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
$$a(\lambda):=[\lambda\_k]\_q!\prod\_{j=1}^{k-1}\,\,[\lambda\_j-\lambda\_{j+1}]\_q! \qquad \text{and} \qquad b(\lambda)=\prod\_{j=1}^k[\lambda\_j]\_q.$$



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> **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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Partitions and $q$-integers
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 $n-im$.


