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Dec 18, 2023 at 19:16 history edited Ira Gessel CC BY-SA 4.0
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Jan 2, 2022 at 5:35 comment added Ira Gessel Incidentally, this other interpretation of $\frac{j}{2(j+k)}\binom {2j}{j}\binom{2k}{k}$ has been published. It appears in my paper Super Ballot Numbers, sciencedirect.com/science/article/pii/0747717192900342, section 7.
Jan 1, 2022 at 14:25 comment added Shannon Starr Thanks! In your slides (Gessel's slides) you also noted this is also ``the number of paths with m+n up steps and m+n-1 down steps whose last return to the x-axis is before the point (2m,0).'' So I think that may give an answer to my question, since the number of non-negative paths ending at a particular spot was also enumerated in Feller. It may give a different sum, partitioning on the last return. Those slides based on your work with your students Aminil Huq (PhD) for the first half and Apratim Roy (Masters) for the second half are wonderful, especially because of the included figures.
Jan 1, 2022 at 14:12 vote accept Shannon Starr
Dec 30, 2021 at 20:56 history edited Ira Gessel CC BY-SA 4.0
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Dec 30, 2021 at 20:40 history answered Ira Gessel CC BY-SA 4.0