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We have a famous partiton identity which states "the number of partitions of $n$ into odd parts equals the number of partitions of $n$ into distinct parts". This is a famous result of Euler on partitions.. I know the proof using generating functions. But how can we prove the number of partitions of $n$ into odd parts equals the number of partitions of $n$ into distinct parts using Ferrers graph

Megha Goyal
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  • @JackD'Aurizio: I think the proof in the question you linked to is precisely the generating functions proof that the OP knows, and not the Ferrers graph one (whatever that is) that is being asked. – Willie Wong Sep 23 '16 at 14:00
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    @WillieWong: sorry, I pasted the wrong link: https://math.berkeley.edu/~mhaiman/math172-spring10/partitions.pdf – Jack D'Aurizio Sep 23 '16 at 14:03
  • @JackD'Aurizio: I am tempted to ask you to post that as an answer; but someone will then come by and tell you to "include the gist of the answer in the body of the answer" and not just post link only. But looking at the Ferrers diagrams, I am not sure if there are easy and good ways to type them here on Math.SE. – Willie Wong Sep 23 '16 at 14:08
  • That link does not answer the question and it is not a duplicate. Mark's notes prove "distinct, odd = self-conjugate" bijectively with diagrams. They prove "distinct = odd" in the traditional algebraic way and say, "An interesting exercise is to find a bijective proof of the above theorem". This "interesting exercise" is what OP was asking about. Nonetheless, the bijection in the question which this is not literally a duplicate of can be artificially be turned into a diagrammatic proof. It's likely my own ignorance, but I don't know of a natural proof with Ferrers diagrams. – Joshua P. Swanson May 04 '20 at 10:12

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