Computing the high infinite. Threshold changes for the 257th to the 384th Laver tables now reordered

The n-th monogenic Laver table is the unique algebraic structure A_n=({1,...,2^n},*) where x*(y*z)=(x*y)*(x*z) and x*1=x+1 mod 2^n for al x,y,z. One may use the thresholds to compute the n+1-th Laver table from the n-th Laver table. Let *_n denote the n-th Laver table operation. Define o_n(x) to be the least natural number where x*_n 2^{o_n(x)}=2^n. If x is in {1,...,2^(n-1)-1}, then we define the threshold \theta_n(x) to be the integer y in {0,...,2^o_n(x)} where x*y\leq 2^(n-1) and x*(y+1)\geq 2^(n-1)+1 and set \theta_n(2^(n-1))=0. The values \theta_n allow us to compute larger Laver tables from smaller ones. In most cases, \theta_(n-1)(x)=\theta_n(x), so we only need to take note of the instances where \theta_(n-1)(x) differs from \theta_n(x). This is a visualization of those cases where \theta_{256}(x) differs from \theta_{256}(x) where x is less than 2^128.This calculation allows us to compute the 257th Laver table up through the 384th Laver table.

In order to improve interpretability or at least give a different perspective, I ordered the elements of the 257th Laver table. In particular, the element x precedes the element y in the visualization if there is some m where x=y mod 2^m but where y=x+2^m mod 2^(m+1) where y is in {2^m+1,...,2^(m+1)} mod 2^(m+1).
 
I computed these values using a conglomeration of techniques similar to machine learning and evolutionary computation to search for instances of non-distributivity. I do not have a proof that this is actually the threshold data that allows us to compute the 384th Laver table.

When I made these calculations, I did not use neural networks, but it seems like neural networks would be useful for making these calculations too or testing their accuracy.

The notion of a Laver table is not my own. Randall Dougherty has computed the 48th Laver table back in the 1990's, and I extended his technique to compute up to the 768th Laver table, but I made a tradeoff since I do not have a proof that my calculations are completely correct.

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