The following identity is presented an exercise:

[2.3] Exercise 3
Show that $$ \begin{align} p(n \mid \text{even parts}) = p(n/2) = p(n \mid \text{even number of each part}) \end{align} $$ That is, the number of integer partitions made of even parts is the same as the number of integer partitions such that we have an even number of each part. These are also the same as the number of integer partitions of \(n/2\).

Example

Take \(n = 6\). Then using only even parts, we can write

$$ \begin{align} 6 &= 6 \\ 6 &= 4 + 2 \\ 6 &= 2 + 2 + 2 \end{align} $$

For \(n/2 = 3\), we can partition it as

$$ \begin{align} 3 &= 3 \\ 3 &= 2 + 1 \\ 3 &= 1 + 1 + 1 \end{align} $$

which correspond to partitions of \(6\) in which each part appears an even number of times

$$ \begin{align} 6 &= 3 + 3 \\ 6 &= 1 + 1 + 1 + 1 + 1 + 1 \\ 6 &= 2 + 2 + 1 + 1 \end{align} $$

\(p(n \mid \text{even parts}) = p(n/2)\)

Note that if \(n\) is odd, then both sides are \(0\), since we can’t write an odd integer as a sum of even parts. Now suppose that we have a partition of \(n\) into even parts so

$$ \begin{align*} n = 2a_1 + 2a_2 + \cdots + 2a_k \end{align*} $$

Consider the following map: starting with even parts, divide each part by \(2\). Thus

$$ \begin{align*} \frac{n}{2} &= a_1 + a_2 + \cdots + a_k \end{align*} $$

This is exactly a partition of \(n/2\). Conversely, suppose we have the above partition of \(n/2\). Then consider multiplying every part by \(2\). Then

$$ \begin{align*} n = 2a_1 + 2a_2 + \cdots + 2a_k \end{align*} $$

This is the original partition of \(n\) we started with. Since the maps are inverses of each other then it’s a bijection and therefore, \(p(n \mid \text{even parts}) = p(n/2)\).

\(p(n \mid \text{even number of each part}) = p(n/2)\)

Next, we show that \(p(n \mid \text{even number of each part}) = p(n/2)\). Suppose that we have a partition of \(n\) such that we have an even number of each part. Hence each part must occur at least two times in the partition and we can write

$$ \begin{align} n = 2(a_1+\cdots+a_1) + 2(a_2+\cdots+a_2) + \cdots + 2(a_k+\cdots+a_k) \end{align} $$

Then replace each pair of \(a_i\)s with a single \(a_i\). Hence we get a partition of \(n/2\) as follows

$$ \begin{align} n/2 = (a_1+\cdots+a_1) + (a_2+\cdots+a_2) + \cdots + (a_k+\cdots+a_k) \end{align} $$

Conversely, the inverse of this map is duplicating each part \(a_i\) in the partition. Therefore, this map is a bijection.

References