|
%I #26 Mar 27 2020 06:57:42
%S 1,1,0,2,0,0,6,0,2,2,2,0,0,6,12,12,0,2,4,8,4,2,0,0,6,24,52,40,18,0,2,
%T 6,18,18,18,6,2,0,0,6,36,120,180,180,84,24,0,2,8,32,48,72,48,32,8,2,0,
%U 0,6,48,216,480,744,672,432,144,30,0,2,10,50,100,200,200,200,100,50,10,2
%N A statistic on orbital systems over n sectors: the number of orbitals which make k turns.
%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 'turn' of an orbital w takes place where signum(w[i]) is not equal to signum(w[i+1]).
%C A152659 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) = 2*C(n/2-1,(k-1+mod(k-1,2))/2)*C(n/2-1,(k-1-mod(k-1,2))/2) for k=0..n-1 (from A152659).
%e Triangle read by rows, n>=0. The length of row n is n for n>=1.
%e [n] [k=0,1,2,...] [row sum]
%e [0] [1] 1
%e [1] [1] 1
%e [2] [0, 2] 2
%e [3] [0, 0, 6] 6
%e [4] [0, 2, 2, 2] 6
%e [5] [0, 0, 6, 12, 12] 30
%e [6] [0, 2, 4, 8, 4, 2] 20
%e [7] [0, 0, 6, 24, 52, 40, 18] 140
%e [8] [0, 2, 6, 18, 18, 18, 6, 2] 70
%e [9] [0, 0, 6, 36, 120, 180, 180, 84, 24] 630
%e T(5,2) = 6 because the six orbitals [-1, -1, 0, 1, 1], [-1, -1, 1, 1, 0], [0, -1, -1, 1, 1], [0, 1, 1, -1, -1], [1, 1, -1, -1, 0], [1, 1, 0, -1, -1] make 2 turns.
%o (Sage) # uses[unit_orbitals from A274709]
%o # Brute force counting
%o def orbital_turns(n):
%o if n == 0: return [1]
%o S = [0]*(n)
%o for u in unit_orbitals(n):
%o L = sum(0 if sgn(u[i]) == sgn(u[i+1]) else 1 for i in (0..n-2))
%o S[L] += 1
%o return S
%o for n in (0..12): print(orbital_turns(n))
%Y Cf. A056040 (row sum), A152659, A232500.
%Y Other orbital statistics: A241477 (first zero crossing), A274706 (absolute integral), A274708 (number of peaks), A274709 (max. height), A274878 (span), A274879 (returns), A274880 (restarts), A274881 (ascent).
%K nonn,tabf
%O 0,4
%A _Peter Luschny_, Jul 10 2016
|
