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%I #18 Mar 23 2020 12:09:58
%S 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,
%T 150,30,0,6,120,90,30,6,0,126,1470,882,252,42,0,7,350,371,147,42,7,0,
%U 196,5292,4508,1568,392,56,0,8,1008,1456,672,224,56,8
%N A statistic on orbital systems over n sectors: the number of orbitals which have an ascent of length k.
%C 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.
%C The ascent of an orbital is its longest up-run.
%H Peter Luschny, <a href="https://oeis.org/wiki/User:Peter_Luschny/Orbitals">Orbitals</a>
%e Triangle read by rows, n>=0. The length of row n is floor((n+2)/2).
%e [ n] [k=0,1,2,...] [row sum]
%e [ 0] [1] 1
%e [ 1] [1] 1
%e [ 2] [0, 2] 2
%e [ 3] [0, 6] 6
%e [ 4] [0, 3, 3] 6
%e [ 5] [0, 18, 12] 30
%e [ 6] [0, 4, 12, 4] 20
%e [ 7] [0, 40, 80, 20] 140
%e [ 8] [0, 5, 40, 20, 5] 70
%e [ 9] [0, 75, 375, 150, 30] 630
%e [10] [0, 6, 120, 90, 30, 6] 252
%e [11] [0, 126, 1470, 882, 252, 42] 2772
%e [12] [0, 7, 350, 371, 147, 42, 7] 924
%e T(6,3) = 4 because four orbitals over six sectors have a maximal up-run of length 3.
%e [-1,-1,-1,1,1,1], [-1,-1,1,1,1,-1], [-1,1,1,1,-1,-1], [1,1,1,-1,-1,-1].
%o (Sage) # uses[unit_orbitals from A274709]
%o # Brute force counting
%o def orbital_ascent(n):
%o if n < 2: return [1]
%o S = [0]*((n+2)//2)
%o for u in unit_orbitals(n):
%o B = [0]*n
%o for i in (0..n-1):
%o B[i] = 0 if u[i] <= 0 else B[i-1] + u[i]
%o S[max(B)] += 1
%o return S
%o for n in (0..12): print(orbital_ascent(n))
%Y Cf. A056040 (row sum), A232500.
%Y 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).
%K nonn,tabf
%O 0,4
%A _Peter Luschny_, Jul 12 2016
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