How many binary operations are associative?

For questions like these you can try out alg. It is a program which takes some axioms (it works best for equations) and outputs, or just counts, non-isomorphic models of a given size. It also provides a link to OEIS for you to check the sequence it got.

The theory of an associative operation looks like this:

Theory associative.
Binary *.
Axiom: (x * y) * z = x * (y * z).

The output says:

./alg.native --size 1-4 --count theories/associative.th 
# Theory associative

    Theory associative.
    Binary *.
    Axiom: (x * y) * z = x * (y * z).

    size | count
    -----|------
       1 | 1
       2 | 5
       3 | 24
       4 | 188

Check the numbers [5, 24, 188](http://oeis.org/search?q=5,24,188) on-line at oeis.org

The point is, you can easily experiment (of course someone has counted these things before me).

Semigroups form a bigger chunk than you might think. Basically you call a symbol 0 and declare xyz=0 for all elements (making associativity trivial). You still have a huge flexibility on how to define the remaining products. This is the content of the paper Michael links. In fact 99% of all semigroups up to isomorphism and anti-isomorphism satisfy xyz=0. A recent paper of Distler and Mitchell count the exact number of these guys up to isomorphism http://www.combinatorics.org/ojs/index.php/eljc/article/view/v19i2p51. I think they also count the number of such multiplication tables.

Here is a guide to the intuition. I will not swear that the numerics are exact, but I will bet that the numerical truth is not far off.

Look at the diagonal for the multiplication table of a (labeled) groupoid on $n>3$ elements. Of the n^n possibilities, only one of them is idempotent, so with one exception aa=b will happen for some a and some b different from a. Now all we need for associativity to fail in this case is that ab and ba are different, which will happen for all but n of the n^2 possibilities. So we are already looking at associativity happening only on a small fraction of all (non-idempotent) tables, especially as there are often several candidates for a, and only one is needed.

Even for idempotent groupoids, one finds a,b,c distinct and needs to consider only d=ab, g=bc, and the ways in which dc and ag can fail to be equal. Again in rough terms we are talking about n^(-2), and this is just by fixing a,b, and c in advance, and that for the 1 out of n^n tables that are idempotent.

I'll let someone else tighten up the numerics. For strengthening Joseph's intuition, I hope this will suffice.

Gerhard "Ask Me About 2-Deficient Groupoids" Paseman, 2012.10.21