%I #24 Jul 06 2022 11:00:05

%S 1401400,28028000,333533200,3073270200,24234675465,172096749825,

%T 1134040872965,7069307049805,42240545297951,244205509154607,

%U 1375458924105651,7586883537988755,41147137237012950,220107145169421510,1164186829638102270,6100518487069916910

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

%F 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).

%F 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).

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

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

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

%e For n=17, the set of vertices of a convex 17-gon can be partitioned into 5 polygons in 333533200 different ways:

%e - 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

%e - 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).

%p A059022 := proc(n,k)

%p option remember;

%p if n<3 then

%p 0;

%p elif n < 6 and k=1 then

%p 1 ;

%p else

%p k*procname(n-1,k)+binomial(n-1,2)*procname(n-3,k-1) ;

%p end if;

%p end proc:

%p A352611 := proc(n)

%p A059022(n,5) ;

%p end proc:

%p seq(A352611(n),n=15..50) ; # _R. J. Mathar_, Apr 08 2022

%t S3[3, 1] = S3[4, 1] = S3[5, 1] = 1;

%t 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];

%t S3[_, _] = 0;

%t a[n_] := S3[n, 5];

%t Table[a[n], {n, 15, 50}] (* _Jean-François Alcover_, Jul 06 2022 *)

%Y Column 5 of A059022.

%K nonn,easy

%O 15,1

%A _Janaka Rodrigo_, Mar 23 2022