Recall the integer partition function $P(n)$ with generating function $$\sum_{n\geq0}P(n)x^n=\prod_{k=1}^{\infty}\frac1{1-x^k}.$$ Let $[n]_q=\frac{1-q^n}{1-q}$ denote the $q$-analogue of the integer $n$ and let $\lambda=(\lambda_1,\lambda_2,\dots)\vdash n$ be a partition of $n$. Now, define the function $$\Psi_q(n)=\sum_{\lambda\vdash n}\,\,\sum_{j\geq1}\,\, [\lambda_j]_q.$$ For example, $\Psi_q(3)=[3]_q+([2]_q+[1]_q)+([1]_q+[1]_q+[1]_q)=q^2+2q+6$. In particular, when $q=1$, we obtain $\Psi_q(n)=nP(n)$ with generating function $$\sum_{n\geq1}nP(n)x^n=x\frac{d}{dx}\prod_{k=1}^{\infty}\frac1{1-x^k}.$$

I would like to ask

Question 1. Is there a generating function for the sequence of $q$-polynomials $\Psi_q(n)$?

Question 2. What is the combinatorial interpretation of the coefficients of $\Psi_q(n)$?

For example, the constant term of $\Psi_q(n)$ enumerates the number of $1$'s in all partitions of $n$.

  • $\begingroup$ Your link is for -Sum of all parts of all partitions of n.- maybe you want oeis.org/A006128 The constant term is the number of parts = number of parts greater than $0$, $\endgroup$ – Aaron Meyerowitz Sep 19 '18 at 19:17
  • $\begingroup$ It might be quite hard to find an ordinary differential equation for $\sum\Psi_q(n)z^n$, because it is already hard to find one for $\sum\Psi_1(n)z^n$. By contrast, it is "easy" (for a computer) to find such an equation for $\sum\Psi_1(n)/n z^n=\sum P(n) z^n$. A q-differential equation might be a completely different story, but I could not find anything quickly. $\endgroup$ – Martin Rubey Sep 20 '18 at 12:44

A generating function of sorts is given by $$ \sum_{n\geq 1}\Psi_q(n)x^n = P(x)\sum_{m\geq 0}q^m\sum_{k\geq m+1} \frac{x^k}{1-x^k}, $$ where $P(x)=\prod_{i\geq 1}(1-x^i)^{-1}$.

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    $\begingroup$ Maybe it's worthwhile to get rid of the second summation: $\frac{P(x)}{1-q}\sum_{k\geq1}\frac{1-q^k}{1-x^k}x^k$ $\endgroup$ – Martin Rubey Sep 20 '18 at 16:02
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    $\begingroup$ @MartinRubey: your formula is certainly more elegant, while mine shows directly what is the coefficient of $q^m$. $\endgroup$ – Richard Stanley Sep 20 '18 at 19:00
  • $\begingroup$ Thank you, but I'm not sure whether the rewriting has any merit. It does have a plethystic feel. $\endgroup$ – Martin Rubey Sep 20 '18 at 19:20
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    $\begingroup$ I just realized that I should have written $P(x)\sum_k [k]_q \frac{x^k}{1-x^k}$, which makes its relation to $\sigma_0$, the number of divisors function (oeis.org/A000005), and $\sigma_1$, the sum of the divisors function (oeis.org/A000203), clear. $\endgroup$ – Martin Rubey Sep 21 '18 at 6:50

I can answer part 2.

The co-efficient of $q^k$ in $\Psi_q(n)$ represents the number of elements greater than $k$ in all partitions of $n$

This can be proved with elementary analysis, mainly each part $k$ provides a contribution of $\{q^k,\dots,q^0\}$ to the q-nomial.

Part 1 is harder - both the differential of the partition function and the integral of the related inverted partition function are unresolved I believe.

  • $\begingroup$ This coefficient sequence is oeis.org/A181187 with your interpretation along a comment about $k$th ranks (but nothing about the OP's $q$-analogues). $\endgroup$ – Brian Hopkins Sep 20 '18 at 5:03
  • $\begingroup$ not sure - do you mean sum k-th parts = number parts >= k? @BrianHopkins $\endgroup$ – JMP Sep 20 '18 at 21:49
  • $\begingroup$ @BrianHopkins; first parts are in 1-1 with partition length? $\endgroup$ – JMP Sep 20 '18 at 21:53

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