I have to prove that $p(n)=p(n-1)+\displaystyle\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}p_{k}(n-k)$ and I am quite stuck on it... My first intuition was that, as $p(n)-p(n-1)$ is the number of partitions of $n$ that don't have $1$ in them, proving the identity was the same as proving that the number of partitions of $n$ without $1$ is $\displaystyle\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}p_{k}(n-k)$. Nevertheless, I wasn't able to continue. I also thought about how, using the known recurrence formulas with partitions, could I simplify it; as $p_{k}(n)=p_{k-1}(n-1)+p_{k}(n-k)$, I could write $\displaystyle\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}p_{k}(n-k)=\displaystyle\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}p_{k}(n)-\displaystyle\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}p_{k-1}(n-1)$. I also know that $p(n)=\displaystyle\sum_{k=1}^{n} p_{k}(n)$ and $p(n-1)=\displaystyle\sum_{k=1}^{n-1} p_{k}(n-1)$... I suppose that, with all this, I could arrive somewhere... But, I haven't been able to do it. A hint, or some help, would help a lot! Thanks in advanced!
Note: $p(n)$ denotes the number of partitions of $n$, and $p_{k}(n)$ denotes the number of partitions of $n$ with exactly $k$ summands.
You are correct that $p(n)-p(n-1)$ is the number of partitions of $n$ with no part equal to $1$. We just need to prove that the number of partitions of $n$ with no part equal to $1$ is also $\sum_{k=1}^{n/2}p_k(n-k)$.
The idea is to find a bijection between these two sets: $$ \left\{\begin{matrix} \text{partitions of $n$ with $k$ parts}\\ \text{with no part equal to $1$.}\end{matrix}\right\} \longleftrightarrow \left\{\begin{matrix} \text{partitions of $n-k$}\\ \text{with exactly $k$ parts.}\end{matrix}\right\} $$ Once you have this, you are done by summing over $k$.
Can you think of a bijection between these two sets? Further hint behind a spoiler in case you need it: