Let $p_0^d (n)$ be the number of partitions of n into distinct odd parts. Show that $$p(n) \equiv p_0^d(n) \pmod 2$$

So I have to show that the total number of partitions has the same parity as the number of partitions of n into distinct odd parts. I know that there are equal amounts of even and odd partitions, but other than that don't really know how to approach this.

  • $\begingroup$ If you want to find out about ideas and tools for treating problems about integer partitions, the book Integer Partitions by G. Andrews and K. Eriksson, (Cambridge U. Press, 2004) is a very nice account. $\endgroup$ Apr 28, 2011 at 12:57
  • 2
    $\begingroup$ This is really an aside, but anyways, here professor Ken Ono talking about integer partitions in a popularized talk youtube.com/watch?v=aj4FozCSg8g $\endgroup$ Apr 28, 2011 at 13:46

1 Answer 1


You can represent a partition $k_1+\dots +k_m=n$, $k_1\geq\dots\geq k_m$ by its Young diagram: a row of $k_1$ squares, below it a row of $k_2$ squares ... and finally a row of $k_m$ squares.

A partition to different odd numbers $l_1>\dots >l_p$ can also be represented by a Young diagram: a hook of $l_1$ squares (if $l_1=2q_1+1$, the hook has one square in the corner and $q_1$ squares horizontally and $q_1$ squares vertically); then another hook of $l_2$ squares, etc. The diagrams obtained in this way are invariant wrt. the diagonal reflection.

The diagrams which are not invariant wrt. the diagonal refection come in pairs (a diagram and it reflection), so the number of all diagrams (with $n$ squares) is $2\times$something plus the number of invariant diagrams.

edit: a picture will say more:


  • $\begingroup$ +1 though I think I might have worded it slightly differently, saying that the number of partitions with Ferrers diagrams which are self-conjugate (invariant under a diagonal reflection, such as $5+4+4+3+1$ in your second picture) is equal to the number of partitions into distinct odd parts (such as $9+5+3$ in your picture), and the number of partitions which are not self-conjugate is even. $\endgroup$
    – Henry
    Apr 28, 2011 at 16:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .