This historical question recalls
Pafnuty Chebyshev's estimates for the prime distribution function. In his derivation
Chebyshev used the factorial ratio sequence
$$
u_n=\frac{(30n)!n!}{(15n)!(10n)!(6n)!}, \qquad n=0,1,2,\dots,
$$
which assumes integer values only. The latter fact can be established with the help of
$$
\operatorname{ord}_p n!
=\biggl\lfloor\frac{n}{p}\biggr\rfloor+\biggl\lfloor\frac{n}{p^2}\biggr\rfloor
+\biggl\lfloor\frac{n}{p^3}\biggr\rfloor+\dots
$$
and routine verification of
$$
\lfloor 30x\rfloor+\lfloor x\rfloor-\lfloor 15x\rfloor-\lfloor 10x\rfloor-\lfloor 6x\rfloor\ge0.
$$
Other Chebyshev-like examples of integer-valued factorial sequences are known;
the complete list of such
$$
u_n=\frac{(a_1n)!\dots(a_rn)!}{(b_1n)!\dots(b_sn)!}
$$
in the case $s=r+1$ was recently tabulated in
[J.W. Bober, J. London Math. Soc. (2) 79 (2009) 422–444].
A motivation for this classification problem is in relation with
a certain approach to Riemann's hypothesis, but I would prefer to
refer everybody interested in to Bober's paper (which could be
found in the arXiv as well). The proofs of $u_n\in\mathbb Z$
make use of the above formula for $\operatorname{ord}_p n!$
There are three 2-parameter families in Bober's list, namely,
$$
\frac{(n+m)!}{n!m!}, \qquad
\frac{(2n)!(2m)!}{n!(n+m)!m!}, \qquad\text{and}\qquad
\frac{(2n)!m!}{n!(2m)!(n-m)!} \quad (n>m);
$$
the first one includes the binomial coefficients, while some
properties of the second family are mentioned in
this question.
For the binomial family, a standard way to establish integrality
purely combinatorially amounts to interpreting the factorial
ratio as coefficients in the expansion
$$
(1+t)^{n+m}=\sum_{k=0}^{n+m}\binom{n+m}{k} t^k,
$$
that is, as the number of $m$-element subsets of an $(n+m)$-set.
There is lack of similar interpretation for the other two 2-parametric
families, although Ira Gessel indicates in
[J. Symbolic Computation
14 (1992) 179–194] that the inductive argument together with identity
$$
\frac{(2n)!(2(n+p))!}{n!(n+(n+p))!(n+p)!}
=\sum_{k=0}^{\lfloor p/2\rfloor}2^{p-2k} \binom{p}{2k}
\frac{(2n)!(2k)!}{n!(n+k)!k!}
\qquad (p\geq 0)
$$
allows one to show that the numbers in question are indeed integers.
A slight modification of the formula can be used for showing that
the third 2-parametric family is integer valued. In these cases
one uses a reduction to binomial sums for which the integrality is
already known. But what about the 1-parametric families, like Chebyshev's
or, say,
$$
\frac{(12n)!n!}{(6n)!(4n)!(3n)!}?
$$
Is there any way to establish the integrality without referring to
the $p$-order formula?
My own motivation is explained in the joint recent
preprint
with Ole Warnaar, where we observe a $q$-version of the integrality
in a "stronger form".
Best Answer
Along with the binomial coefficients, the other two infinite families each enjoy a fairly simple recurrence. for example $$f(n,m)=\frac{(2n)!(2m)!}{n!(n+m)!m!}.$$ has $f(0,t)=\binom{2t}{t}$ and $f(i+1,j)=4f(i,j)-f(i,j+1).$
Consider the one parameter family $$\frac{(2n)!(6n)!}{n!(4n)!(3n)!}.$$ Viewed in isolation it seems hard to establish integrality without referring to the p-order formula. However as the case m=3n of the family $f(n,m)$ it is the values in a line of cells with slope 3.
I've wondered if any of the various "sporadic" one parameter families can be embedded in a similar manner in a 2 parameter family defined by a recurrence. Evidently the entire table would not all be given by a formula exactly of that form.