# Result on partitions with distinct odd parts

Let $$pdo(n)$$ be the number of partitions of n into distinct odd parts. Then $$p(n)$$ is odd if and only if $$pdo(n)$$ is odd.

I am well aware that a proof of this is available here but I want to do it algebraically using generating functions.

So $$pdo(n)$$ $$=$$ $$\prod_{i=odd} (1+x^i)}$$

I know that I can do this by showing that $$pdo(n)$$ $$\equiv$$ $$p(n)$$ $$\mod 2$$ but I am not sure how I can do that. Any hint or help would be appreciated.

## 1 Answer

You have g.f. $$\prod_{i=0}^\infty (1 + x^{2i+1})$$ for partitions into distinct odd parts. The g.f. for general partitions is $$\frac{1}{\prod_{i=1}^\infty(1 - x^i)}$$ So $$pdo(n) \equiv p(n) \pmod 2$$ iff $$\prod_{i=0}^\infty (1 + x^{2i+1}) - \frac{1}{\prod_{i=1}^\infty(1 - x^i)}$$ has only even coefficients.

Now, $$\begin{eqnarray} \prod_{i=0}^\infty (1 + x^{2i+1}) - \frac{1}{\prod_{i=1}^\infty(1 - x^i)} &=& \frac{\prod_{i=1}^\infty (1 + x^i)}{\prod_{i=1}^\infty (1 + x^{2i})} - \frac{1}{\prod_{i=1}^\infty(1 - x^i)} \\ &=& \frac{\prod_{i=1}^\infty (1 + x^i)\prod_{i=1}^\infty(1 - x^i) - \prod_{i=1}^\infty (1 + x^{2i})}{\prod_{i=1}^\infty (1 + x^{2i})\prod_{i=1}^\infty(1 - x^i)} \\ &=& \frac{\prod_{i=1}^\infty (1 - x^{2i}) - \prod_{i=1}^\infty (1 + x^{2i})}{\prod_{i=1}^\infty (1 + x^{2i})(1 - x^i)} \\ \end{eqnarray}$$

Is that enough to say something about the parity of the coefficients?