

A274881


A statistic on orbital systems over n sectors: the number of orbitals which have an ascent of length k.


10



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

0,4


COMMENTS

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 uprun.


LINKS

Table of n, a(n) for n=0..63.
Peter Luschny, Orbitals


EXAMPLE

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 uprun 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].


PROG

(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..n1):
B[i] = 0 if u[i] <= 0 else B[i1] + u[i]
S[max(B)] += 1
return S
for n in (0..12): print(orbital_ascent(n))


CROSSREFS

Cf. A056040 (row sum), A232500.
Other orbital statistics: A241477 (first zero crossing), A274706 (absolute integral), A274708 (peaks), A274709 (max. height), A274710 (number of turns), A274878 (span), A274879 (returns), A274880 (restarts).
Sequence in context: A274878 A050821 A076257 * A303638 A162974 A275325
Adjacent sequences: A274878 A274879 A274880 * A274882 A274883 A274884


KEYWORD

nonn,tabf


AUTHOR

Peter Luschny, Jul 12 2016


STATUS

approved



