## Is the limit of classical Laver tables connected anywhere?

Let $A_{n}=(\{1,\dots,2^{n}\},*_{n})$ be the $n$-th classical Laver table. Then $*_{n}$ is the unique operation on $\{1,\dots,2^{n}\}$ where $x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$ and $x*_{n}1=x+1\mod 2^{n}$ whenever $x,y,z\in A_{n}$. Let $L_{n}:\{0,\dots,2^{n}-1\}\rightarrow\{0,\dots,2^{n}-1\}$ be the mapping that reverses the ordering of the digits in the binary expansion of the number in $\{0,\dots,2^{n}-1\}$. More explicitly, $$L_{n}(\sum_{k=0}^{n-1}2^{k}a_{k})=\sum_{k=0}^{n-1}2^{n-1-k}a_{k}$$ whenever $a_{0},\dots,a_{n-1}\in\{0,1\}$. Let $L_{n}^{\sharp}:\{1,\dots,2^{n}\}\rightarrow\{1,\dots,2^{n}\}$ be the mapping … Read more