Prove that $p_2(n) = \left \lfloor{\frac{n}{2}}\right \rfloor+1$ using identity

Question

Prove that $p_2(n) = \left \lfloor{\frac{n}{2}}\right \rfloor+1$ using the identity $$\frac{1}{(1-x)(1-x^2)}=\frac{1}{2}\left(\frac{1}{(1-x)^2}\right)+\frac{1}{2}\left(\frac{1}{1-x^2}\right)$$

where $p_k(n)$ is the number of partitions of an integer $n$ into a most $k$ parts. The generating function $P_k(x)$ of $\{p_k(n)\}$ is $$P_k(x) = \frac{1}{\prod_{r=0}^{k}(1-x^k)}$$ Therefore, the generating function $P_2(x)$ of ${p_2(n)}$ is $$P_2(x) = \frac{1}{(1-x)(1-x^2)}$$ Now I see here that we can use the identity given above, but I am confused of how to apply it to prove the desired statement.

Answer 1

Note that $$\begin{align}\frac{4}{(1-x)(1-x^2)}&=\frac{1}{1+x}+\frac{1}{1-x}+\frac{2}{(1-x)^2}\\ &=\frac{1}{1+x}+\frac{1}{1-x}+\frac{d}{dx}\left(\frac{2}{1-x}\right)\\ &=\sum_{n=0}^{\infty}(-x)^n+\sum_{n=0}^{\infty}x^n +2\sum_{n=1}^{\infty}nx^{n-1}\\ &=\sum_{n=0}^{\infty}((-1)^n+1+2(n+1))x^n \end{align}$$ Can you take it from here?

Answer 2

You have $$P_2(x) = \frac{1}{2}\left(\frac{1}{(1-x)^2}\right)+\frac{1}{2}\left(\frac{1}{1-x^2}\right) = \frac12 (1-x)^{-2} + \frac12 (1-x^2)^{-1}$$ Now, the generalised binomial expansion is $(1+x)^a = 1 + \frac{a}{1!}x + \frac{a(a-1)}{2!}x^2 + \cdots$

Therefore $$\begin{eqnarray} (1 - x)^{-2} &=& 1 + \frac{(-2)}{1!}(-x) + \frac{(-2)(-3)}{2!}(-x)^2 + \cdots \\ &=& 1 + 2x + 3x^2 + \cdots \end{eqnarray}$$

and $$\begin{eqnarray}(1-x^2)^{-1} &=& 1 + \frac{(-1)}{1!}(-x^2) + \frac{(-1)(-2)}{2!}(-x^2)^2 + \cdots \\ &=& 1 + x^2 + x^4 + \cdots \end{eqnarray}$$

Combine and you're done. It may be helpful to consider separately the cases where $n$ is even vs odd.