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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...

draks ...
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  • I do not know if i understand it quite well. But it seems to be $n$ sets. See the partitions that looks like $\underbrace{1+\ldots+1}_{\text{$k$ times}}+(n-k)$. – Phicar Dec 15 '14 at 21:11
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    There is no nice closed form for the partition function, if that's what you're wondering. There are many equivalent expressions, including the generating function. There are approximate estimates, and other known properties. – vadim123 Dec 15 '14 at 21:16
  • @Brian Do you have any idea towards a solution? – draks ... Dec 20 '14 at 19:49

1 Answers1

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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.

Ahaan S. Rungta
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