I have found a way to enlarged the Erdos-Szekeres Conjecture. I have computationally confirmed what might be the simplest unknown case for some small values, and was reaching out for how to prove this case in general.

**Definitions** Given a set of points in
the general position on the plane,
a subset is a "convex polygon" if each point is
a vertex of the subset's convex hull.
A cap is a sequence of points with increasing $x$-coordinate
such that the slope of consecutive pairs is decreasing. A cup
is defined similarly but the slopes are increasing.
A $n$-polygon (size $n$ convex polygon) can be decomposed into a
$k$-cap and an $k'$-cup where $k' = n + 2 - k$.
We refer to this as a $(k, k')$-polygon.

For example in the image below, the yellow and red points form a $4$-cap, the yellow and green form $6$-cup, collectively the red, yellow, and green points form both a $(4, 6)$-polygon and a $8$-polygon.

**Erdos-Szekeres Conjecture**
The tooling in the next section are used to generalize the following
famous conjecture. If a set contains $2^{n-2}+1$ points
in the general position on the plane,
does it contain an $n$-polygon?

**Polygon Forcing Theorems**
Define $C(X; k, k')$ to be true if and only
the set of points $X$ contain a $(k, k')$-polygon.
We define a *forced polygon theorem* is one of the form
$$|X| > N \implies \bigcup_i C(X; k_i, k_i')$$
We now show some results and problems that
are forced polygon theorems.
If the Erdos-Szekeres Conjecture is correct then
$$|X| > 2^{n-2} \implies \bigcup_{i = 2}^{n} C(X; i, n+2-i)$$
Another example, Erdos and Szekeres show that
$$|X| > \binom{k + k' - 4}{k - 2} \implies C(X; 2, k') \cup C(X; k, 2)$$
which they use to provide an upper bound to their problem.
See "A Combinatorial Problem in Geometry" by Erdos and Szekeres. And in "Ramsey-remainder", by Erdos, Tuza, and Valtr showed that the Erdos-Szekeres conjecture is equivalent to
$$|X| > \sum_{i = n - k'}^{k-2} \binom{n - 2}{i}
\implies C(X; k, 2) \cup C(X; 2, k') \cup \bigcup_{i = n+2 - (k'-1)}^{k-1} C(X; i, n+2-i)$$
And when $k, k' = n$ they are equal.

**My Problem**
I have found a computational technique for creating polygon forcing theorems. Using it I have observed for all $k$ and $k'$ in cases when the program halts:

$$|X| > \frac{(k-1) (k'-1)}{2} \implies C(X; 2, k') \cup C(X; 3, 3) \cup C(X; k, 2)$$

I suspect there is a labeling argument similar to another related problem but I haven't be able to find it and was wondering if I could have help. Thank you for reading this post, as it is quite long.

**Edit**

Here are all the cases I can compute, in case someone can come up with a comprehensive conjecture. Note that these are upper bounds: If `21 -> [(2, 7), (3, 3), (8, 2)]`

it is still possible that `20 -> [(2, 7), (3, 3), (8, 2)]`

. I am curious if there are better bounds for any of these cases.

```
2 -> [(2, 3), (3, 2)]
3 -> [(2, 3), (4, 2)]
4 -> [(2, 3), (5, 2)]
5 -> [(2, 3), (6, 2)]
3 -> [(2, 4), (3, 2)]
6 -> [(2, 4), (4, 2)]
10 -> [(2, 4), (5, 2)]
15 -> [(2, 4), (6, 2)]
4 -> [(2, 5), (3, 2)]
10 -> [(2, 5), (4, 2)]
20 -> [(2, 5), (5, 2)]
5 -> [(2, 6), (3, 2)]
15 -> [(2, 6), (4, 2)]
4 -> [(2, 4), (3, 3), (4, 2)]
6 -> [(2, 4), (3, 3), (5, 2)]
7 -> [(2, 4), (3, 3), (6, 2)]
7 -> [(2, 4), (4, 3), (5, 2)]
9 -> [(2, 4), (4, 3), (6, 2)]
11 -> [(2, 4), (5, 3), (6, 2)]
6 -> [(2, 5), (3, 3), (4, 2)]
8 -> [(2, 5), (3, 3), (5, 2)]
10 -> [(2, 5), (3, 3), (6, 2)]
7 -> [(2, 5), (3, 4), (4, 2)]
12 -> [(2, 5), (3, 4), (5, 2)]
17 -> [(2, 5), (3, 4), (6, 2)]
12 -> [(2, 5), (4, 3), (5, 2)]
14 -> [(2, 5), (4, 3), (6, 2)]
14 -> [(2, 5), (4, 4), (5, 2)]
7 -> [(2, 6), (3, 3), (4, 2)]
10 -> [(2, 6), (3, 3), (5, 2)]
12 -> [(2, 6), (3, 3), (6, 2)]
9 -> [(2, 6), (3, 4), (4, 2)]
14 -> [(2, 6), (3, 4), (5, 2)]
11 -> [(2, 6), (3, 5), (4, 2)]
15 -> [(2, 7), (3, 3), (6, 2)]
21 -> [(2, 7), (3, 3), (8, 2)]
8 -> [(2, 5), (3, 4), (4, 3), (5, 2)]
11 -> [(2, 5), (3, 4), (4, 3), (6, 2)]
13 -> [(2, 5), (3, 4), (5, 3), (6, 2)]
11 -> [(2, 6), (3, 4), (4, 3), (5, 2)]
14 -> [(2, 6), (3, 4), (4, 3), (6, 2)]
13 -> [(2, 6), (3, 5), (4, 3), (5, 2)]
```

Explaining the notation `12 -> [(2, 6), (3, 3), (6, 2)]`

means
$$|X| > 12 \implies C(X;2,6) \cup C(X;3,3) \cup C(X;6,2)$$