Does the length of the period of the first row in the odd bad laver tables grow without bound?

If $n$ is a natural number, then the $n$-th bad Laver table is the algebra $B_{n}=(\{1,...,n\},*)$ where

$n*x=x$

$x*1=x+1$ whenever $x<n$

$x*(y+1)=(x*y)*(x+1)$ whenever $x<n,y<n$.

If $n=2^{N}$ for some $n$, then the algebra $B_{2^{n}}$ satisfies the self-distributivity law $x*(y*z)=(x*y)*(x*z)$ and $B_{2^{n}}$ is known as the $n$-th classical Laver table. If $1<x\leq n$ and $B_{n}=(\{1,...,n\},*)$, then there is a unique natural number $p_{n,x}$ called the period length of $x$ in $B_{n}$ where the sequence $x*1,...,x*p_{n,x}$ is strictly increasing with $x*p_{n,x}=n$ but where $x*y=x*z$ whenever $y=z\mod p_{n,x}$.

Suppose that $N\geq 0$ and $m=2^{N}$. Is $\sup_{n}p_{m\cdot(2\cdot n+1),1}=\infty$?

Under the assumption of very large cardinal hypotheses, we know that $\sup_{n}p_{2^{n},1}=\infty$, but the function $n\mapsto p_{2^{n},1}$ is extremely slow growing, so I wonder if the unboundedness of the periods $p_{n,1}$ also holds for bad Laver tables $B_{n}$ with $n$ odd.