Questions on integer-valued polynomials

Question

An integer-valued polynomial or numerical polynomial is a polynomial $f \in \mathbb Q[x]$ with the property that $f(\mathbb Z)\subseteq \mathbb Z$. The set of numerical polynomials forms a subring $\mathcal N$ of $\mathbb Q[x]$. Obviously $\mathbb Z[x]\subseteq \mathcal N$, but $\mathcal N$ is much larger than that.

The binomial polynomials, given for each $n\geq 0$ by

$${x \choose n} = \frac{x(x-1)\cdots(x-n+1)}{n!}.$$

They are numerical polynomials because the product of any $n$ consecutive integers is divisible by $n!$. In fact, one can show that the ${x \choose n}$ form a basis of $\mathcal N$ as a $\mathbb Z$-module:

Any $f\in \mathcal N$ can be written in a unique way as a finite linear combination $\sum_n a_n {x \choose n}$ of binomial polynomials, where the $a_n$'s are integers.

Now if $n,m$ are non-negative integers, then the product

$${x \choose n}{x \choose m},$$

being a numerical polynomial, has an expansion as above. For instance,

$${x \choose 2}^2 = \left(\frac{x(x-1)}{2}\right)^2 = 6{x \choose 4} + 6 {x \choose 3} + {x \choose 2}$$

or

$${x \choose 5}{x \choose 8} = 1287{x \choose 13} + 3960 {x \choose 12} + 4620{x \choose 11} + 2520 {x \choose 10} + 630{x \choose 9} +56 {x \choose 8}.$$

I am interested in knowing more about these coefficients, as they determine completely the structure of the ring $\mathcal N$. Is there a nice formula for the coefficient of ${x \choose d}$ in the expansion of ${x \choose n}{x \choose m}$?

With a computer I looked at quite a few examples, and noticed that:

1. The coefficient of ${x \choose d}$ in the expansion of ${x \choose n}{x \choose m}$, call it $c_d(m,n)$, seems to vanish for $d<\max\{m,n\}$. You can see this in the examples above.

2. The coefficients $c_d(m,n)$ all seem to be positive.

Can we prove any one of these observations, or better even provide a useful formula for $c_d(m,n)$ which would explain these observations?

Answer 1

For the first remark, take $f\in\mathcal{N}$ then if $d:=deg(f)$ then there exists integers $a_0,...,a_d$ such that :

$$f(x)=\sum_{k=0}^da_k\begin{pmatrix}x\\k\end{pmatrix} $$

Now, it follows that for any $l\leq d$ we have :

$$f(l)=\sum_{k=0}^la_k\begin{pmatrix}l\\k\end{pmatrix} $$

Now with this formula it is a trivial fact to show the following (by induction) :

If there exists $l_0$ a positive integer such that $f(l)$ is null for each integer $0\leq l<l_0$ and $f(l_0)\neq 0$ then $a_0=...=a_{l_0-1}=0$ and $a_{l_0}\neq 0$.

When $f(x):=\begin{pmatrix}x\\n\end{pmatrix}\begin{pmatrix}x\\m\end{pmatrix}$ then $l_0:=max(m,n)$ verifies the above statement proving your first statement.

For the second, from what is written above (with $n>m$) :

$$\begin{pmatrix}x\\n\end{pmatrix}\begin{pmatrix}x\\m\end{pmatrix}=\sum_{k=n}^{n+m}a_k\begin{pmatrix}x\\k\end{pmatrix} $$

Now evaluating the above in $x:=n$ you have :

$$a_n=\begin{pmatrix}n\\m\end{pmatrix} $$

Then evaluating it in $x:=n+1$ you have :

$$a_{n+1}=(n+1)\begin{pmatrix}n\\m-1\end{pmatrix} $$

And so on. Now use an induction to prove the following :

If $f(x):=\begin{pmatrix}x\\n\end{pmatrix}\begin{pmatrix}x\\m\end{pmatrix}$ then let $a_n,...,a_{n+m}$ be its coefficients then you have $a_{n+k}=\begin{pmatrix}n+k\\k\end{pmatrix}\begin{pmatrix}n\\m-k\end{pmatrix}$ for $0\leq k\leq m$.

In particular all the coefficients are strictly positive (you can see that my formula works for your examples).

Answer 2

Here's another way to derive the formula for $c_d(m,n)$, using generating functions. By the binomial theorem, $\binom{x}{n}\binom{x}{m}$ is the coefficient of $u^n v^m$ in $(1+u)^x (1+v)^x$. We have \begin{align*} (1+u)^x (1+v)^x &= ((1+u)(1+v))^x = (1+u+v+uv)^x\\ &= \sum_{d=0}^\infty \binom{x}{d} (u+v+uv)^d. \end{align*} So $c_d(m,n)$ is the coefficient of $u^n v^m$ in $(u+v+uv)^d$. To find an explicit formula for $c_d(m,n)$, we expand $(u+v+uv)^d$ by the trinomial theorem: $$(u+v+uv)^d = \sum_{i+j+k=d}\binom{d}{i,j,k} u^i v^j (uv)^k,$$ where the trinomial coefficient $\binom{d}{i,j,k}$ is $\frac{d!}{i!\,j!\,k!}$ if $i+j+k=d$ and $i$, $j$, and $k$ are nonnegative integers, and is 0 otherwise. Setting $i+k=n$ and $j+k=m$, with $i+j+k=d$, and solving for $i$, $j$, and $k$ in terms of $n$, $m$, and $d$, we find that $i=d-m$, $j=d-n$, and $k=m+n-d$. Thus $$c_d(m,n) = \binom{d}{d-n,d-m, n+m-d}.$$