TITLE: What is the largest Laver table which has been computed?
QUESTION [21 upvotes]: Richard Laver proved that there is a unique binary operation $*$ on $\{1,\ldots,2^n\}$ which satisfies $$a*1 \equiv a+1 \mod 2^n$$
$$a* (b* c) = (a* b) * (a * c).$$
This is the $n$th Laver table $(A_n,*)$.
There is an algorithm for computing $a * b$ in $A_n$, but in general (and especially for small values of $a$), this requires one to compute much of the rest of $A_n$.
What is the largest value for $n$ for which someone can, in a modest amount of time, compute an arbitrary entry in $A_n$?  I am able to compute entries in $A_{27}$.
I should note that the map which sends $a$ to $a\ \mathrm{mod}\ 2^m$ defines a homomorphism from $A_n$ to $A_m$ for $m < n$ and hence the problem becomes strictly harder for larger $n$.
Edit: I have actually been able to compute $A_{28}$, not just $A_{27}$.

REPLY [16 votes]: On Azimuth, on May 6, 2016, Joseph van Name wrote:

The largest classical Laver table computed is actually $A_{48}$. The 48th table was computed by Dougherty and the algorithm was originally described in Dougherty's paper here. With today's technology I could imagine that one could compute $A_{96}$ if one has access to a sufficiently powerful computer.
One can compute the classical Laver tables up to the 48th table on your computer here at my website.