## How distributive are the bad Laver tables?

Suppose that $n\in\omega\setminus\{0\}$. Then define $(S_{n},*)$ to be the algebra where $S_{n}=\{1,…,n\}$ and $*$ is the unique operation on $S_{n}$ where $n*x=x$ $x*1=x+1\,\text{mod}\, n$ and if $y<n$, then $x*(y+1)=(x*y)*(x*1)$. The algebra $(S_{n},*)$ satisfies the self-distributivity law $x*(y*z)=(x*y)*(x*z)$ if and only if $n$ is a power of $2$, and if $n$ is a power of $2$, … Read more