A semigroup is **nilpotent of degree 3** if every product of 3 elements gives the same result. In 2012, Andreas Distler and James D. Mitchell wrote that:

It is part of the folklore of semigroup theory that almost all finite semigroups are nilpotent of degree 3.

Here's a way to make this precise:

**Conjecture.** If $S(n)$ is the number of isomorphism classes of semigroups with $n$ elements, and $S_3(n)$ is the number of isomorphism classes of semigroups with $n$ elements that are nilpotent of degree 3, then

$$ \lim_{n \to \infty} \frac{S_3(n)}{S(n)} = 1. $$

**Has anyone proved this, or a similar result?**

Semigroup theorists seem to enjoy counting semigroups up to **equivalence**, meaning up to isomorphism or anti-isomorphism, so maybe someone has proved a similar result but counting semigroups up to equivalence rather than up to isomomorphism.

In 2012, Andreas Distler, Chris Jefferson, Tom Kelsey, and Lars Kottho counted the semigroups with 10 elements and found

$$ 12,418,001,077,381,302,684 $$

of them. Of these, all but

$$ 718,981,858,383,872 $$

were nilpotent of degree 3. So, about 99.994% were nilpotent of degree 3.

By the way, this calculation still took a lot of work.