# Prove that the number of self-conjugate partitions of $n$ equals the number of partitions of $n$ into distinct odd parts

First, I would love if someone can provide some clarification of this problem. Then possibly help me map out/begin a proof.

So If I were taking the number $$6$$ and partitioning for example (just to make sure I understand what the question is asking):

The only partition with distinct odd parts would be $$6=5+1$$. However, for self-conjugate partitions I understand when I flip over the middle diagonal the picture should look exactly the same? That would also only happen once.

How would I go about formulating a proof?

• Any thoughts on the answers that have been posted? Commented Dec 24, 2019 at 5:27
• Are you still here, Lil? Commented Dec 25, 2019 at 14:56
• It is rude to post a question and then ignore the answers. Commented Dec 27, 2019 at 17:53

Let's look at an example. It should be possible to work out the general case by careful inspection of this example. $$\matrix{A&A&A&A&A&A\cr A&B&B\cr A&B&C\cr A\cr A\cr A\cr}$$ This is the self-conjugate partition $$15=6+3+3+1+1+1$$, and it is also the partition into distinct odd parts $$15=11+3+1$$, $$11$$ copies of $$A$$, $$3$$ of $$B$$, $$1$$ of $$C$$.

The Wikipedia article is quite good on a proof.

You can see that $$x^{2n+1}=x^n\cdot x\cdot x^n$$ to form both 'legs' of a self-conjugate partition in a Ferrers diagram.

Or, if you travel along the main diagonal and read only to the right, we are looking at the number of partitions into distinct parts, $$\prod 1+x^k$$. We need two of these - $$\prod 1+x^{2k}$$ - to form the reflection when travelling downwards, and we also need to supply the diagonal - $$\prod 1+x^{2k}\cdot x$$.

It has been quite long, but I will like to turn Gerry's answer into a proof of the post.

Consider $$A_1$$ to be the set of all partitions of $$n$$ into distinct odd parts, and let $$A_2$$ be the set of all partitions of $$n$$ into self conjugate partitions.

Then, consider any $$\lambda \in A_2,$$ and let $$[ \lambda ]$$ be the Young diagram of $$\lambda.$$ Then, fill in the first row and the first column of $$[\lambda]$$ with colour $$c_1,$$ fill the uncoloured cells of second row and the second column of $$[ \lambda ]$$ with colour $$c_2$$ and so on. Then, notice that since $$\lambda$$ is self-conjugate, so for any $$i,$$ the size of the $$i$$ th row and the $$i$$ th column of $$[\lambda]$$ must be the same.

Thus, each colour is used an odd (even-1 [due to the corners]) number of times, and clearly no two colours are used the same number of times. (Why ? Think about it yourself, from the very structure a Young diagram has.) Now counting the number of cells of each colour, gives us a partition of $$n$$ into distinct odd parts.

The other direction (obtaining a self conjugate partition from a partition into distinct odd parts) is very similar. Just try to reach the configuration with the colours as described, and you will see how it happens.