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I wanted to put this online long ago but somehow never came to actually doing it. Now is a good occasion:

http://www.cip.ifi.lmu.de/~grinberg/index.html#hyperfactorial

or, directly, the PDF file: http://www.cip.ifi.lmu.de/~grinberg/hyperfactorialBRIEF.pdfhttp://mit.edu/~darij/www/hyperfactorialBRIEF.pdf

This is about a theorem by MacMahon stating that for any three nonnegative integers $a$, $b$, $c$, the number $\frac{H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)}{H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)}$ is an integer, where $H\left(m\right)$ means $0!\cdot 1!\cdot ...\cdot \left(m-1\right)!$. There are various proofs of this now, some of them combinatorial (see the references in the note), but the simplest one is probably the one I give using the Vandermonde determinant (I don't think it's new...).

The note is a bit long (10 pages), but the proof ends at page 6. Also note that I prove Vandermonde itself, which takes up some space as well. Another application of Vandermonde appears on page 9: If $a_1$, $a_2$, ..., $a_m$ are $m$ integers, then $\prod\limits_{1\leq i < j\leq m}\left(a_i-a_j\right)$ is divisible by $H\left(m\right)$. This is very well-known (and so is the proof).

Finally, a little question - slightly offtopic, I know. Back to the $\frac{H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)}{H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)}$ problem, one might try proving that this is an integer by showing that every prime $p$ divides $H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)$ at least as often as it divides $H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)$. This can be easily shown equivalent to the following: Any nonnegative integers $a$, $b$, $c$ satisfy

$\sum\limits_{k=0}^{a-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{b-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{c-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{a+b+c-1} \lfloor \frac{k}{p} \rfloor$ $\geq \sum\limits_{k=0}^{b+c-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{c+a-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{a+b-1} \lfloor \frac{k}{p} \rfloor$.

(Yes, it can be shown that the $p^2$, $p^3$, ... terms can be ignored.) Is there an easy way to see this? Or any way at all, without going back to the Vandermonde determinant?

2 deleted 10 characters in body

I wanted to put this online long ago but somehow never came to actually doing it. Now is a good occassionoccasion:

http://www.cip.ifi.lmu.de/~grinberg/index.html#hyperfactorial

or, directly, the PDF file: http://www.cip.ifi.lmu.de/~grinberg/hyperfactorialBRIEF.pdf

This is about a theorem by MacMahon stating that for any three nonnegative integers $a$, $b$, $c$, the number $\dfrac{H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)}{H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)}$ \frac{H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)}{H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)}$is an integer, where$H\left(m\right)$means$0!\cdot 1!\cdot ...\cdot \left(m-1\right)!$. There are various proofs of this now, some of them combinatorial (see the references in the note), but the simplest one is probably the one I give using the Vandermonde determinant (I don't think it's new...). The note is a bit long (10 pages), but the proof ends at page 6. Also note that I prove Vandermonde itself, so it which takes up some space as well. Another application of Vandermonde appears on page 9: If$a_1$,$a_2$, ...,$a_m$are$m$integers, then$\prod\limits_{1\leq i < j\leq m}\left(a_i-a_j\right)$is divisible by$H\left(m\right)$. This is very well-known (and so is the proof). Finally, a little question - slightly offtopic, I know. Back to the$\dfrac{H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)}{H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)}$\frac{H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)}{H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)}$ problem, one might try proving that this is an integer by showing that every prime $p$ divides $H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)$ at least as often as it divides $H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)$. This can be easily shown equivalent to the following: Any nonnegative integers $a$, $b$, $c$ satisfy

$\sum\limits_{k=0}^{a-1} \lfloor \dfrac{k}{p} frac{k}{p} \rfloor + \sum\limits_{k=0}^{b-1} \lfloor \dfrac{k}{p} frac{k}{p} \rfloor + \sum\limits_{k=0}^{c-1} \lfloor \dfrac{k}{p} frac{k}{p} \rfloor + \sum\limits_{k=0}^{a+b+c-1} \lfloor \dfrac{k}{p} frac{k}{p} \rfloor$ $\geq \sum\limits_{k=0}^{b+c-1} \lfloor \dfrac{k}{p} frac{k}{p} \rfloor + \sum\limits_{k=0}^{c+a-1} \lfloor \dfrac{k}{p} frac{k}{p} \rfloor + \sum\limits_{k=0}^{a+b-1} \lfloor \dfrac{k}{p} frac{k}{p} \rfloor$.

(Yes, it can be shown that the $p^2$, $p^3$, ... terms can be ignored.) Is there an easy way to see this? Or any way at all, without going back to the Vandermonde determinant?

1 [made Community Wiki]

I wanted to put this online long ago but somehow never came to actually doing it. Now is a good occassion:

http://www.cip.ifi.lmu.de/~grinberg/index.html#hyperfactorial

or, directly, the PDF file: http://www.cip.ifi.lmu.de/~grinberg/hyperfactorialBRIEF.pdf

This is about a theorem by MacMahon stating that for any three nonnegative integers $a$, $b$, $c$, the number $\dfrac{H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)}{H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)}$ is an integer, where $H\left(m\right)$ means $0!\cdot 1!\cdot ...\cdot \left(m-1\right)!$. There are various proofs of this now, some of them combinatorial (see the references in the note), but the simplest one is probably the one I give using the Vandermonde determinant (I don't think it's new...).

The note is a bit long (10 pages), but the proof ends at page 6. Also note that I prove Vandermonde itself, so it takes up some space as well. Another application of Vandermonde appears on page 9: If $a_1$, $a_2$, ..., $a_m$ are $m$ integers, then $\prod\limits_{1\leq i < j\leq m}\left(a_i-a_j\right)$ is divisible by $H\left(m\right)$. This is very well-known (and so is the proof).

Finally, a little question - slightly offtopic, I know. Back to the $\dfrac{H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)}{H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)}$ problem, one might try proving that this is an integer by showing that every prime $p$ divides $H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)$ at least as often as it divides $H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)$. This can be easily shown equivalent to the following: Any nonnegative integers $a$, $b$, $c$ satisfy

$\sum\limits_{k=0}^{a-1} \lfloor \dfrac{k}{p} \rfloor + \sum\limits_{k=0}^{b-1} \lfloor \dfrac{k}{p} \rfloor + \sum\limits_{k=0}^{c-1} \lfloor \dfrac{k}{p} \rfloor + \sum\limits_{k=0}^{a+b+c-1} \lfloor \dfrac{k}{p} \rfloor$ $\geq \sum\limits_{k=0}^{b+c-1} \lfloor \dfrac{k}{p} \rfloor + \sum\limits_{k=0}^{c+a-1} \lfloor \dfrac{k}{p} \rfloor + \sum\limits_{k=0}^{a+b-1} \lfloor \dfrac{k}{p} \rfloor$.

(Yes, it can be shown that the $p^2$, $p^3$, ... terms can be ignored.) Is there an easy way to see this? Or any way at all, without going back to the Vandermonde determinant?