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Suppose that $n\in\omega\setminus\{0\}$. Then define $(S_{n},*)$ to be the algebra where $S_{n}=\{1,...,n\}$ and $*$ is the unique operation on $S_{n}$ where

  1. $n*x=x$

  2. $x*1=x+1\,\text{mod}\, n$ and

  3. if $y<n$, then $x*(y+1)=(x*y)*(x*1)$.

The algebra $(S_{n},*)$ satisfies the self-distributivity law $x*(y*z)=(x*y)*(x*z)$ if and only if $n$ is a power of $2$, and if $n$ is a power of $2$, then the algebra $(S_{n},*)$ is called a Laver table.

If $n$ is not a power of 2, then one would still expect the algebra $(S_{n},*)$ to have a certain amount of self-distributivity and a computer verification shows that in $S_{n}$ we have $x*(y*z)=(x*y)*(x*z)$ for many triples $(x,y,z)$.

Define $f(n)=\frac{1}{n^{3}}\cdot|\{(x,y,z)\in\{1,...,n\}^{3}:S_{n}\models x*(y*z)=(x*y)*(x*z)\}|$ for each positive integer $n$.

Let $\ell=\liminf_{n\in\omega}f(n)$. Let $\mathfrak{m}=\liminf_{n\in\omega}\frac{1}{n}(f(1)+...+f(n))$. Then is $\ell=0$? If not, then can one give an explicit value of $\ell$? What is the value of $\mathfrak{m}$? Does $\mathfrak{m}=\lim_{n\in\omega}\frac{1}{n}(f(1)+...+f(n))$? Is $\mathfrak{m}=\frac{1}{2}$? Is there a proof that for all natural numbers $n$, we have $\liminf_{m\in\omega}f(n2^{m})=1$?

The first few values of $f(n)$ are 1., 1., 0.851852, 1., 0.696, 0.856481, 0.612245, 1., 0.615912, 0.713, 0.489106, 0.930556, 0.516158, 0.715015.

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  • $\begingroup$ Are these algebras rigid? Primal? Can you "nearly" embed S_n into a larger S_m ? What literature on Laver tables have you not read yet? Gerhard "Always More Questions Than Answers" Paseman, 2015.08.08 $\endgroup$ – Gerhard Paseman Aug 9 '15 at 3:42
  • $\begingroup$ If $m,n$ are integers, then the mapping $\phi:S_{2^{n}\cdot m}\rightarrow S_{2^{n}}$ where $\phi(x)=x\,\text{mod}\, 2^{n}$ is a homomorphism, so the algebra $S_{n}$ is not simple whenever $n$ is even. The algebras $S_{n}$ are have no nontrivial automorphisms since $1$ is the unique generator for each of the algebras $S_{n}$. There are many ways to embed Laver tables $S_{2^{n}}$ into larger Laver tables $S_{2^{m}}$ (Drapal has proven some results about embedding smaller Laver tables into larger Laver tables). $\endgroup$ – Joseph Van Name Aug 9 '15 at 17:12
  • $\begingroup$ I am familiar with most of the literature on the distributive Laver tables, but there does not appear to be much literature on the "bad" (non-distributive) Laver tables since they do not appear to have much use besides being a construction that includes the distributive Laver tables $S_{2^{n}}$. $\endgroup$ – Joseph Van Name Aug 9 '15 at 17:55
  • $\begingroup$ If one were able to nearly cover a large S_m by embeddings of a smaller S_{2^n}, that would help give you some idea as to how distributive S_m can be. Even knowing how S_2 embeds in S_5 might help in figuring out f(n) in general. Gerhard "Also Look At Their Clones" Paseman, 2015.08.09 $\endgroup$ – Gerhard Paseman Aug 10 '15 at 5:43

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