Recall that the classical Laver table $A_{n}$ is the unique algebraic structure $(\{1,\ldots,2^{n}\},*_{n})$ where
$x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$, and
$x*_{n}1=x+1\mod n$ for all $x,y,z\in A_{n}$.
If $x\in A_{n}$, then let $o_{n}(x)$ denote the least natural number $m$ such that $x*_{n}2^{m}=2^{n}$.
There are typically very few elements $x\in A_{n}$ where $o_{n}(x)=5$.
The following function maps $m$ to the number of elements $x\in A_{20}$ with $o_{n}(x)=m$. $(0\rightarrow 1,1\rightarrow 20, 2\rightarrow 555085,3\rightarrow 107010,4\rightarrow 316545,5\rightarrow 55,6\rightarrow 37235,7\rightarrow 7255,8\rightarrow 21230,9\rightarrow 24, 10\rightarrow 2193,11\rightarrow 462,12\rightarrow 1191,13\rightarrow 12,14\rightarrow 144,15\rightarrow 41,16\rightarrow 58,17\rightarrow 4,18\rightarrow 8,18\rightarrow 2,20\rightarrow 1)$
Observe that the only elements $x\in A_{n}$ with $o_{n}(x)=2$ are of the form $x=2^{n}-2^{m}-1$ for $0\leq m<n$ and the only element $x\in A_{n}$ with $o_{n}(x)=1$ is $x=2^{n}-1$. Observe also that as $m$ grows larger there are fewer and fewer elements $x\in A_{n}$ with $o_{n}(x)=m$. The next most prominent trend in the data for $o_{n}(x)$ is that it is very rare for $o_{n}(x)=5$ or $o_{n}(x)=9$. Why is it so rare for $o_{n}(x)=5$ but it is much more common for $o_{n}(x)=6$ or $o_{n}(x)=8$?
Here are the hexadecimal representations of the numbers $x\in A_{20}$ with $o_{n}(x)=5$.
e,1000e,2000e,3000e,4000e,5000e,6000e,7000e,70fff,7f0ff,7ff0f,7fff0,8000e,9000e,a000e,b000e,b0fff,bf0ff,bff0f,bfff0,c000e,d000e,d0fff,df0ff,dff0f,dfff0,e000e,e0fff,ef0ff,eff0f,efff0,f70ff,f7f0f,f7ff0,fb0ff,fbf0f,fbff0,fd0ff,fdf0f,fdff0,fe0ff,fef0f,feff0,ff70f,ff7f0,ffb0f,ffbf0,ffd0f,ffdf0,ffe0f,ffef0,fff70,fffb0,fffd0,fffe0
