# Does the period of the first row in the odd size bad Laver tables grow without bound?

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

1. $$n*x=x$$

2. $$x*1=x+1$$ whenever $$x

3. $$x*(y+1)=(x*y)*(x+1)$$ whenever $$x.

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 $$1 and $$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.