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Prove that the number of ways to partition $n\in\Bbb N$ into distinct even summands is equal to the number of ways of partitioning $n$ into (not necessarily) distinct summands of the form $4k-2,k\in\Bbb N$.

My thoughts:

It appeared to me, if $$n=\sum_{k=1}^pa_k,$$ where $a_1,\ldots,a_p$ are distinct even numbers, then $$\frac{n}2=\displaystyle\sum_{k=1}^pb_k, \quad b_k=\frac{a_k}2,$$ and $b_1,\ldots,b_p$ should be distinct summands.

On the other hand, $$n=\sum_{k=1}^r c_k,$$ where $c_k=4m_k-2,m_k\in\Bbb N,$ so $$\frac{n}2=\sum_{k=1}^rd_k,d_k=\frac{c_k}2=2m_k-1,m_k\in\Bbb N,$$ so we might use the fact that the number of ways to partition $n\in\Bbb N$ into distinct summands is equal to the number of ways to partition $n$ into odd summands, but this doesn't seem convincing enough.

I thought we could compare the generating functions $$f_1(X)=\prod_{k=1}^\infty(1+x^{2k})=\prod_{k=1}^\infty(1+x^{4k-2})(1+x^{4k})$$ and maybe $f_2(X)=\displaystyle\prod_{k=1}^\infty\frac1{1-x^{4k-2}},$ but I couldn't get any further.

I also checked this answer about partitions of $n$ where no part is divisible by $d$ and partitions of $n$ in which no part occurs more than $d-1$ times, hoping it could help.

How can I approach this problem?

Matcha Latte
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  • This is equivalent (after division by $2$) to the statement that the number of partitions of $n$ into distinct parts is equal to the number of number of partitions of $n$ into odd parts. A bijective proof is at https://math.stackexchange.com/questions/54961/the-number-of-partitions-of-n-into-distinct-parts-equals-the-number-of-partiti though there are other approaches which work too including some mentioned in Wikipedia and OEIS A000009 – Henry Jan 10 '22 at 19:53
  • Note that the generating functions approach is almost immediate. You just have to manipulate the series accordingly. – Calvin Lin Jan 10 '22 at 20:10
  • @Henry, yes, I included that in one part of the post. – Matcha Latte Jan 10 '22 at 20:14

2 Answers2

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We obtain \begin{align*} \color{blue}{\prod_{k=1}^{\infty}\left(1+x^{2k}\right)} &=\prod_{k=1}^{\infty}\frac{\left(1+x^{2k}\right)\left(1-x^{2k}\right)} {\left(1+x^{k}\right)\left(1-x^{k}\right)}\tag{1}\\ &=\prod_{k=1}^{\infty}\frac{\left(1+x^{2k}\right)\left(1-x^{2k}\right)} {\left(1+x^{2k}\right)\left(1+x^{2k-1}\right)\left(1-x^{2k}\right)\left(1-x^{2k-1}\right)}\tag{2}\\ &=\prod_{k=1}^{\infty}\frac{1} {\left(1+x^{2k-1}\right)\left(1-x^{2k-1}\right)}\tag{3}\\ &\,\,\color{blue}{=\prod_{k=1}^{\infty}\frac{1} {1-x^{4k-2}}}\\ \end{align*} and the claim follows.

Comment:

  • In (1) we use $1-x^{2k}=(1+x^k)(1-x^k)$.

  • In (2) we use $\prod_{k=1}^{\infty}\left(1+x^k\right)=\prod_{k=1}^{\infty}\left(1+x^{2k}\right)\left(1+x^{2k-1}\right)$.

  • In (3) we do some cancellation.

Markus Scheuer
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I think I have the answer using generating functions:

As mentioned before, if $p_1(n)$ is the number of partitions of $n\in\Bbb N$ into distinct summands, and $p_2(n)$ number of partitions of $n\in\Bbb N$ into odd summands, then their respective generating functions $f_1(X)=\displaystyle\prod_{k=1}^\infty(1+X)$ and $f_2(X)=\displaystyle\prod_{k=1}^\infty\frac1{1-X^{2k-1}}$ are equal.

As in my post, $g_1(X)=\displaystyle\prod_{k=1}^\infty(1+X^{2k})$ is the generating functions for partitions of $n\in\Bbb N$ into distinct even summands and $g_2(X)=\displaystyle\prod_{k=1}^\infty\frac1{1-X^{4k-2}}$ is the generating function for partitions of $n\in\Bbb N$ into summands of the form $4k-2.$

Let $h(X)=X^2$

$$g_1(X)=\prod_{k=1}^\infty(1+(X^2)^k)=(f_1\circ h)(X)\\g_2(X)=\prod_{k=1}^\infty\frac1{1-(X^2)^{2k-1}}=(f_2\circ h)(X).$$

Now, $f_1=f_2\implies g_1=f_1\circ h=f_2\circ h=g_2.$

Matcha Latte
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