-run of length 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].
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PROG
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(Sage) # uses[unit_orbitals from A274709]
# Brute force counting
def orbital_ascent(n):
if n < 2: return [1]
S = [0]*((n+2)//2)
for u in unit_orbitals(n):
B = [0]*n
for i in (0..n-1):
B[i] = 0 if u[i] <= 0 else B[i-1] + u[i]
S[max(B)] += 1
return S
for n in (0..12): print(orbital_ascent(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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