A352611 a(n) is the number of different ways to partition the set of vertices of a convex n-gon into 5 polygons.

1401400, 28028000, 333533200, 3073270200, 24234675465, 172096749825, 1134040872965, 7069307049805, 42240545297951, 244205509154607, 1375458924105651, 7586883537988755, 41147137237012950, 220107145169421510, 1164186829638102270, 6100518487069916910

Offset: 15

Author

Janaka Rodrigo, Mar 23 2022

Keywords

• nonn
• easy

Examples

			For n=17, the set of vertices of a convex 17-gon can be partitioned into 5 polygons in 333533200 different ways:
- as 4 triangles and one pentagon ((1/4!)*C(17,3)*C(14,3)*C(11,3)*C(8,3)*C(5,5) = 95295200 different ways) or
- as 3 triangles and 2 quadrilaterals ((1/3!)*(1/2!)*C(17,3)*C(14,3)*C(11,3)*C(8,4)*C(4,4) = 238238000 different ways).
		

Crossrefs

Column 5 of A059022.

Programs

Maple

A059022 := proc(n,k)
    option remember;
    if n<3 then
        0;
    elif n < 6 and k=1 then
        1 ;
    else
        k*procname(n-1,k)+binomial(n-1,2)*procname(n-3,k-1) ;
    end if;
end proc:
A352611 := proc(n)
    A059022(n,5) ;
end proc:
seq(A352611(n),n=15..50) ; # R. J. Mathar, Apr 08 2022

Mathematica

S3[3, 1] = S3[4, 1] = S3[5, 1] = 1;
S3[n_, k_] /; 1 <= k <= Floor[n/3] := S3[n, k] = k*S3[n-1, k] + Binomial[n-1, 2]*S3[n-3, k-1];
S3[, ] = 0;
a[n_] := S3[n, 5];
Table[a[n], {n, 15, 50}] (* Jean-François Alcover, Jul 06 2022 *)

Formula

Let S(n,k) be the number of different ways to partition the set of vertices of a convex n-gon into k polygons, where each partition contains at least 3 objects (vertices).

By the k-associated Stirling numbers of second kind, it can be deduced that S(n,k) = k*S(n-1,k) + C(n-1,2)*S(n-3,k-1).

When k = 5 this gives the required formula for this particular case,

a(n) = S(n,5) = 5*S(n-1,5) + C(n-1,2)*S(n-3,4)

where n > 14 and S(14,5) = 0.