Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathcal{A}$ be the set of partitions in which each part may occur 0, 1, 4, or 5 times and let $\mathcal{B}$ be the set of partitions which have no parts congruent to 2mod4, and in which parts divisible by four occur at most once each. Then for any $n \in \mathbb{N}$, I need to show that $|\mathcal{A}_n| = |\mathcal{B}_n|$. In other words, I need to show that the number of partitions in $\mathcal{A}$ of size n equals the number of partitions $\mathcal{B}$ of size n.

What I've done so far was the following. I've managed to derive both generating functions, namely:

$$\begin{equation*}\Phi_\mathcal{A} = \prod_{j=1}^{\infty}(1 + x^j + x^{4j} + x ^{5j})\end{equation*}$$

$$\begin{equation*}\Phi_\mathcal{B} = \prod_{j=1}^{\infty}(1 + x^{4j})\left(\frac{1}{1-x^{4j-3}}\right)\left(\frac{1}{1-x^{4j-1}}\right)\end{equation*}$$

Furthermore, in order to show that $|\mathcal{A}_n| = |\mathcal{B}_n|$ for any $n \in \mathbb{N}$, it suffices to show that the two generating functions are equal, $\Phi_\mathcal{A} = \Phi_\mathcal{B}$. However, I am unsure how to go about to manupulate them in order to show this final desired result. Any help would be appreciated.


share|cite|improve this question
Try using the Taylor series expansion for $\frac{1}{1-u}$ to simplify $\Phi_B$.… – Austin Mohr Nov 13 '12 at 22:11
I've also noticed that $(1+x^j+x^{4j}+x^{5j})$ factors to $(1+x^j)(1+x^{4j})$. So it would seem to be that it would be sufficient to show that $(1+x^j)$ is equal to the product of the two fractional portions of $\Phi_\mathcal{B}$, which doesn't seem to be the case at all (even if one does consider the Taylor expansion)? – Nizbel99 Nov 13 '12 at 22:33
up vote 3 down vote accepted

Yes, your own observation nicely works. Start with ${\cal B}$.

First use the trick also mentioned in wikipedia (for partitions in odd parts) to show that

$\frac{1}{(1-x)(1-x^3)(1-x^5)(1-x^7)\dots} =$ $ (1+x)(1+x^2)(1+x^3) \dots$; this is the product of the fractions of odd powers $\prod(\frac{1}{1-x^{4j-3}})(\frac{1}{1-x^{4j-1}})$.

Now you have to multiply that result with the multiples of $4$, i.e, $(1+x^4)(1+x^8)(1+x^{12}) \dots$, combining $(1+x^j)(1+x^{4j})$ -- here $(1+x^j)$ is a term from the previous computation and $(1+x^{4j})$ one of the multiples of $4$. Voila.

share|cite|improve this answer
I looked at the article you mentioned, and I understand the trick with regards to the example on Wikipedia. However, I don't see how I would use it here. Could you please explain a little more? Say I consider $\frac{1}{1-x^{4j-3}}$, you're suggesting that I multiple top and bottom by $(1+x^4)(1+x^8)...$, but I don't see how that all cancels out with the other fractional portion to become $(1+x^j)$. – Nizbel99 Nov 14 '12 at 1:05
I feel like I may be missing something obvious. – Nizbel99 Nov 14 '12 at 1:10
OK. I tried to add some explanation where the factors are coming from. You take both $4j-3$ and $4j-1$ together, and you get all odd numbers, at least that seems to bother you? (But now I go to bed.) – Hendrik Jan Nov 14 '12 at 1:30
Thanks! - I follow now :) – Nizbel99 Nov 14 '12 at 1:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.