A241543 -id:A241543 - cp's OEIS frontend

A241477 Triangle read by rows, number of orbitals classified with respect to the first zero crossing, n>=1, 1<=k<=n.

1, 0, 2, 2, 2, 2, 0, 4, 0, 2, 6, 12, 4, 2, 6, 0, 12, 0, 4, 0, 4, 20, 60, 12, 12, 12, 4, 20, 0, 40, 0, 12, 0, 8, 0, 10, 70, 280, 40, 60, 36, 24, 40, 10, 70, 0, 140, 0, 40, 0, 24, 0, 20, 0, 28, 252, 1260, 140, 280, 120, 120, 120, 60, 140, 28, 252, 0, 504, 0

Offset: 1

Peter Luschny, Apr 23 2014

For the combinatorial definitions see A232500. An orbital w over n sectors has its first zero crossing at k if k is the smallest j such that the partial sum(1<=i<=j, w(i))) = 0, where w(i) are the jumps of the orbital represented by -1, 0, 1.

			[1], [ 1]
[2], [ 0,  2]
[3], [ 2,  2,  2]
[4], [ 0,  4,  0,  2]
[5], [ 6, 12,  4,  2,  6]
[6], [ 0, 12,  0,  4,  0, 4]
[7], [20, 60, 12, 12, 12, 4, 20]

Row sums: A056040.

Cf. A232500.

A241477 := proc(n, k)
  if n = 0 then 1
elif k = 0 then 0
elif irem(n, 2) = 0 and irem(k, 2) = 1 then 0
elif k = 1 then (n-1)!/iquo(n-1,2)!^2
else 2*(n-k)!*(k-2)!/iquo(k,2)/(iquo(k-2,2)!*iquo(n-k,2)!)^2
  fi end:
for n from 1 to 9 do seq(A241477(n, k), k=1..n) od;

T[n_, k_] := Which[n == 0, 1, k == 0, 0, Mod[n, 2] == 0 && Mod[k, 2] == 1,  0, k == 1, (n-1)!/Quotient[n-1, 2]!^2, True, 2*(n-k)!*(k-2)!/Quotient[k, 2]/(Quotient[k-2, 2]!*Quotient[n-k, 2]!)^2];
Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 20 2018, from Maple *)

def A241477_row(n):
    if n == 0: return [1]
    Z = [0]*n; T = [0] if is_odd(n) else []
    for i in (1..n//2): T.append(-1); T.append(1)
    for p in Permutations(T):
        i = 0; s = p[0]
        while s != 0: i += 1; s += p[i];
        Z[i] += 1
    return Z
for n in (1..9): A241477_row(n)

If n is even and k is odd then T(n, k) = 0 else if k = 1 then T(n, 1) = A056040(n-1) else T(n, k) = 2*A057977(k-2)*A056040(n-k).
T(n, n) = A241543(n).
T(n+1, 1) = A126869(n).
T(2*n, 2*n) = |A002420(n)|.
T(2*n+1, 1) = A000984(n).
T(2*n+1, n+1) = A241530(n).
T(2*n+2, 2) = A028329(n).
T(4*n, 2*n) = |A010370(n)|.
T(4*n, 4*n) = |A024491(n)|.
T(4*n+1, 1) = A001448(n).
T(4*n+1, 2*n+1) = A002894(n).

A274879 A statistic on orbital systems over n sectors: the number of orbitals with k returns.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 4, 6, 12, 12, 4, 8, 8, 20, 40, 48, 32, 10, 20, 24, 16, 70, 140, 180, 160, 80, 28, 56, 72, 64, 32, 252, 504, 672, 672, 480, 192, 84, 168, 224, 224, 160, 64, 924, 1848, 2520, 2688, 2240, 1344, 448, 264, 528, 720, 768, 640, 384, 128

Offset: 0

Peter Luschny, Jul 11 2016

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.
When a segment of an orbital starts at a point on the central circle this point is called a 'return' of the orbital if it is not the origin.
If an orbital touches the central circle only in the origin it is called a prime orbital. Column 0 counts the prime orbitals over n sectors.
A108747 is a subtriangle.

			Triangle read by rows, n>=0. The length of row n is floor((n+1)/2) for n>=1.
[ n] [k=0,1,2,...] [row sum]
[ 0] [1] 1
[ 1] [1] 1
[ 2] [2] 2
[ 3] [2, 4] 6
[ 4] [2, 4] 6
[ 5] [6, 12, 12] 30
[ 6] [4, 8, 8] 20
[ 7] [20, 40, 48, 32] 140
[ 8] [10, 20, 24, 16] 70
[ 9] [70, 140, 180, 160, 80] 630
[10] [28, 56, 72, 64, 32] 252
[11] [252, 504, 672, 672, 480, 192] 2772
T(6,0) = 4 because the following 4 orbitals stay above or below the central
circle: [-1, -1, -1, 1, 1, 1], [-1, -1, 1, -1, 1, 1], [1, 1, -1, 1, -1, -1],
[1, 1, 1, -1, -1, -1].

