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I am working through the wonderful book Proofs and Confirmations by David Bressoud. In the section 2.2, I came across the following exercise, which has me scratching my head.

(2.2.8) ~ Let $a(m,n)$ be the number of representations of $m$ as the sum of integers in two partitions into distinct parts for which the first partition must have exactly $n$ more elements than the second, and the second is allowed to have $0$s. If $n$ is negative, then the second parition has exactly $-n$ more elements than the first. […] Find a bijection between the partitions counted by $a(m,n)$ and those counted by $a(m-n,-n)$.

[emphasis added]

My best guess at what this is saying is that we are looking at pairs of partitions of $m$ with some properties, and the relation between them is that one has $n$ more parts than the other. But then the statement is not true:

  • $a(4,1)=1$ because only $(\{3,1\}, \{4\})$ has both partitions with distinct parts such that the second contains more than the first.
  • but $a(3,-1)>1$ because both $(\{3\}, \{3,0\})$ and $(\{3\}, \{2,1\})$ have the same properties.

That said, there is a lot of language here that I'm not used to.

  • Does the word "representations" mean something special here?
  • What word is "into" modifying?
  • What is an "element" of a partition?

Probably this is really obvious, but right now I'm not seeing it. Any help is appreciated.

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  • $\begingroup$ I read this as: count the number of pairs of partitions of $m$ each of which has distinct parts and where the number of elements in the first is $n$ more than the number in the second. But I haven't checked whether that satisfies the relation claimed. $\endgroup$ – rogerl Jun 22 '14 at 18:24
  • $\begingroup$ @rogerl: I think this is the same version I was using: if I understand you correctly, the counterexample shows that the relation does not hold. $\endgroup$ – Eric Stucky Jun 22 '14 at 19:01
  • $\begingroup$ Actually, I still don't know what "number of elements" means. Particularly confusing because it also says "…two partitions into distinct parts…". $\endgroup$ – Eric Stucky Jun 22 '14 at 19:17
  • $\begingroup$ Is it possible that what he meant to say is that the number is the sum of all the numbers in both partitions? For example: a(4,1)=2 because 4 can be written as ({2,1},{1}) and ({3,1},{0}) and a(3,-1)=2 because 3 can be written as ({1},{2,0}) and ({2},{1,0}). If not, then I have no idea what could be the case here... $\endgroup$ – ryagami Jun 23 '14 at 14:52
  • $\begingroup$ I read this as counting pairs of partitions ($\lambda$ is a partition into distinct parts of $k$, $\mu$ is a partition into distinct parts of $l$) such that $k+l=m$, $\lambda$ has $n$ more parts than $\mu$ with the complication of the $0$s. $\endgroup$ – Peter Taylor Jul 9 '14 at 7:27

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