most recent 30 from http://mathoverflow.net 2013-05-21T19:58:41Z http://mathoverflow.net/feeds/question/57980 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57980/what-is-the-largest-laver-table-which-has-been-computed Justin Moore 2011-03-09T18:48:41Z 2011-03-24T19:52:01Z

<p>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 $(A_n,*)$.</p> <p>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 $A_{27}$.</p> <p>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 $m &lt; n$ and hence the problem becomes strictly harder for larger $n$.</p> <p>Edit: I have actually been able to compute $A_{28}$, not just $A_{27}$.</p>

http://mathoverflow.net/questions/57980/what-is-the-largest-laver-table-which-has-been-computed/59467#59467 Justin Moore 2011-03-24T19:52:01Z 2011-03-24T19:52:01Z

<p>I've been in contact with Patrick Dehornoy and Ales Drapal and both thought that $A_{28}$ is likely the current record for a Laver table computation.</p>