How to show that there are $1121505$ conjugacy classes in $S_{61}$?

Question

How to show that there are $1121505$ conjugacy classes in $S_{61}$?

Can someone help me with this question. I have no idea how to count them.

Second Question: When is a conjugacy class a subgroup?

Answer 1

You want to show that $p(61)=1121505$ where $p(n)$ is the partition function, OEIS sequence A000041, tabulated here. You can calculate the values recursively using Euler's recurrence: $p(n)=0$ for $n\lt0;\ $ $p(0)=1;$ and, for $n\gt0,$ $$p(n)=p(n-1)+p(n-2)-p(n-2-3)-p(n-3-4)+p(n-3-4-5)+p(n-4-5-6)-p(n-4-5-6-7)-p(n-5-6-7-8)+\cdots$$$$=p(n-1)+p(n-2)-p(n-5)-p(n-7)+p(n-12)+p(n-15)-p(n-22)-p(n-26)+\cdots$$

So $p(1)=p(0)=1,$
$p(2)=p(1)+p(0)=1+1=2,$
$p(3)=p(2)+p(1)=2+1=3,$
$p(4)=p(3)+p(2)=3+2=5,$
$p(5)=p(4)-p(3)-p(0)=5+3-1=7,$
$p(6)=p(5)+p(4)-p(1)=7+5-1=11,$
$p(7)=p(6)+p(5)-p(2)-p(0)=11+7-2-1=15,$
$p(8)=p(7)+p(6)-p(3)-p(1)=15+11-3-1=22,$
$p(9)=p(8)+p(7)-p(4)-p(2)=22+15-5-2=30,$
$p(10)=p(9)+p(8)-p(5)-p(3)=30+22-7-3=42,$
$p(11)=p(10)+p(9)-p(6)-p(4)=42+30-11-5=56,$
$p(12)=p(11)+p(10)-p(7)-p(5)+p(0)=56+42-15-7+1=77,$
$p(13)=p(12)+p(11)-p(8)-p(6)+p(1)=77+56-22-11+1=101,$
etc.