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Use the notation $(n_1,n_2,\ldots,n_k) \vdash n$ to denote that $(n_1,n_2,\ldots,n_k)$ is a partition of the positive integer $n$, that is, $n_1+n_2+\ldots+n_k = n$ and $n_1 \ge n_2 \ge \ldots \ge n_k \ge 1$. Let $f=f(x_1,x_2,\ldots)$ be a multivariate function of countably many arguments.

Is there a way to simplify the following generating function over partitions: \begin{equation}\tag{$*$} \sum_{n=1}^{\infty} \left( \sum_{k=1}^n \sum_{(n_1,n_2,\ldots,n_k) \vdash n} f(n_1,n_2,\ldots,n_k) 1^{n_1} 2^{n_2} \ldots k^{n_k} \right) z^n, \end{equation} say, by making use of generating function of partition functions, or some magic identities in partition theory?

Thank you in advance for any suggestions or references.

Edit: Thanks to @SamHopkins and @Gro-Tsen for their observations, I put the term $z^n$ in the summand which makes the sum $(*)$ the form of a generating function. Also thanks to @Gro-Tsen's hint, the sum $(*)$ can be simplified for a special class of $f$ by making use of the conjugation of partitions. Let $(n'_1,n'_2,\ldots,n'_l)$ denote the conjugate partition of $(n_1,n_2,\ldots,n_k)$, that is, $n'_j := \# \{i: n_i\ge j \}$. Define $$ \hat f(n'_1,n'_2,\ldots,n'_l) := f(n_1,n_2,\ldots,n_k).$$ Then the sum $(*)$ becomes \begin{equation} (*) = \sum_{n=1}^{\infty} \left( \sum_{l=1}^n \sum_{(n'_1,n'_2,\ldots,n'_l) \vdash n} \hat f(n'_1,n'_2,\ldots,n'_l) n'_1! n'_2! \ldots n'_l! \right) z^n. \end{equation} Now if the new function $\hat f$ is a tensor power, that is, $$ \hat f(n'_1,n'_2,\ldots,n'_l) = a(n'_1) \ldots a(n'_l),$$ for some function $a$, then due to the first comment of this post, \begin{equation} \begin{split} 1+(*) &= \sum_{n=0}^{\infty} \left( \sum_{l=1}^n \sum_{(n'_1,n'_2,\ldots,n'_l) \vdash n} a(n'_1) n'_1! \ldots a(n'_l) n'_l! \right) z^n \\ &= \prod_{i=1}^\infty \frac{1}{1- a(i) i! z^i}. \end{split} \end{equation} In the special case that $f\equiv 1$, we have $a=1$ and the resulting function is the generating function of the OEIS sequence A077365. Another special case is that $a(i) = 1/i!$, where the resulting function is the generating function of the basic partition function.

But what about the general case where $\hat f$ is not a tensor power, e.g., $\hat f$ is a general tensor product like $ \hat f(n'_1,n'_2,\ldots,n'_l) = a_1(n'_1) \ldots a_l(n'_l)$?

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    $\begingroup$ Do you know the answer (omitting the sum on $n$) for some special cases, like $f=1$ or $n_1+\cdots+n_k$ or $n_1\cdots n_k$? (Are the corresponding sequences of $n$ in the OEIS? If you don't know the answer offhand, searching for the numerical values there might be a start.) This might give a clue as to what can be hoped for in the general case. $\endgroup$
    – Gro-Tsen
    Commented Aug 6, 2023 at 12:24
  • $\begingroup$ @Gro-Tsen Thank you very much for your suggestion. To be honest, I don't know the answer for the case $f=1$. I'm really a beginner of combinatorics (even worse). I've been reading some books on combinatorics. But so far I didn't find something related to the summation in my question. Could you please provide a reference, say, for the special case that $f=1$? $\endgroup$
    – Dreamer
    Commented Aug 6, 2023 at 14:29
  • $\begingroup$ No, I don't know! But I suggest you try computing the first few values and look them up in the OEIS, to see if this gives a known sequence. This might provide helpful pointers. $\endgroup$
    – Gro-Tsen
    Commented Aug 6, 2023 at 14:32
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    $\begingroup$ @Gro-Tsen Very nice suggestion! Thank you! I found the sequence for the case $f=1$ in OEIS. It is A077365. I'm trying to find more clues from this sequence. $\endgroup$
    – Dreamer
    Commented Aug 6, 2023 at 15:21
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    $\begingroup$ Attaching some terminology to your question: The product $1^{n_1} \cdots k^{n_k}$ is called the norm of the partition by Robert Schneider and Andrew Sills in a 2020 paper (Integers 20A, A13). You might look at it and the handful of papers that reference it. $\endgroup$
    – Brian Hopkins
    Commented Aug 14, 2023 at 0:31

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