A274710 A statistic on orbital systems over n sectors: the number of orbitals which make k turns.

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, 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, 0, 6, 48, 216, 480, 744, 672, 432, 144, 30, 0, 2, 10, 50, 100, 200, 200, 200, 100, 50, 10, 2

Offset: 0

Author

Peter Luschny, Jul 10 2016

Keywords

• nonn
• tabf

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.

A 'turn' of an orbital w takes place where signum(w[i]) is not equal to signum(w[i+1]).

A152659 is a subtriangle.

Examples

			Triangle read by rows, n>=0. The length of row n is n for n>=1.
[n] [k=0,1,2,...]                      [row sum]
[0] [1]                                    1
[1] [1]                                    1
[2] [0, 2]                                 2
[3] [0, 0, 6]                              6
[4] [0, 2, 2,  2]                          6
[5] [0, 0, 6, 12,  12]                    30
[6] [0, 2, 4,  8,   4,   2]               20
[7] [0, 0, 6, 24,  52,  40,  18]         140
[8] [0, 2, 6, 18,  18,  18,   6,  2]      70
[9] [0, 0, 6, 36, 120, 180, 180, 84, 24] 630
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.
		

Links

• Peter Luschny, Orbitals

Crossrefs

Cf. A056040 (row sum), A152659, A232500.

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).

Programs

Sage

# uses[unit_orbitals from A274709]
# Brute force counting
def orbital_turns(n):
    if n == 0: return [1]
    S = [0]*(n)
    for u in unit_orbitals(n):
        L = sum(0 if sgn(u[i]) == sgn(u[i+1]) else 1 for i in (0..n-2))
        S[L] += 1
    return S
for n in (0..12): print(orbital_turns(n))

Formula

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).