The classical Laver table $A_{n}$ is the unique algebraic structure $(\{1,\dots,2^{n}\},*_{n})$ where $$x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$$ and where $$x*_{n}1=x+1\mod 2^{n}$$ for $x,y,z\in\{1,\dots,2^{n}\}$. For each $n$, and each $x\in\{1,\dots,2^{n}\}$, let $o_{n}(x)$ be the least natural number $m$ such that $x*_{n}2^{m}=2^{n}$.
Suppose that for each $n$, there exists an $n$-huge cardinal. Then $o_{n}(1)\leq o_{n}(2)$ for each $n\geq 1.$
We have $o_{n}(1)<o_{n}(2)$ for $n\in\{1,7,8\}$. Are there any other $n$ where $o_{n}(1)<o_{n}(2)$? Are there infinitely many $n$ with $o_{n}(1)<o_{n}(2)$?
If $o_{n}(1)<o_{n}(2)$ and $n>8$, then $n$ must be extremely large.
