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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? $$\sum_{n\geq0}cp_k(n+1)x^n=\frac{A_k(x)}{(1-x)^{k+1}}\prod_{j\geq1}\frac1{1-x^j}.$$

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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

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  • $\begingroup$ I found the same answer, but 27 minutes after you. My sum ran from $i=1$ to $i=n$, and I simply said (in my deleted answer) that $i$ participates in exactly $p(n-i)$ stripped partitions, so the identity is clear. The notation $A_k(x)$ is irrelevant to the question and the answer. $\endgroup$ – GH from MO Aug 25 '18 at 16:33
  • $\begingroup$ @GHfromMO: I've changed summation to start with $i=1$ to avoid confusion. The answer is indeed quite straightforward. I do not quite follow your comment about $A_k(x)$ being irrelevant. $\endgroup$ – Max Alekseyev Aug 25 '18 at 16:39
  • $\begingroup$ What I meant is: you can ask the question and answer it without ever mentioning $A_k(x)$. The identity you state is equivalent to the OP's identity. $\endgroup$ – GH from MO Aug 25 '18 at 16:45

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