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 $0$, $\endgroup$ Sep 19, 2018 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. $\endgroup$ Sep 20, 2018 at 12:44

2 Answers 2


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 $P(x)=\prod_{i\geq 1}(1-x^i)^{-1}$.

  • 3
    $\begingroup$ Maybe it's worthwhile to get rid of the second summation: $\frac{P(x)}{1-q}\sum_{k\geq1}\frac{1-q^k}{1-x^k}x^k$ $\endgroup$ Sep 20, 2018 at 16:02
  • 1
    $\begingroup$ @MartinRubey: your formula is certainly more elegant, while mine shows directly what is the coefficient of $q^m$. $\endgroup$ Sep 20, 2018 at 19:00
  • $\begingroup$ Thank you, but I'm not sure whether the rewriting has any merit. It does have a plethystic feel. $\endgroup$ Sep 20, 2018 at 19:20
  • 3
    $\begingroup$ 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. $\endgroup$ Sep 21, 2018 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). $\endgroup$ Sep 20, 2018 at 5:03
  • $\begingroup$ not sure - do you mean sum k-th parts = number parts >= k? @BrianHopkins $\endgroup$
    – JMP
    Sep 20, 2018 at 21:49
  • $\begingroup$ @BrianHopkins; first parts are in 1-1 with partition length? $\endgroup$
    – JMP
    Sep 20, 2018 at 21:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.