Largest conjugacy class of $S_n$

Question

I am familiar with the formula for computing the size of a conjugacy class of $S_n$ given its cycle type, but other than guessing and checking I do not know how to find either:

1. The size of the largest conjugacy class
2. The cycle type of the largest conjugacy class.

Is there any simple way of computing these?

Answer 1

For each cycle type, there is a formula of how many elements are in this conjugacy class, e.g., for $(123\cdots n)$ we see that its conjugacy class has size $(n-1)!$. However, the conjugacy class of $(123\cdots n-1)(n)$ is bigger for $n\ge 3$, namely $$ (n-1)!+(n-2)!=\frac{n!}{n-1}. $$

This is the largest size, if you look at the Conjugacy class size formula in symmetric group. I don't know a simple way of computing it, since we have $p(n)$ conjugacy classes, i.e., cycle types, which grows rapidly with $n$. As for the computation of the size of the largest conjugacy class, see OEIS. It is mentioned, too, that this is the "the maximum entry in each row of A036039".