Partitions and $q$-integers

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

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

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 $$n-im$$.