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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?

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

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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 $C$.

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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 - $\prod 1+x^{2k}\cdot x$.

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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.

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