# Why does $p_{n}(i,1)=1$ so often where the polynomials $p_{n}$ are obtained from the classical Laver tables

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

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 $$5\leq n? Why are the outputs $$p_{n}(i,1)$$ so simple compared to the horrendous polynomials $$p_{n}(x,y)$$?

• 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 $(a_{1},\dots,a_{s})\in\{1,2\}^{s}.$ – Joseph Van Name Mar 16 at 15:07
• So this phenomenon also holds for various fake Laver tables. – Joseph Van Name Mar 24 at 13:28