> [!NOTE] > Let $n$ be a natural number. The number of [[Integer Partition|partitions]] of $n$ into odd parts is equal to the number of partitions of $n$ into distinct parts. ###### Proof \[MA241\] By this same [[Ordinary Generating Function of Partition Function|reasoning]], the generating functions of $n$ into odd parts and distinct parts are given by $\prod_{i \text{ odd}} \frac{1}{1-x^i} \quad \text{ and } \quad \prod_{i=1}^\infty (1+x^i)$respectively. Applying difference of two squares yields $1+x^i = \frac{1-x^{2i}}{1-x^i}$, we have $\begin{align} \prod_{i \geq 1} (1 + x^i) &= \prod_{i\geq 1} \frac{1-x^{2i}}{1-x^i} \\ &= \frac{(1-x^2)(1-x^4)(1-x^6)\cdots}{(1-x)(1-x^2)(1-x^3)\cdots} \\ &= \frac{1}{(1-x)(1-x^3)(1-x^5)\cdots} \\ &= \prod_{i \text{ odd}} \frac{1}{1-x^i}. \end{align}$ ###### Sylvester's Bijective Proof \[MA241\]