A001762 Number of labeled n-vertex dissections of a ball.

1, 1, 10, 180, 4620, 152880, 6168960, 293025600, 15990004800, 984647664000, 67493121696000, 5094263446272000, 419688934689024000, 37465564582397952000, 3601861863990534144000, 370962724717928318976000, 40744403224500159055872000

Offset: 3

Author

N. J. A. Sloane

Keywords

• nonn

Comments

This is the number of labeled Apollonian networks (planar 3-trees). - Allan Bickle, Feb 20 2024

Examples

			There is one maximal planar graph with 4 vertices, and one way to label it, so a(4) = 1.
		

References

• L. W. Beineke and R. E. Pippert, Enumerating labeled k-dimensional trees and ball dissections, pp. 12-26 of Proceedings of Second Chapel Hill Conference on Combinatorial Mathematics and Its Applications, University of North Carolina, Chapel Hill, 1970. Reprinted in Math. Annalen, 191 (1971), 87-98.
• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• T. D. Noe, Table of n, a(n) for n = 3..100
• L. W. Beineke and R. E. Pippert, The Number of Labeled Dissections of a k-Ball, Math. Annalen, 191 (1971), 87-98.
• Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.

Crossrefs

Cf. A000407, A001763, A001764, A027610.

Programs

Julia

using Combinatorics
a(n) = n < 4 ? 1 : binomial(BigInt(n),3)*factorial(BigInt(3*n-9))÷factorial(BigInt(2*n-4))
print([a(n) for n in 3:28]) # Paul Muljadi, Mar 27 2024

Mathematica

Join[{1}, Table[Binomial[n, 3]*(3*n - 9)!/(2*n - 4)!, {n, 4, 25}]] (* T. D. Noe, Aug 10 2012 *)

Python

from math import factorial
from sympy import binomial
def a(n):
    if n < 4:
        return 1
    else:
        return binomial(n, 3) * factorial(3*n-9) // factorial(2*n-4)
print([a(n) for n in range(3, 21)]) # Paul Muljadi, Mar 05 2024

Formula

a(n) = binomial(n,3)*(3*n-9)!/(2*n-4)!, n >= 4; a(3) = 1.

a(n) ~ 3^(3*n - 19/2) * n^(n-2) / (2^(2*n - 5/2) * exp(n)). - Vaclav Kotesovec, Mar 14 2024

Extensions

More terms from Wolfdieter Lang

Name clarified by Andrey Zabolotskiy, Mar 15 2024