<s>[INST] Number of Sets of Partitions I looked at the partitions of numbers, like let's say $n=5$. You get
$$
\begin{eqnarray}
5&=&5\\
\hline
&=&4+1\\
&=&3+2\\
\hline
&=&3+1+1\\
&=&1+2+2\\
\hline
&=&2+1+1+1\\
\hline
&=&1+1+1+1+1\\
\end{eqnarray}
$$
where I grouped the partitions according to their distribution (i.e. appearance) of summands. So you get $5$ sets.
Is it possible to get the number of sets for general $n$?
ANOTHER EDIT, thanks to Brian
If $\pi$ is a partition of $n$, let $M_\pi$ be the multiset of pieces, and let $\sigma_\pi$ be the sequence of multiplicities of $M_\pi$ listed in non-decreasing order. Then partitions $\pi$ and $\pi'$ are in the same set if $\sigma_\pi=\sigma_{\pi'}$. Thus, $\pi=1+3+3$ and $\pi'=2+2+3$ are in the same set, because $M_\pi=[1,3,3]$, so $\sigma_\pi=\langle 1,2\rangle$, and $M_{\pi'}=[2,2,3]$, so $\sigma_{\pi'}=\langle 1,2\rangle$ as well.
EDIT 
Here's another example for $n=6$:
$$
\begin{eqnarray}
6 = 6\\
\hline
5 + 1 = 6\\
4 + 2 = 6\\
\hline
3 + 3 = 6\\
\hline
4 + 1 + 1 = 6\\
\hline
3 + 2 + 1 = 6\\
\hline
2 + 2 + 2 = 6\\
\hline
3 + 1 + 1 + 1 = 6\\
\hline
2 + 2 + 1 + 1 = 6\\
\hline
2 + 1 + 1 + 1 + 1 = 6\\
\hline
1 + 1 + 1 + 1 + 1 + 1 = 6
\end{eqnarray}
$$
so we have $10$ sets...
 [/INST] As pointed out, there is no easy closed form for the partition function without any restrictions. However, given some restrictions, there are some nicer closed forms. For example, if we let a composition of $n$ be a partition of $n$ in which order matters, then let $P(n)$ be the number of compositions of $n$ which only use $1$ and $2$ as its parts and let $Q(n)$ be the number of compositions of $n$ which only use odd parts. Then, it can be easily proven that $P(n)=F_{n+1}$ and $Q(n)=F_n$, where $F_n$ is the $n$th Fibonacci number. 
</s>