# Growth rate of the critical points of the Fibonacci terms $t_{n}(x,y)$ vs $t_{n}(1,1)$ in the classical Laver tables

The classical Laver table $$A_{n}$$ is the unique algebra $$(\{1,\dots,2^{n}\},*_{n})$$ where $$x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$$ and $$x*_{n}1=x+1\mod 2^{n}$$ for all $$x,y,z\in A_{n}$$.

Define the Fibonacci terms $$t_{n}$$ for $$n\geq 1$$ by letting $$t_{1}(x,y)=y,t_{2}(x,y)=x,t_{n+2}(x,y)=t_{n+1}(x,y)*t_{n}(x,y)$$.

Then do there exist natural numbers $$N,n,x,y$$ with $$x,y\in A_{N}$$ and where

1. $$\gcd(2^{N},x)\leq\gcd(2^{N},y)$$, and

2. $$\gcd(2^{N},t_{2n+1}(x,y))<\gcd(2^{N},t_{2n+1}(1,1))$$?

I have tested this conjecture empirically and I do not expect any computer experiments to find such $$N,n,x,y$$ since the classical Laver tables tend to produce extremely very fast growing functions.