Adopt the standard notation for integer partitions, writing $\lambda\_1^{a\_1} \cdots \lambda\_k^{a\_k}$ as shorthand for the partition $a\_1 \lambda\_1 + \cdots + a\_k \lambda\_k$ with parts $\lambda\_1 > \cdots > \lambda\_k \geq 1$ and multiplicities $a\_i \geq 1$. 


If $\lambda = \lambda\_1^{a\_1} \cdots \lambda\_k^{a\_k} \vdash n$, its *collapsed image* is the partition with multiple occurrences of parts removed, and is denoted by $\underline{\lambda} := \lambda\_1 \cdots \lambda\_k$. Introduce the set $\underline{\mathcal{P}\_n}=\{\,\underline{\lambda}: \,\lambda\vdash n\}$ as well as the all-familiar 
$\mathcal{P}\_n=\{\,\lambda: \,\lambda\vdash n\}$ 



> 
> **QUESTION.** Is this true? If so, is there a combinatorial proof?
>  $$\#\underline{\mathcal{P}\_n}=\sum\_{j=0}^{n-1}\#\mathcal{P}\_j.$$
> **CORRECTED.**
> $$\sum\_{\underline{\lambda}\in\underline{\mathcal{P}\_n}}\text{length}(\underline{\lambda})=\sum\_{j=0}^{n-1}\#\mathcal{P}\_j.$$
> 
> 
> 


**Remark.** For a related problem, [see here](https://mathoverflow.net/questions/309098/collapsed-partitions-and-generating-functions):



Collapsed partitions and ordinary partitions
The result does not seem to be stated correctly. For instance, $\underline{\mathcal{P}\_3}=\{ 3,21,1\}$, yet $p(0)+p(1)+p(2)=4$, where $p(n)=\#\mathcal{P}\_n$. It is true that $$\sum\_{\lambda\in\underline{\mathcal{P\_n}}}\ell(\lambda)=p(0)+p(1)+\cdots+ p(n-1),$$ where $\ell(\lambda)$ is the number of parts of $\lambda$. This is equivalent to the case $k=1$ of *Enumerative Combinatorics*, vol. 1, second ed., Exercise 1.80, which has a simple combinatorial proof.


