A108747 -id:A108747 - cp's OEIS frontend

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

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).

A068551 a(n) = 4^n - binomial(2*n,n).

Original entry on oeis.org

0, 2, 10, 44, 186, 772, 3172, 12952, 52666, 213524, 863820, 3488872, 14073060, 56708264, 228318856, 918624304, 3693886906, 14846262964, 59644341436, 239532643144, 961665098956, 3859788636664, 15488087080696, 62135313450064

Offset: 0

N. J. A. Sloane, Mar 23 2002

Number of rooted two-face n-edge maps in the plane (planar with a distinguished outside face). - Valery A. Liskovets, Mar 17 2005
Total number of returns to the x axis in all lattice paths using steps (1,1) and (1,-1) from the origin to (2n,0). Cf. A108747. - Geoffrey Critzer, Jan 30 2012
Total depth of all leaves in all binary trees on 2n+1 nodes. - Marko Riedel, Sep 10 2016

H. W. Gould, Combinatorial Identities, Morgantown, WV, 1972. p. 32.
Hojoo Lee, Posting to Number Theory List, Feb 18 2002.
V. A. Liskovets and T. R. Walsh, Enumeration of unrooted maps on the plane, Rapport technique, UQAM, No. 2005-01, Montreal, Canada, 2005.

Vincenzo Librandi, Table of n, a(n) for n = 0..175
Dennis E. Davenport, Lara K. Pudwell, Louis W. Shapiro, and Leon C. Woodson, The Boundary of Ordered Trees, Journal of Integer Sequences, Vol. 18 (2015), Article 15.5.8.
Nicolle González, Pamela E. Harris, Gordon Rojas Kirby, Mariana Smit Vega Garcia, and Bridget Eileen Tenner, Pinnacle sets of signed permutations, arXiv:2301.02628 [math.CO], 2023.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Marko R. Riedel, Average depth of a leaf in a binary tree, Math.Stackexchange.com.
V. A. Liskovets and T. R. Walsh, Counting unrooted maps on the plane, Advances in Applied Math., 36(4) (2006), 364-387.

Cf. A000108, A000346, A000984, A005470, A045621.

[4^n - Binomial(2*n,n): n in [0..35]]; // Vincenzo Librandi, Jun 07 2011

A068551:=n->4^n - binomial(2*n,n): seq(A068551(n), n=0..30); # Wesley Ivan Hurt, Mar 22 2014

nn=20;c=(1-(1-4x)^(1/2))/(2x); D[CoefficientList[ Series[ 1/(1-2y x c), {x,0,nn}], x], y]/.y->1 (* Geoffrey Critzer, Jan 30 2012 *)

PARI
```
a(n)=if(n<0,0,4^n-binomial(2*n,n))
```

x='x+O('x^100); concat(0, Vec(1/(1-4*x)-1/sqrt(1-4*x))) \\ Altug Alkan, Dec 29 2015

G.f.: 1/(1 - 4*x) - 1/sqrt(1 - 4*x) = C(x)*2*x/(1 - 4*x) where C(x) = g.f. for Catalan numbers A000108.
a(n) = Sum_{k >= 1} binomial(2*m-2*k, m-k) * binomial(2*k, k).
a(n+1) = 4*a(n) + 2*C(n), where C(n) = Catalan numbers.
a(n) = 2*A000346(n-1) for n > 0.
a(n) = A045621(2*n).
Conjecture: n*a(n) + 2*(3-4*n)*a(n-1) + 8*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Apr 01 2012
Recurrence (an alternative): n*a(n) = 2^9*(2*n - 9)*a(n-5) + 2^8*(18 - 5*n)*a(n-4) + 2^6*(10*n - 27)*a(n-3) + 2^5*(9 - 5*n)*a(n-2) + 2*(10*n - 9)*a(n-1), n >= 5. - Fung Lam, Mar 22 2014
Asymptotics: a(n) ~ 2^(2*n)*(1 - 1/sqrt(n*Pi)). - Fung Lam, Mar 22 2014
E.g.f.: (exp(2*x) - BesselI(0, 2*x))*exp(2*x). - Ilya Gutkovskiy, Sep 10 2016
a(n) = (-1)^(n+1)*binomial(-n, n + 1)*hypergeom([1, 2*n + 1], [n + 2], 1/2). - Peter Luschny, Nov 29 2023

cp's OEIS Frontend

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

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A068551 a(n) = 4^n - binomial(2*n,n).

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