A185113 Number of dissections of a convex (3n+3)-sided polygon into n pentagons and one triangle (up to equivalence).

1, 3, 18, 130, 1020, 8379, 70840, 610740, 5340060, 47187580, 420412278, 3770221338, 33991902308, 307826695050, 2798052616800, 25514463687720, 233296537299228, 2138295980859588, 19639886707062280, 180724535020583400, 1665767679910654320, 15376467276901980315

Offset: 0

Author

F. Chapoton, Feb 03 2011

Keywords

• nonn

Comments

This sequence counts dissections of a convex 3n+3-sided polygon into one triangle and n pentagons, modulo a simple equivalence relation. This equivalence relation is defined by moving the triangle according to a simple rule (not detailed here).

(The equivalence relation is not defined by a group, but by local moves. Consider the hexagon formed by a pentagon adjacent to the triangle. The local move is half-rotation of such hexagons.)

The terms seem to be odd exactly for indices in A002450. - F. Chapoton Mar 08 2020

Examples

			For n=0, there is just one triangle, so that a(0)=1. For n=1, one can dissect an hexagon in 6 ways into a pentagon and a triangle. In this case, the equivalence relation just relates every such dissection to its half rotated image, so that a(1)=3.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A001700, A002293, A174687, A069271, A000217, A000260.

Programs

Maple

f := RootOf(81*x - 8 - (75*x - 8)*Z + (288*x^2 - 30*x)*_Z^2 + (256*x^3 - 27*x^2)*_Z^4): seq(coeff(series(f, x, 20), x, n), n = 0..19); # _Peter Luschny, Apr 06 2023

Mathematica

Table[Binomial[4*n + 1, n]*(n + 2)/(3*n + 2), {n, 0, 50}] (* G. C. Greubel, Jun 23 2017 *)

PARI

for(n=0,25, print1(binomial(4*n+1,n)*(n+2)/(3*n+2), ", ")) \\ G. C. Greubel, Jun 23 2017

Sage

def A185113(n):
    return binomial(4*n+1, n) * (n+2) / (3*n+2)

Formula

a(n) = binomial(4*n+1,n-1)*(n+2)/n = binomial(4*n+1,n)*(n+2)/(3*n+2).

a(n) = binomial(n+2,2) * A000260(n). - F. Chapoton Feb 22 2024