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Jul 21, 2022 at 20:57 comment added Peter Taylor @HughDenoncourt, the "standard" $q,t-$analogue doesn't work except in the special case $n = m+1$. (See pages 11 and 12 of Sam's first link). And my experiments suggest that the g.f. for fixed $m,n$ don't tend to have many factors.
Jul 20, 2022 at 20:01 comment added Sam Hopkins Ah, sorry, I see. Then I think there is no super nice closed-form formula. There are ways, going back to MacMahon, that you can use determinants to get at the answer (see e.g. the expository part of my question mathoverflow.net/questions/350445/… - noting that $m=1$ there just gives subshapes).
Jul 20, 2022 at 19:59 comment added Yly @SamHopkins You missed the $x$. I'm asking a harder question than enumerating lattice paths under the diagonal.
Jul 20, 2022 at 19:58 comment added Sam Hopkins With the coprime condition it is the rational Catalan number $(m+n-1)!/(m! n!)$; see e.g. math.ucdavis.edu/~egorskiy/Presentations/slides_badmath.pdf. But this question is probably not suited for MathOverflow...
Jul 20, 2022 at 19:54 history asked Yly CC BY-SA 4.0