Given $n\in\Bbb{N}$, the number of (unrestricted) [integer partitions](https://en.wikipedia.org/wiki/Partition_(number_theory)) 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](http://oeis.org/wiki/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?
>  $$\sum\_{n\geq0}cp\_k(n+1)x^n=\frac{A\_k(x)}{(1-x)^{k+1}}\prod\_{j\geq1}\frac1{1-x^j}.$$
> 
> 
> 



Collapsed partitions and generating functions
It is easy to see that
$$cp\_k(n) = \sum\_{i=1}^n p(n-i)\cdot i^k,$$
where $p(n-i)$ stands for the number of (collapsed) partitions of $n$ that contain $i$ as a part. 


Since $i^k$ is the coefficient of $x^{i-1}$ in $\frac{A\_k(x)}{(1-x)^{k+1}}$, we conclude that $cp\_k(n)$ equals the coefficient of $x^{n-1}$ in $\frac{A\_k(x)}{(1-x)^{k+1}}\cdot \prod\_{j\geq 1}\frac{1}{1-x^j}$. QED


