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