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Recently I got interested in plane partitions and the following formula by MacMahon, which counts the number of plane partitions $\pi \in B(r,s,t)$ fitting in an $(r,s,t)$-box:

$$ \binom{r+s+t}{r,s,t}_q:=\sum_{\pi \in B(r,s,t)}\, q^{|\pi|}=\prod_{i=1}^r\prod_{j=1}^s\prod_{k=1}^t\,\frac{1-q^{i+j+k-1}}{1-q^{i+j+k-2}}. $$

This is the $q$-analog version of the formula, which remembers of how often a unit cube had to be swapped in order to reach a certain plane partition.

Now let $\phi_q(n):=[1]_q! \cdot \dots \cdot [n-1]_q!$ be the $q$-analog of the superfactorial (the -1 is by intention, but maybe I did a mistake in the calculation). Then it is quite easy to see that

$$ \binom{r+s+t}{r,s,t}_q=\frac{\phi_q(r+s+t)\phi_q(r)\phi_q(s)\phi_q(t)}{\phi_q(r+s)\phi_q(r+t)\phi_q(s+t)}. $$

Here I noticed that this is quite similar to the $q$-binomial coefficient (actually it's a generalization):

$$ \binom{r+s}{r,s}_q=\frac{[r+s]_q!}{[r]_q![s]_q!}. $$ I think of the $q$-factorial $[n]_q!$ as the number of elements in $S_n$, remembering the number of transpositions in a reduced word for every element $\sigma \in S_n$. The $q$-binomial coefficient is the number of $(r,s)$-shuffles in $S_{r+s}$, again remembering the number of transpositions.

On the other hand, the number $\binom{r+s+t}{r,s,t}_q$ can be understood as the number of reduced words of the permutation $$(r+s+1,\dots,r+s+t,r+1,\dots,r+s,1,\dots,s) \in S_{r+s+t}$$ taking into account the number of Reidemeister moves from one word to another (at least I think so). So far this seemes to me as a nice higher analog of $q$-binomial calculus. I thought I had good arguments that the natural analogue of the $q$-factorial $[n]_q!$ should be then the ($q$-analog of the) number of all commutation classes of reduced words of the longest element in $S_n$. However, by the previous formula it actually seems to be $\phi_q(n)$, of which I have no interpretation at all. Which leads me to my question: Do you have an interpretation of $\phi_q(n)$ in terms of plane partitions, string diagrams, counting stuff in $n$-cubes or whatever? And do you know any relation to the number of commutation classes of reduced words of the longest element in $S_n$?

EDIT: I think I got it now. If somenone is interested, I will provide an interpretation of $\phi_q(n)$ in terms of reduced words of the longest element in $S_n$.

EDIT: False alarm, sorry.. the problem is that I don't even understand how to interpret $[n]_q!$ in terms of reduced words, only $[n]_q$. It was not my intention to clickbait with the previous announcement, so feel free to downvote again if you just upvoted because of that.

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  • $\begingroup$ I don't quite understand the very end of your question (because you don't define $sf_q(n)$ anywhere as far as I can tell); but oeis.org/A006245 is the number of commutation classes for reduced words for the longest word in the symmetric group, and there is no "nice" formula for this number, as far as I am aware. $\endgroup$ Feb 15, 2022 at 23:29
  • $\begingroup$ @SamHopkins Whoops, I changed the notation for the superfactorial in between, not a good idea :) Thank you, I edited it $\endgroup$ Feb 15, 2022 at 23:32
  • $\begingroup$ And yes, counting the commutation classes for reduced words for the longest element in the symmetric group is probably hopeless, but I don't really need a formula $\endgroup$ Feb 15, 2022 at 23:34
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    $\begingroup$ I calculate $\phi_q(4) = q^{10} + 6q^9 + 18q^8 + 36q^7 + 53q^6 + 60q^5 + 53q^4 + 36q^3 + 18q^2 + 6q + 1$ and note that this doesn't appear to be in OEIS. $\endgroup$ Feb 16, 2022 at 8:38
  • $\begingroup$ If writing up your explanation does not take you too much time then, sure, I'd be interested to see it $\endgroup$ Feb 16, 2022 at 18:48

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