Fix $m.$ For each $n$ consider the the question: among all the conjugacy classes for permutations in $S_n$ with exactly $m+1$ cycles, which has the largest size? I will write $\prod_{i \geq 1}i^{a_i}$ to describe a partition with $a_i$ cycles of length $i.$ Below is some data. It is the result of rather unsophisticated exhaustive calculation. It strongly suggests the following unsurprising result: > There is a $k=k(m)$ integers $a_1 \geq a_2 \geq \cdots \geq a_k=1$ and $t=t(m)=\sum_1^ka_kk$ such that, for $n$ large enough relative to $m,$ the unique maximum occurs for type $1^{a_1}2^{a_2}\cdots k^{a_k}(n-t)$ **later** > Based on the calculations one might speculate further that the $a_i$ decrease until they are $1$ (perhaps no later than $\sqrt{k}$ or some such $j \ll k$) The conjecture from the question is that $a_1$ is about $\frac{m}2$, $a_2$ is about $\frac{m}4$ **and so on**. That is a bit vague. Does it mean that $a_j$ is about $\frac{m}{2^j}$ (perhaps for $1 \leq j \ll k?.$ An now deleted answer speculated that $a_1=2a_2=3a_3=\cdots=ja_j$ again perhaps for $1 \leq j \ll k.$ The present calculations are not sufficient to test if either of these is likely. - For $m=1$ the maximum occurs for $1\ (n-1)$ - For $m=2$ and $n \gt 5$ the maximum occurs for $1\ 2\ (n-3)$ - For $m=3$ and $n \gt 10$ the maximum occurs for $1^2\ 2\ (n-4)$ - For $m=4$ and $n \gt 10$ the maximum occurs for $1^2\ 2\ \ 3 (n-7)$ - For $m=5$ and $n \gt 20$ the maximum occurs for $1^3\ 2\ 3\ (n-8)$ - For $m=6$ and $n \gt 17$ the maximum occurs for $1^3\ 2\ 3\ 4\ (n-12)$ **except that** - for $n=22$ there is a tie between $1^3\ 2\ 3\ 4\ (10)$ and $1^2\ 2\ 3\ 4\ 5\ 6$ - For $m=7$ and $n \gt 29$ the maximum occurs for $1^3\ 2^2\ 3\ 4\ (n-14)$ This might allow a more refined conjecture. Certainly data for larger $m$ would be suggestive. That would justify taking the time to do design a more intelligent search. Here are more detailed results on $m=7$ to show exactly what I know for sure: For $30 \leq n \leq 60$ the maximum occurs only for $1^3\ 2^2\ 3\ 4\ (n-14).$ Presumably the same is true for all $30 \leq n.$ It is also true for $n=19,20,21,22. $ the other cases are: - $1^8$ - $1^72$ - $1^62^2$ - $1^62\,3$ - $1^52^23$ - $1^42^33$ - $1^5\,2\,3\,4$ - $1^42^23\,4$ - $1^42^23\,5$ - For $n=17$ a tie between $1^42^23\,6$ and $1^32^23^24$ - $1^4\,2\,3\,4\,5$ for $n=18.$ - $1^32\,3\,4\,5\,(n-17)$ for $23 \leq n \leq 28$ - for $n=29$ a tie between $1^3\,2\,3\,4\,5\,12$ and $1^32^23\,4\,15$