Number of Partitions of  No Part Appears Exactly Once So I am having a little trouble here. The question is 
"Let $r_n$ be the Number of Partitions of $n$ No Part Appears Exactly Once. Find the generating function."
I realize there is already an answer here, giving the generating function. But I don't understand how this person got the generating function. The answer can be found:
Number of Partitions of $n$ No Part Appears Exactly Once Equals to $n$ Partitioned into 0,2 ,3 or 4 (Mod 6 )
But, it is somewhat asserted without showing this. I also am having some trouble finding some resources online as well. So, usually I would like to find some type of recurrence but I don't think that would be efficient in this case. So I am a little confused. Furthermore, this isn't a homework question. This is an old qual question I found. 
 A: 
I realize there is already an answer here, giving the generating function. But I don't understand how this person got the generating function.

It might help to add some variables to the generating function:
$((z_1x)^0 + (z_1x)^2 + (z_1x)^3 + \cdots)
 ((z_2x^2)^0 + (z_2x^2)^2 + (z_2x^2)^3 + \cdots)
 ((z_3x^3)^0 + (z_3x^3)^2 + (z_3x^3)^3 + \cdots) \cdots \\
= (1 + (z_1x)^2 + (z_1x)^3 + \cdots)
 (1 + (z_2x^2)^2 + (z_2x^2)^3 + \cdots)
 (1 + (z_3x^3)^2 + (z_3x^3)^3 + \cdots) \cdots$
So, for example, we have a term $(z_1x)^3 (z_3x^3)^2 = z_1{}^3 z_3{}^2 x^9$ corresponding to the partition $1 + 1 + 1 + 3 + 3 = 9$.
Now set all of the $z_i$ to $1$ to get the univariate generating function.
Then we can express it more elegantly using $\prod$ notation and the well-known expression for a geometric sum: $$\begin{eqnarray}
\prod_{i=1}^\infty (1 + (x^i)^2 + (x^i)^3 + \cdots)
&=& \prod_{i=1}^\infty \left(1 + (x^i)^2 \frac{1}{1 - x^i}\right) \\
&=& \prod_{i=1}^\infty \left(1 +  \frac{x^{2i}}{1 - x^i}\right) \\
&=& \prod_{i=1}^\infty \frac{1 - x^i + x^{2i}}{1 - x^i} \\
&=& \prod_{i=1}^\infty \frac{(1 - x^i + x^{2i})(1+x^i)}{(1 - x^i)(1+x^i)} \\
&=& \prod_{i=1}^\infty \frac{1 + x^{3i}}{1 - x^{2i}} \\
\end{eqnarray}$$

So, usually I would like to find some type of recurrence but I don't think that would be efficient in this case.

Quite a lot of partition-related sequences have nice recurrences, essentially all due to Euler's pentagonal theorem (and its generalisation, Jacobi's triple product).
Euler's theorem is $$\prod_{n=1}^\infty (1 - x^n) = 1 + \sum_{k=1}^\infty (-1)^k x^{k(3k \pm 1)/2}$$
This is useful because the partition numbers $P(n)$ have g.f $\prod_{n=1}^\infty \frac{1}{1 - x^n}$, or in other words $$\frac{1}{\sum_n P(n) x^n} = 1 + \sum_{k=1}^\infty (-1)^k x^{k(3k \pm 1)/2}$$ which rearranges to $$\sum_n P(n) x^n = 1 - \left(\sum_n P(n) x^n\right) \left(\sum_{k=1}^\infty (-1)^k x^{k(3k \pm 1)/2}\right)$$ and gives rise to the famous recurrence $$P(n) = P(n-1) + P(n-2) - P(n-5) - P(n-7) + \cdots$$
Now, partitions into distinct parts, $Q(n)$, have g.f. $\prod_{n=1}^\infty (1 + x^n)$ and various recurrences: $$\begin{eqnarray}
Q(n) &=& (-1)^j \left[j \in \mathbb{N}, n = \frac{j(3j\pm 1)}{2}\right] - 2 \sum_{k=1}^{\lfloor \sqrt n \rfloor} (-1)^k Q(n - k^2) \\
&=& P(n) + \sum_{k=1}^\infty (-1)^k P(n - k(3k \pm 1)) \\
&=& (-1)^j [j \in \mathbb{N}, n = j(3j\pm1)] - \sum_{k=1}^\infty (-1)^k Q\left(n - \frac{k(3k \pm 1)}{2} \right)
\end{eqnarray}$$ where $[\textrm{expr}]$ is an Iverson bracket, evaluating to $1$ if the expression is true and $0$ otherwise. The first two recurrences are in OEIS; the third I derived, but is probably not original.
The point of bringing up $Q$ is that our generating function of interest is a product of an inflated version of $Q$'s generating function and an inflated version of $P$'s generating function. I'm going to call it $A$ for "At least two". Then $$\begin{eqnarray}
\sum_{n=0}^\infty A(n) x^n &=& \prod_{i=1}^\infty \frac{1 + x^{3i}}{1 - x^{2i}} \\
\left(\sum_{n=0}^\infty A(n) x^n\right)\left(\prod_{i=1}^\infty (1 - x^{2i})\right) &=& \prod_{i=1}^\infty (1 + x^{3i}) \\
\left(\sum_{n=0}^\infty A(n) x^n\right)\left(1 + \sum_{k=1}^\infty (-1)^k x^{k(3k \pm 1)}\right) &=& \sum_{k=0}^\infty Q(k)x^{3k} \\
\sum_{n=0}^\infty A(n) x^n &=& \sum_{k=0}^\infty Q(k)x^{3k} - \left(\sum_{j=0}^\infty A(j) x^j\right)\left(\sum_{k=1}^\infty (-1)^k x^{k(3k \pm 1)}\right) \\
A(n) &=& Q(\tfrac{n}{3}) [\tfrac{n}{3} \in \mathbb{N}] - \sum_{k=1}^\infty (-1)^k A(n - k(3k \pm 1))
\end{eqnarray}$$
