For each natural number $ n $ we consider the equation $$x_{1}+2x_{2}+\dots+nx_{n}=n$$ Where $x_{1},\dots,x_{n}$ are nonnegative integers. Prove that this equation has the same number of solutions that satisfy condition 1 as solutions that satisfy condition 2.
- For each $k\in\{1,2,\dots,n-1\}$, $x_{k} >0$ or $x_{k+1}=0$
- For each $k\in\{1,2,\dots,n-1,n\}$, $x_{k}=0$ or $x_{k}=1$
I proved this using Ferrers diagram and showing bijection, however, I'm curious if this can be solved by the method of generating functions. Thanks in advance for all your help.
Here's a start. The generating function for the partitions that satisfy condition 2 (also known as partitions into distinct parts) is $\prod_{k=1}^\infty (1+z^k)$.
For condition 1, conditioning on the largest part $k$ yields generating function $$ \sum_{k=0}^\infty \prod_{i=1}^k \left(z^i + z^{2i} + z^{3i} + \cdots \right) = \sum_{k=0}^\infty \prod_{i=1}^k \frac{z^i}{1-z^i}. $$