Given $n\in\Bbb{N}$, the number of (unrestricted) integer partitions of $n$ are given by $$\sum_{n\geq0}p(n)x^n=\prod_{j\geq1}\frac1{1-x^j}.$$ Define the collapsed partitions of $n$ to be the partitions of $n$ with multiplicities removed. For example, if $n=4$ then its partitions are $4, 31, 22, 211, 1111$. The collapsed partitions become $4, 31, 2, 21, 1$.

Denote the sum of the $k$-th powers of the collapsed partitions of $n$ by $cp_k(n)$. For example, $cp_1(4)=4+3+1+2+2+1+1=14$. The first few values of $cp_1(n)$ are: $$cp_1(1)=1, cp_1(2)=3, cp_1(3)=7, cp_1(4)=14, cp_1(5)=26.$$

Recall the Eulerian polynomials of type $A$ defined by $$\sum_{n\geq0}(n+1)^kx^n=\frac{A_k(x)}{(1-x)^{k+1}}.$$

Experiments prompt me to ask: is this true? $$\sum_{n\geq0}cp_k(n+1)x^n=\frac{A_k(x)}{(1-x)^{k+1}}\prod_{j\geq1}\frac1{1-x^j}.$$