Up to duality, there are two known families of finite (thick) generalised hexagons:

1. the Split Cayley hexagons of order $(q,q)$ related to the Dickson exceptional groups $G_2(q)$.
2. the Twisted Triality hexagons of order $(q,q^3)$ related to the Steinberg exceptional groups $\,^3D_4(q)$.

Since I will only consider generalised hexagons up to duality, I will assume that the order $(s,t)$ of a given one has $s\le t$. The parameter $s$ is one less than the number of points on a line, and $t+1$ is the number of lines on a point. To date, we do not know much about the possible values of these positive integers $s$ and $t$. Here is what we know:

• $s\le t^3$ and $t\le s^3$ (Haemers and Roos, 1981)
• $s^2+st+t^2$ divides $s^3(s^2t^2+st+1)$ (from the multiplicities of the eigenvalues of the point graph).
• There are two (up to duality) generalised hexagons with $s=2$ and they have $t\in\{2,8\}$ respectively (Cohen and Tits 1985).

We do not know that $s+1$ divides $t+1$, even though the known examples satisfy this simple divisibility relation.

Here is a cool number theoretic thing I’ve observed recently, but I can’t (quite yet) prove that it is true:

Claim: If $s,t$ are integers greater than 1, satisfying $s+1 \mid t+1$ and $s^2+st+t^2\mid s^3(s^2t^2+st+1)$, then apart from a small finite number of exceptions, we have $t=s$ or $t=s^3$.

I claim that the only exception is $(s,t)=(14,224)$.

We know that the known examples satisfy the extra relation $s+1\mid t+1$, and it is conceivable that such a simple looking relation holds for generalised hexagons and that it arises for some combinatorial reason. It’s simplicity is the only reason to believe that it might hold, but wouldn’t it be cool if every generalised hexagon satisfied it?