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Proving that a ratio of factorials is an integer

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tc...@lsa.umich.edu

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Dec 14, 2006, 9:26:30 PM12/14/06
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In Problem 6514 of the American Mathematical Monthly, Richard Askey asked
for a proof that (3m+3n)!(3n)!(2m)!(2n)!/(2m+3n)!(m+2n)!(m+n)!m!n!n! is
always an integer if m and n are nonnegative integers. The solution by
Gregg Patruno gave a general method for attacking problems of this type.
Reference: Amer. Math. Monthly 94 (1987), 1012-1014, or see JSTOR:

http://www.jstor.org/view/00029890/di991727/99p0111d/0

What I'm wondering is, is there a theorem of the form, "Whenever an
expression like this is always an integer, then there is always an
expression for it in terms of binomial coefficients and polynomials
in m and n that makes it obvious that it is an integer"? For a slightly
different example of the kind of thing I'm after, the Catalan number
(2n)!/n!(n+1)! is not obviously an integer when you write it in that
form, but (2n)!/n!(n+1)! = (2n choose n) - (2n choose n-1), which is
obviously an integer. Patruno's method does not seem to answer my
question directly.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences

William Shanley

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Dec 15, 2006, 2:05:13 AM12/15/06
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It is an open problem to find a combinatorial interpretation for the
numbers (2n)!(2m)!/n!m!(n+m)!, which are known to be integers. There
is an article by Ira M. Gessel and Guoce Xin, "A combinatorial
interpretation of the numbers 6(2n)! /n! (n+2)!" J. Integer Seq. 8
(2005), no. 2, Article 05.2.3, 13 pp.
This paper is available at
http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Gessel/xin.html

tc...@lsa.umich.edu

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Dec 15, 2006, 3:06:34 PM12/15/06
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In article <1166148313.8...@t46g2000cwa.googlegroups.com>,

William Shanley <wsha...@gmail.com> wrote:
>It is an open problem to find a combinatorial interpretation for the
>numbers (2n)!(2m)!/n!m!(n+m)!, which are known to be integers. There
>is an article by Ira M. Gessel and Guoce Xin, "A combinatorial
>interpretation of the numbers 6(2n)! /n! (n+2)!" J. Integer Seq. 8
>(2005), no. 2, Article 05.2.3, 13 pp.

Thanks...that's certainly a relevant paper, but I'm not convinced that it
answers my question. I'm asking for something weaker than a "combinatorial
interpretation," which is a term usually reserved for a *natural* meaning
for an integer. Even for the relatively trivial case of the Catalan
numbers, the equation (2n)!/n!(n+1)! = (2n choose n) - (2n choose n-1)
would not be considered to give a "combinatorial interpretation" of the
number until you did some extra work to describe a natural set of size
(2n choose n) and a natural subset of size (2n choose n-1). But all I'm
asking for is the expression in terms of binomial coefficients and other
"obviously integral" quantities like polynomials in m and n with integer
coefficients.

For example, Gessel and Xin quote the recurrence

sum_n 2^(p-2n) (p choose 2n) T(m,n) = T(m,m+p)

where T(m,n) = (2n)!(2m)!/2(n!m!(n+m)!), which makes it obvious that
T(m,n) is an integer. For this particular case, the above recurrence
is pretty close to a positive answer to my question (although the form
of the expression is not quite what I had naively had in mind, suggesting
that it may be a little tricky to properly formulate my question precisely).

So it seems to me that (a suitably cleaned-up version of) my original
question might still be a tractable one.

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