# Why do highly composite rows on the bad Laver tables have longer periods?

For all natural numbers $$n$$, let $$(B_{n},*_{n})$$ be the algebraic structure with underlying set $$\{1,\dots,n\}$$ where

1. $$x*_{n}1=x+1\mod n$$,

2. $$n*_{n}y=y$$, and

3. $$x*_{n}(y+1)=(x*_{n}y)*_{n}(x+1)$$ for $$x.

The algebra $$(B_{n},*_{n})$$ is called the $$n$$-th bad Laver table. For each $$n$$ and $$x\in\{1,\dots,n\}$$, there exists some number $$\pi_{n}(x)$$ such that $$x*_{n}1<\dots and where $$x*_{n}y=x*_{n}z$$ whenever $$y=z\mod \pi_{n}(x)$$. If $$n=2^{N}$$, then $$(B_{n},*_{n})$$ is known as the $$N$$-th classical Laver table.

Let $$\sigma_{0}(x)$$ denote the number of divisors of the number $$x$$. Here are a few unexplained observations about the bad Laver tables.

1. For all $$n$$, the linear correlation coefficient between $$\sigma_{0}(x)$$ and $$\log(\pi_{n}(x))$$ is quite high. For example, if $$n=339849$$, then the linear correlation coefficient between $$\sigma_{0}(x)$$ and $$\log(\pi_{n}(x))$$ is $$0.436963$$. The linear correlation coefficient between $$\sigma_{0}(x)$$ and $$\log(\pi_{n}(x))$$ seems to depend mainly on $$\gcd(2^{\infty},n)$$.

2. For all $$n$$, if $$\alpha$$ is a highly composite number, then the linear correlation coefficient between $$\log(\pi_{n}(x))$$ and $$\log(\pi_{n}(x+\alpha))$$ is also quite high. For example, if $$\alpha=7200$$ and $$n=367611$$, then the linear correlation coefficient between $$\log(\pi_{n}(x))$$ and $$\log(\pi_{n}(x+\alpha))$$ is $$0.526511$$.

3. If $$x and $$\pi_{n}(x)=\max(\{\pi_{n}(y)|y, then $$\sigma_{0}(x)$$ is often exceptionally large.

What are some mathematical explanations for these phenomena?

• Let $\ell_{n}$ denote the linear correlation coefficient between $\sigma_{0}(x)$ and $\log(\pi_{n}(x))$. Then the distribution of the values $\ell_{n}$ does not follow a normal distribution where the sample is taken from the odd numbers from n=30000 to n=36000. – Joseph Van Name Feb 10 at 23:21