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: