Denote an integer partition of $n$ by $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq\lambda_k)$ where $\lambda_k>0$. Also recall the $q$-analogues of integer $n$ given by $[n]_q=\frac{1-q^n}{1-q}$. Further, let $$[n]_q!=[n]_q[n-1]_q\cdots[2]_q[1]_q \qquad \text{and} \qquad [0]_q!=1.$$ If $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq\lambda_k)\vdash n$, define $$a(\lambda):=[\lambda_k]_q!\prod_{j=1}^{k-1}\,\,[\lambda_j-\lambda_{j+1}]_q! \qquad \text{and} \qquad b(\lambda)=\prod_{j=1}^k[\lambda_j]_q.$$

Question. The following appears to be true. Is it? $$\prod_{\lambda\vdash n}a(\lambda)=\prod_{\lambda\vdash n}b(\lambda).$$

  • 1
    $\begingroup$ Does this possibly follow just by transposing $\lambda$? $\endgroup$ Oct 7, 2018 at 18:09
  • 3
    $\begingroup$ This does not follow by transposing $\lambda$. However, it is equivalent to Enumerative Combinatorics, vol. 1, 2nd ed., Exercise 1.80, after conjugating $\lambda$. Note that the solution to this exercise gives another instance in which a proof was claimed to follow by conjugation. $\endgroup$ Oct 7, 2018 at 18:48
  • 2
    $\begingroup$ Compare with this identity: mathoverflow.net/questions/99271 $\endgroup$
    – Igor Pak
    Oct 8, 2018 at 6:21

1 Answer 1


As expected, it is not about $q$-analogs: the multisets $\cup_{\lambda\vdash n} \{\lambda_1,\lambda_2,\dots\}$ and $\cup_{\lambda\vdash n}\cup_i \{1,2,\dots,\lambda_i-\lambda_{i+1}\}$ coincide. To see this compute the multiplicity of a given integer $m$ in them both. For the first, it equals $p(n-m)+p(n-2m)+p(n-3m)+\dots$: we have $p(n-m)$ partitions containing $m$, $p(n-2m)$ partitions containing $2m$ and so on. For the second it equals to the same thing: for given $i$, $m\in \{1,2,\dots,\lambda_i-\lambda_{i+1}\}$ exactly for $p(n-im)$ partitions $\lambda$. To see this, subtract $m$ from $\lambda_1,\dots,\lambda_i$ and get a partition of $n-im$.


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.