# Do the classical Laver tables induce periodic systems of jump Laver tables?

Let $A_{n}=(\{1,...,2^{n}\},*_{n})$ where $*_{n}$ is the binary operation on $A_{n}$ such that $x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$ for $x,y,z\in\{1,...,2^{n}\}$, $x*1=x+1$ for $x<2^{n}$ and $2^{n}*1=1$.

In $A_{n}$, we have $\text{crit}(x)\leq\text{crit}(y)$ if and only if $2^{m}|x\Rightarrow 2^{m}|y$ for all $m\leq n$. In particular, $\text{crit}(x)\geq\text{crit}(2^{m})$ if and only if $x$ is a multiple of $2^{m}$.

Since $\text{crit}(x*y)\geq\text{crit}(y)$ for all $x,y$, if $y$ is a multiple of $2^{m}$, then $x*y$ is also a multiple of $2^{m}$. Therefore, $\{x|\text{$x$is a multiple of$2^{m}$}\}$ is a subalgebra of $(A_{n},*_{n})$.

Now suppose that $m$ is a natural number and $n\geq m$. Then define a binary operation $*_{m,n}$ on $\{1,...,2^{m}\}$ by letting $x*_{m,n}y=\frac{1}{2^{n-m}}\cdot((x\cdot 2^{n-m})*_{n}(y\cdot 2^{n-m}))$. I suggest calling the table $(\{1,...,2^{m}\},*_{m,n})$ a jump Laver table since we sometimes have $x*_{m,n}1>x+1$ (i.e. we jump from $x$ up to $x*_{m,n}1$ instead of stepping from $x$ to $x+1$).

For all $m$, is the sequence $(*_{m,n})_{m\in\omega,n\geq m}$ of binary operations on $\{1,...,2^{m}\}$ periodic? If not, then is the sequence $(*_{m,n})_{m\in\omega,n\geq m}$ eventually periodic? If the answer to either of these questions is yes, then what is its period?