%I #29 Jan 31 2022 06:46:11
%S 0,0,0,3,7,12,39,105,231,577,1482,3549,8603,21340,52122,126777,310859,
%T 761199,1859014,4549215,11141085,27266225,66760855,163567911,
%U 400786617,982265827,2408361144,5906499136,14489105190,35553445788,87264949808,214241203801
%N Total number of polygons left out in all partitions of the set of vertices of a convex n-gon into nonintersecting polygons.
%e a(3) = a(4) = a(5) = 0 since the only partition of the vertices of a triangle, quadrilateral or pentagon into polygons is the full polygon so nothing is left out.
%e a(6) = 3 since the vertices of a hexagon can be partitioned into two non-intersecting triangles in A350248(6,2) = 3 ways and in each of these cases a quadrilateral is left over.
%e When partitioning the set of vertices of a convex 13-gon into 1 polygon, the number of polygons remaining is 0.
%e When partitioning it into 2 polygons, the remaining polygons are 52 quadrilaterals.
%e When partitioning it into 3 polygons, the remaining polygons are 65 hexagons + 650 quadrilaterals.
%e When partitioning it into 4 polygons, the remaining polygons are 13 octagons + 117 hexagons + 585 quadrilaterals.
%e This gives the total as 1482 polygons.
%o (PARI) seq(n)={my(p=O(x)); while(serprec(p,x)<=n, p = x + x*y*(1/(1 - x*p^2/(1 - p)) - 1)); Vec(subst(deriv(O(x*x^n) + p^3/(1-p), y), y, 1), 2-n) } \\ _Andrew Howroyd_, Jan 30 2022
%Y Partitioning into 3 polygons A350116.
%Y Total number of different ways to partition the set of vertices of a convex polygon into nonintersecting polygons A350248.
%K nonn
%O 3,4
%A _Janaka Rodrigo_, Jan 24 2022
%E More terms from _Andrew Howroyd_, Jan 30 2022