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Identity involving partitions of even and odd parts.

How would I go about to show the following: Let $pe(n)$ be the number of partitions of size n with an even number of parts and let $po(n)$ be the number of partitions of size n with an odd number of parts. If $od(n)$ is the number of partitions of size n with odd and distinct parts, show that for all $n$, $pe(n) - po(n) = (-1)^nod(n)$.

My idea was the following: Getting the generating functions for the 3 sets of partitions above, then try extracting the coefficient of $x^n$, denoted by $[x^n]$, then simply manipulate them to get the desired result. However, the generating functions are all of the form: $\prod_{i=1}^{\infty}\frac{1}{1-x^i}$ (generating function for the set of all partitions), and I have no idea how to extract the coefficients for such a function (and whether or not it's possible using relatively basic principles).

Can anyone point me in the right direction? I'm sure that there must be some better way to do this. Thanks!

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Nizbel99
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    You’ll find two complete solutions at the older question, one with generating functions and one of a more combinatorial nature. – Brian M. Scott Nov 11 '12 at 06:44

1 Answers1

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A few hints to get you started.

There is actually a direct generating function for $p_e(n) - p_o(n)$. It is given by $$\prod_{n\ge1}\frac{1}{1 + x^n} = \sum_{n\ge0}(p_e(n)-p_o(n))x^n$$ Can you see why? (If you are familiar with a proof of Euler's pentagonal formula, then this should be somewhat familiar)

Similarly, there is a generating function for the number of partitions into odd distinct parts. It is given by $$\prod_{k\ge 0}\left(1+x^{2k+1}\right) = \sum_{n\ge 0}p_{od}(n)x^n$$ Now, every partition of an odd number into odd parts necessarily contains an odd number of terms. Every partition of an even number into odd parts necessarily contains an even number of terms. So if we modify the above slightly by introducing a negative sign, there will be no unwanted cancellation and the net effect is $$\prod_{k\ge 0}\left(1-x^{2k+1}\right) = \sum_{n\ge 0}(-1)^np_{od}(n)x^n$$ Your job now is to prove that the two generating functions are equal. Algebraic manipulations will be enough for this purpose, there is no need to extract the coefficients.

EuYu
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  • I'm not familiar with Euler's Pentagonal Theorem. How might I go about to change this product into a sum? For example, in the first formula provided. – Nizbel99 Nov 11 '12 at 05:15
  • I've looked at that theorem actually, and it looks very simliar to what I need... but the proof seems quite complex. I've found another proof in a chapter of my textbook that we didn't cover, but it gives a 5 and a half page bijective proof =( - I didn't know this was such a complicated problem. I will try to see if I can understand it though :) - Thanks for the point of reference. – Nizbel99 Nov 11 '12 at 05:20
  • Euler's pentagonal formula is more or less irrelevant, there is just one step in there which is reminiscent of the first generating function. If you expand the first generating function, you have $$\prod_{k\ge 1}\left(1 - x^k + x^{2k} - x^{3k} + \cdots\right)$$ Every partition with an even number of parts contributes $+1$ to the coefficient of $x^n$ and every partition with an odd number of parts contributes a $-1$ to the coefficient. Can you see why we have then sum then? – EuYu Nov 11 '12 at 05:35
  • Sort of... I'm still trying to figure out how you got that product to begin with as well. I think I should be okay to figure things out though... not sure. Trying to derive that result right now. – Nizbel99 Nov 11 '12 at 05:55
  • It's just a product of geometric series. – EuYu Nov 11 '12 at 05:59
  • I'm trying to generate the generating function for $p_e(n)$ and $p_o(n)$, then subtract them from one another to see if I get the product you got. – Nizbel99 Nov 11 '12 at 06:02