Peter Luschny, Orbitals

Cf. A056040 (row sum), A108747, A232500, A241543 (col. 0).

Other orbital statistics: A241477 (first zero crossing), A274706 (absolute integral), A274708 (peaks), A274709 (max. height), A274710 (number of turns), A274878 (span), A274880 (restarts), A274881 (ascent).

# uses[unit_orbitals from A274709]
from itertools import accumulate
# Brute force counting
def orbital_returns(n):
    if n == 0: return [1]
    S = [0]*((n+1)//2)
    for u in unit_orbitals(n):
        L = list(accumulate(u))
        Z = len(list(filter(lambda z: z == 0, L)))
        S[Z-1] += 1  # exclude origin
    return S
for n in (0..10): print(orbital_returns(n))

For even n>0: T(n,k) = 2^(k+1)*(k+1)*binomial(n-k-1,n/2)/(n-k-1) for k=0..n/2-1 (from A108747).

A275325 Triangle read by rows: number of orbitals over n sectors which have a Catalan decomposition into k parts.

Original entry on oeis.org

1, 0, 1, 0, 2, 0, 6, 0, 4, 2, 0, 20, 10, 0, 10, 8, 2, 0, 70, 56, 14, 0, 28, 28, 12, 2, 0, 252, 252, 108, 18, 0, 84, 96, 54, 16, 2, 0, 924, 1056, 594, 176, 22, 0, 264, 330, 220, 88, 20, 2, 0, 3432, 4290, 2860, 1144, 260, 26, 0, 858, 1144, 858, 416, 130, 24, 2

Offset: 0

Peter Luschny, Aug 15 2016

The definition of an orbital system is given in A232500.
The Catalan decomposition of an orbital w is a list of orbitals which are alternately entirely above or below the main circle ('above' and 'below' in the weak sense) such that their concatenation equals w. If a zero is on the border of two orbitals then it is allocated to the first one. By convention T(0,0) = 1.
The number of orbitals over n sectors is counted by the swinging factorial A056040.

			Table starts:
[ n] [k=0,1,2,...] [row sum]
[ 0] [1] 1
[ 1] [0, 1] 1
[ 2] [0, 2] 2
[ 3] [0, 6] 6
[ 4] [0, 4, 2] 6
[ 5] [0, 20, 10] 30
[ 6] [0, 10, 8, 2] 20
[ 7] [0, 70, 56, 14] 140
[ 8] [0, 28, 28, 12, 2] 70
[ 9] [0, 252, 252, 108, 18] 630
[10] [0, 84, 96, 54, 16, 2] 252
[11] [0, 924, 1056, 594, 176,  22] 2772
[12] [0, 264, 330, 220, 88, 20, 2] 924
For example T(2*n, n) = 2 counts the Catalan decompositions
[[-1, 1], [1, -1], [-1, 1], ..., [(-1)^n, (-1)^(n+1)]] and
[[1, -1], [-1, 1], [1, -1], ..., [(-1)^(n+1), (-1)^n]].

Peter Luschny, Orbitals

Cf. A056040 (row sum), A241543, A232500, A266722, A275326.

Other orbital statistics: A241477 (first zero crossing), A274706 (absolute integral), A274709 (max. height), A274710 (number of turns), A274878 (span), A274879 (returns), A274880 (restarts), A274881 (ascent).

# uses[unit_orbitals from A274709]
# Brute force counting
from itertools import accumulate
def catalan_factors(P):
    def bisect(orb):
        i = 1
        A = list(accumulate(orb))
        if orb[1] > 0 if orb[0] == 0 else orb[0] > 0:
            while i < len(A) and A[i] >= 0: i += 1
        else:
            while i < len(A) and A[i] <= 0: i += 1
        return i
    R = []
    while P:
        i = bisect(P)
        R.append(P[:i])
        P = P[i:]
    return R
def orbital_factors(n):
    if n == 0: return [1]
    if n == 1: return [0, 1]
    S = [0]*(n//2 + 1)
    for o in unit_orbitals(n):
        S[len(catalan_factors(o))] += 1
    return S
for n in (0..9): print(orbital_factors(n))

T(n,1) = 2*floor((n+2)/2)*n!/floor((n+2)/2)!^2 = A241543(n+2) for n>=2.
For odd n>1 T(n,1) = Sum_{k>=0} T(n+1,k).
A056040(n) - T(n,1) = A232500(n) for n>=2.
Main diagonal: T(n, floor(n/2)) = A266722(n) for n>1.
A275326(n,k) = ceiling(T(n,k)/2).

cp's OEIS Frontend

A241477 Triangle read by rows, number of orbitals classified with respect to the first zero crossing, n>=1, 1<=k<=n.

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A274879 A statistic on orbital systems over n sectors: the number of orbitals with k returns.

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A275325 Triangle read by rows: number of orbitals over n sectors which have a Catalan decomposition into k parts.

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