Integrality criterion for inverses of reciprocal binomial Hankel matrices

Proof

Use zero-based indices $0\leq a,b<n$, and write $H=R_n(b_k(r))$. Then

$$H_{ab} =\frac{1}{\binom{a+b+r}{r}} =r\int_0^1 x^{a+b}(1-x)^{r-1}\,dx. \qquad (1)$$

Thus $H$ is the moment matrix of the weight $r(1-x)^{r-1}$ on $[0,1]$.

For $k\geq0$, define the shifted Jacobi polynomial

$$Q_k(x)=P_k^{(r-1,0)}(2x-1) =\sum_{a=0}^k(-1)^{k+a} \binom{k}{a}\binom{k+r+a-1}{a}x^a. \qquad (2)$$

Every coefficient of $Q_k$ is an integer. Its constant coefficient is $(-1)^k$, and its leading coefficient is

$$\lambda_k=\binom{2k+r-1}{k}. \qquad (3)$$

The corresponding Rodrigues formula is

$$Q_k(x)=\frac{(-1)^k}{k!}(1-x)^{1-r} \frac{d^k}{dx^k}\left(x^k(1-x)^{k+r-1}\right).$$

Integration by parts shows that $Q_k$ is orthogonal to every polynomial of degree less than $k$. All boundary terms vanish because of the factors $x^k$ and $(1-x)^{k+r-1}$. Using $Q_k^{(k)}=k!\lambda_k$ also gives

$$r\int_0^1(1-x)^{r-1}Q_k(x)^2\,dx =\frac{r}{2k+r}. \qquad (4)$$

Let $A_{ka}=[x^a]Q_k(x)$. Equations (1) and (4) imply

$$AHA^{\mathsf T} =\operatorname{diag}\left( \frac r r,\frac r{r+2},\ldots,\frac r{r+2n-2} \right).$$

Since $A$ is lower triangular with nonzero diagonal, it is invertible. Therefore

$$(H^{-1})_{ab} =\frac1r\sum_{k=0}^{n-1}(2k+r)A_{ka}A_{kb}. \qquad (5)$$

Define the integer polynomial

$$K_n(x,y)=\sum_{k=0}^{n-1}(2k+r)Q_k(x)Q_k(y).$$

Equation (5) says that $H^{-1}$ is the coefficient matrix of $K_n/r$. Hence $H^{-1}$ is integral exactly when every coefficient of $K_n$ is divisible by $r$.

The Christoffel-Darboux identity, using the leading coefficients (3) and norms (4), gives

$$K_n(x,y) =\frac{n(n+r-1)}{2n+r-1} \frac{ Q_n(x)Q_{n-1}(y)-Q_{n-1}(x)Q_n(y) }{x-y}. \qquad (6)$$

Indeed, the required scalar follows from

$$\frac{\lambda_{n-1}}{\lambda_n}\frac{2n+r-2}{r} =\frac{n(n+r-1)}{r(2n+r-1)}.$$

Put

$$L_n(x,y)= \frac{ Q_n(x)Q_{n-1}(y)-Q_{n-1}(x)Q_n(y) }{x-y}.$$

The numerator is an integer polynomial that vanishes when $x=y$. Since $x-y$ is monic, $L_n$ belongs to $\mathbb Z[x,y]$. Thus (6) is equivalently

$$(2n+r-1)K_n=n(n+r-1)L_n. \qquad (7)$$

Suppose first that

$$r\mid n(n-1).$$

Then $r\mid n(n+r-1)$, since

$$n(n+r-1)=n(n-1)+nr.$$

Moreover,

$$\gcd(r,2n+r-1)=1. \qquad (8)$$

To see this, any common divisor divides both $2n-1$ and $n(n-1)$. But

$$(2n-1)^2=4n(n-1)+1,$$

so that common divisor must divide $1$.

Taking coefficients in (7), the right-hand side is divisible by $r$. Equation (8) allows cancellation of $2n+r-1$, proving that every coefficient of $K_n$ is divisible by $r$. Hence every entry of $H^{-1}$ is an integer.

Conversely, the constant coefficient of each $Q_k(x)Q_k(y)$ is $1$. Therefore

$$[x^0y^0]K_n =\sum_{k=0}^{n-1}(2k+r) =n(n+r-1). \qquad (9)$$

If $H^{-1}$ is integral, equations (5) and (9) imply

$$r\mid n(n+r-1).$$

Subtracting the term $nr$ gives

$$r\mid n(n-1).$$

We have therefore proved

$$H^{-1}\text{ is integral} \quad\Longleftrightarrow\quad r\mid n(n-1). \qquad (10)$$

Finally, write $r=\prod_p p^{e_p}$. Since $\gcd(n,n-1)=1$,

$$p^{e_p}\mid n(n-1)$$

holds exactly when

$$n\equiv0\pmod{p^{e_p}} \quad\text{or}\quad n\equiv1\pmod{p^{e_p}}.$$

This is equivalent to the stated condition for every prime power dividing $r$. Thus (10) proves Conjecture 5.1.

Original source: https://arxiv.org/abs/math/9905079