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 $A_{27}$.

b = m and ac =n ? Also, can you give a reference for Laver's result that left self-distributive * is unique up to isomorphism? Gerhard "Ask Me About System Design" Paseman, 2011.03.11 $\endgroup$ – Gerhard Paseman Mar 11 '11 at 20:56