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%I #22 Mar 23 2020 17:33:32
%S 1,1,2,5,1,4,2,18,11,1,10,8,2,65,57,17,1,28,28,12,2,238,252,116,23,1,
%T 84,96,54,16,2,882,1050,615,195,29,1,264,330,220,88,20,2,3300,4257,
%U 2915,1210,294,35,1,858,1144,858,416,130,24,2,12441,17017,13013,6461,2093,413,41,1
%N A statistic on orbital systems over n sectors: the number of orbitals with k restarts.
%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 A 'restart' of an orbital is a raise which starts from the central circle.
%C A118920 is a subtriangle.
%H Peter Luschny, <a href="https://oeis.org/wiki/User:Peter_Luschny/Orbitals">Orbitals</a>
%F For even n>0: T(n,k) = 4*(k+1)*binomial(n,n/2-k-1)/n for k=0..n/2-1 (from A118920).
%e Triangle read by rows, n>=0. The length of row n is floor((n+1)/2) for n>=1.
%e [n] [k=0,1,2,...] [row sum]
%e [ 0] [1] 1
%e [ 1] [1] 1
%e [ 2] [2] 2
%e [ 3] [5, 1] 6
%e [ 4] [4, 2] 6
%e [ 5] [18, 11, 1] 30
%e [ 6] [10, 8, 2] 20
%e [ 7] [65, 57, 17, 1] 140
%e [ 8] [28, 28, 12, 2] 70
%e [ 9] [238, 252, 116, 23, 1] 630
%e [10] [84, 96, 54, 16, 2] 252
%e [11] [882, 1050, 615, 195, 29, 1] 2772
%e T(6, 2) = 2 because there are two orbitals over 6 segments which have 2 ascents:
%e [-1, 1, 1, -1, 1, -1] and [1, -1, 1, -1, 1, -1].
%o (Sage) # uses[unit_orbitals from A274709]
%o from itertools import accumulate
%o # Brute force counting
%o def orbital_restart(n):
%o if n == 0: return [1]
%o S = [0]*((n+1)//2)
%o for u in unit_orbitals(n):
%o A = list(accumulate(u))
%o L = [1 if A[i] == 0 and A[i+1] == 1 else 0 for i in (0..n-2)]
%o S[sum(L)] += 1
%o return S
%o for n in (0..12): print(orbital_restart(n))
%Y Cf. A056040 (row sum), A118920, 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), A274881 (ascent).
%K nonn,tabf
%O 0,3
%A _Peter Luschny_, Jul 11 2016
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