By replacing the Laver tables with nearly distributive fake Laver tables, can one produce algebras where the period of 1 grows fast?

I am now generalizing the notion of the classical Laver table and the fake Laver table to a larger class of algebras. A sequence of algebras $(\{1,...,2^{n}\},*_{n})_{n\in\omega}$ is said to be a Laver-table like sequence if

1. $2^{n}*_{n}x=x$ for each $x\in\{1,...,2^{n}\}$

2. $x*_{n}1=x+1$ for all x such that $1\leq x<2^{n}$

3. $x*_{n}2^n=2^n$ for all $x$

4. The mapping $\phi_{n}:(\{1,...,2^{n+1}\},*_{n+1})\rightarrow(\{1,...,2^{n}\},*_{n})$ defined by $\phi_{n}(x)=x$ for $1\leq x\leq 2^{n}$ and $\phi_{n}(x)=x-2^{n}$ for $x>2^n$ is a homomorphism.

5. For each $x\in\{1,...,2^{n}\}$ there is some $m$ with $m\leq n$ so that the sequence $(x*_{n}1,x*_{n}2,x*_{n}3,...,x*_{n}2^{m})$ is strictly increasing and where $x*_{n}y=x*_{n}(y+2^m)$ for all $y$ satisfying $1\leq y<y+2^m\leq 2^n$. We shall write $o(*_{n},x)$ for such an $m$.

Statement 5 simply means that these the rows of the multiplication tables of these Laver like tables are periodic with a period that divides $2^n$ and the $x$-th row is an increasing sequence of numbers that begins at $x+1$ and ends at $2^n$.

For example, the classical Laver tables and the fake laver tables are both Laver-table like sequences.

If $\delta<1$ and $v$ is a natural number, then does there exist a Laver-table like sequence $(\{1,...,2^{n}\},*_{n})_{n\in\omega}$ such that

i. $\frac{1}{8^{n}}|\{(x,y,z)\in\{1,...,2^{n}\}^{3}|x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)\}|\geq\delta$ and

ii. there exists a primitive recursive function $f:\mathbb{N}\rightarrow\mathbb{N}$ so that $o(*_{f(n)},v)\geq n$ for all $n$?

What is the answer to the above question if we replace self-distributivity with some non-trivial identity which is a consequence of self-distributivity?

The motivation behind this question is that if we let $\delta=1$, then the answer to the above question is false since if $\delta=1$, then the only possible choice of $*_{n}$ is the classical Laver table and the above question is false for the classical Laver tables (see chapter 11 in the Handbook of Set Theory or chapter 13 in the book Braids and Self-Distributivity by Patrick Dehornoy).

I also wonder if for $\delta<1$ we can form a Laver-table like sequence $(\{1,...,2^{n}\},*_{n})_{n\in\omega}$ where if $t_{n}=\frac{1}{8^{n}}|\{(x,y,z)\in\{1,...,2^{n}\}^{3}|x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)\}|$, then $t_{n}<1$ for some $n$ but where $t_{n}>\delta$ for all $n$?

The notion of a Laver-table like sequence has not been studied by anyone anywhere before.

• For clarity, add after $(x)_{2^n}$ ( $x - 2^n$ for $x \gt 2^n$), to make sure that something "mod"-ish is wanted. For 4, I think you want only $*_n$, not a mix of $*$ and $*_n$. Also, is the sequence of length $m$ or of length $2^m$? Also, "for all $y$ satisfying" is better than "whenever", as it is unclear to me if $y$ is restricted to the increasing sequence or not. Some visual or algebraic motivation for 4 would be appreciated. Gerhard "How About Fake Laver Trees?" Paseman, 2015.10.20 – Gerhard Paseman Oct 20 '15 at 19:07
• Gerhard Pasemen. Thanks for the corrections. – Joseph Van Name Oct 21 '15 at 3:05