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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? $$\prod_{\lambda\vdash n}a(\lambda)=\prod_{\lambda\vdash n}b(\lambda).$$

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    $\begingroup$ Does this possibly follow just by transposing $\lambda$? $\endgroup$ – Sam Hopkins Oct 7 '18 at 18:09
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    $\begingroup$ 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. $\endgroup$ – Richard Stanley Oct 7 '18 at 18:48
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    $\begingroup$ Compare with this identity: mathoverflow.net/questions/99271 $\endgroup$ – Igor Pak Oct 8 '18 at 6:21
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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 $n-im$.

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