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So I was doing some computer calculations with the classical Laver tables and I found polynomials $p_{n}(x,y)$ such that $p_{n}(i,1)=1$ for many $n$.

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

Define polynomial $p_{n}(x_{1},x_{2})$ for all $n\in\omega$ by letting $$p_{n}(x,y)=1+\sum\{x_{a_{1}}\dots x_{a_{s}} \mid s\leq 2^{n},(a_{1},...,a_{s})\in \{1,2\}^{s},$$ $$a_{1}*_{n+1}...*_{n+1}a_{s}=2^{n},a_{1}*_{n+1}...*_{n+1}a_{r}<2^{n}\,\text{for}\,1\leq r<s\}.$$

Then we have $$(p_{0}(i,1),\dots,p_{13}(i,1))=(1+i, 1, 1+i, 1, 0, 1, 1, 1, 1, 1, 1, 1 ,1, 1).$$

Why is $p_{n}(i,1)=1$ whenever $5\leq n<14$? What is the largest ordinal $N\in\omega+1$ where $p_{n}(i,1)=1$ whenever $5\leq n<N$? Why are the outputs $p_{n}(i,1)$ so simple compared to the horrendous polynomials $p_{n}(x,y)$?

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  • $\begingroup$ So this phenomenon also holds in many but not all cases when we choose an even $a$ and select $(a_{1},\dots,a_{s})\in\{1,a\}^{s}$ instead of $(a_{1},\dots,a_{s})\in\{1,2\}^{s}.$ $\endgroup$
    – Joseph Van Name
    Commented Mar 16, 2019 at 15:07
  • $\begingroup$ So this phenomenon also holds for various fake Laver tables. $\endgroup$
    – Joseph Van Name
    Commented Mar 24, 2019 at 13:28

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