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Question: Given positive, coprime integers $m<n$, consider the Young diagram $Y$ formed by the lattice points in the Cartesian plane lying below the line from $(0,0)$ to $(m,n)$ and within the rectangle $(0,m)\times (0,n)$. The case $m=5$, $n=7$ is shown below. For positive integer $x$, how many integer partitions of $x$ are there whose Young diagram is $\leq Y$ in the Young lattice?

Equivalently, how many lattice paths are there from $(0,0)$ to $(m,n)$, moving only up and right and lying below the diagonal, whose integral over $[0,m]$ is precisely $x$?

Young diagram for m=5, n=7.

This problem originated from this MSE question. The problem considered in that question was the following: Given positive integers $m<n$ (not necessarily coprime), how many binary sequences $(s_j)_{j=0}^\infty$ with exactly $x$ 1's are there which have the property that $s_j=0 \implies s_{j+m}=0 \land s_{j+n}=0$. That problem can be reduced to the one I give above.

The problem posed above seems at least as hard as enumerating integer partitions, since for $x<m$ the number of such Young diagrams/lattice paths is precisely the number of integer partitions of $x$. Probably the best I could hope for is a convenient expression for the generating function.

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  • $\begingroup$ 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... $\endgroup$ Commented Jul 20, 2022 at 19:58
  • $\begingroup$ @SamHopkins You missed the $x$. I'm asking a harder question than enumerating lattice paths under the diagonal. $\endgroup$
    – Yly
    Commented Jul 20, 2022 at 19:59
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    $\begingroup$ 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). $\endgroup$ Commented Jul 20, 2022 at 20:01
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    $\begingroup$ @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. $\endgroup$ Commented Jul 21, 2022 at 20:57

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