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

Question

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 $x<n,y<n$.

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 $x*_{n}1<\dots<x*_{n}\pi_{n}(x)$ 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 $x<n$ and $\pi_{n}(x)=\max(\{\pi_{n}(y)|y<n\})$, then $\sigma_{0}(x)$ is often exceptionally large.

What are some mathematical explanations for these phenomena?

Answer 1

This phenomenon has a simple explanation that I probably should have observed while asking this question.

Suppose that $n$ is a natural number and $x<n$ where $x$ is a highly composite number. Observe that if $a=b\mod\pi_{n}(c)$, then $c*_{n}a=c*_{n}b$. In particular, since $x$ is a highly composite number, it is most likely that $\pi_{n}(x*_{n}y)$ is a factor of $x$, and if $\pi_{n}(x*_{n}y)$ is a factor of $x$, then $x*_{n}(y+1)=(x*_{n}y)*_{n}(x+1)=(x*_{n}y)*_{n}1=(x*_{n}y)+1$. Therefore, since the $x$-th row increments by 1 in most cases, the $x$-th row will usually take a while to go all the way to $n$ and then start again at $x+1$.

Now, in the case that $\pi_{n}(x*_{n}y)$ is not a factor of $x$, it turns out that $\pi_{n}(x*_{n}y)$ is probably large. In this case, it is most likely that $x*_{n}(y+1)=(x*_{n}y)*_{n}(x+1)$ is much smaller than $n$. This will also contribute to a long length of the $x$-th row.