Let $l_n$ be the number of partitions of $n$ which have exactly two parts of size $1$. Let $L(x) = \sum l_n x^{n}$ be the generating function for $(l_n)$. (Let $l_0 = 1$ and $l_1 = l_2 = l_3 = 0.)$ Give an argument using only generating functions for why for $n \geq 3$, $l_n = p_{n−2} − p_{n−3}.$ where $p_n$ is the number of partitions of $n$.
Attempt: I've tried drawing Ferrer's shapes but could not arrive at a conclusion. Any help is appreciated.
Let $P(x)=\sum_{n\ge 0}p_nx^n$ be the generating function for the partition numbers. I expect that you know that
$$P(x)=\prod_{k\ge 1}\frac1{1-x^k}\;.$$
If we restrict ourselves to partitions that don’t have any parts of size $1$, the generating function becomes
$$P_{\ge 2}(x)=\prod_{k\ge 2}\frac1{1-x^k}\;;$$
we just lose the first factor. Thus, $P(x)=\frac1{1-x}P_{\ge 2}(x)$, and $P_{\ge 2}(x)=(1-x)P(x)$.
There is an easy bijection between the set of partitions of $n$ with exactly two parts of size $1$ and the set of partitions of $n-2$ with no parts of size $1$, so $\ell_n$ is the coefficient of $x^{n-2}$ in $P_{\ge 2}(x)$. That is the coefficient of $x^n$ in
$$x^2P_{\ge 2}(x)=x^2(1-x)P(x)=x^2(1-x)\sum_{n\ge 0}p_nx^n\;.$$
Move the factor of $x^2$ inside the final summation, and you’re almost there.