On determinants formed by binomial coefficients

Question

Let $q$ be a number. Let us consider the $q^2-1$-th line of the Pascal triangle (i.e. numbers ${{q^2-1} \choose i}$, $i=0,1,...q^2-1$). We have $q^2$ numbers.

Let us form naively a $q \times q$ matrix from them. Example for $q=2$:

\[\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\]

And let us take its determinant: $-8$ if $q=2$.

How to prove that if $q$ is prime then $q$ enters in this determinant with $(q+4)(q-1)/2 $-th exponent?

Example: $q=2$, $(q+4)(q-1)/2 =3$. Really, $8=2^3$.

Case $q=3$: the binomial coefficients are $1,8,28,56,70,56,28,8,1$.

The matrix:

\[\begin{pmatrix} 1 & 8 & 28 \\ 56 & 70 & 56 \\ 28 & 8 & 1 \end{pmatrix}\]

The determinant is $2 \cdot 3^7 \cdot 7$. Really, $(q+4)(q-1)/2 =7$.

Answer 1

Looking at Krattenthaler's famous "determinant calculus" survey (also, suggested by Per Alexandersson) we find Theorem 26 in there, which answers a more general question, and should (most likely) yield the claim in the OP.

Theorem Let $q$ be a nonnegative integer, and let $L_1,\ldots,L_q$ and $A, B$ be indeterminates. Then, \begin{equation*} \det\left( \binom{BL_i+A}{L_i+j}\right) = \frac{\prod_{i < j}(L_i-L_j)}{\prod_{i=1}^q(L_i+q)!}\prod_{i=1}^q\frac{(BL_i+A)!}{((B-1)L_i+A-1)!}\prod_{i=1}^q(A-Bi+1)_{i-1}, \end{equation*} where $(a)_k$ is the usual Pochhammer symbol.

The matrix in the OP is given by \begin{equation*} M_{ij} = \binom{q^2-1}{q(i-1)+j-1}, \end{equation*} thus we can select $A=q^2-1$, $B=0$, $L_i = q(i-1)-1$, and apply the above result, and hopefully things fall into place.

Answer 2

For the specific determinant in the question there is a slightly more direct proof than Lemma 3 in Krattenthaler. We can evaluate it using the Jacobi-Trudi identity. Let's denote $\lambda _i=(q-i+1)(q-1)$. The determinant in question, $\det\left(\binom{q^2-1}{qi-q+j}\right)$, is equal (up to sign) to the determinant $\det\left(e_{\lambda_i-i+j}\right)$ when you put all variables equal to $1$. But this is equal to the Schur polynomial of the conjugate partition $\lambda^{\star}=(q^{q-1}(q-1)^{q-1}\cdots 1^{q-1}0^{q-1})$ evaluated at $1,1,\dots,1$. (I added a bunch of zeros to the partition so that the number of parts matches the number of variables.)

There is a simple product formula for Schur polynomials evaluated at $1,1,\dots,1$, and in this case it gives $$s_{\lambda^{\star}}(1,1,\dots,1)=\prod_{i<j}\frac{\lambda^{\star}_i-\lambda^{\star}_j+j-i}{j-i}.$$ The denominator has the form $1!2!\cdots (q^2-2)!$ so the highest power of $q$ dividing it is $q^{\frac{(q-1)(q^2-2)}{2}}$. For the numerator, notice that $\lambda^{\star}_i-\lambda^{\star}_j+j-i$ is divisible by $q$ iff $j-i$ is divisible by $q-1$. This means that exactly $(q-1)\binom{q+1}{2}$ factors in the numerator are divisible by $q$. Out of these the pairs $(i,j)=(r,q^2-q+r)$ for $r\in\{1,2,\dots,q-1\}$ are divisible by $q^2$. So the power of $q$ dividing your determinant is exactly $$(q-1)\binom{q+1}{2}+(q-1)-\frac{(q-1)(q^2-2)}{2}=\frac{(q-1)(q+4)}{2}$$ as desired.

Answer 3

Some more numbers, all without too big prime factors:

sage: factor(pasc(2).determinant())
-1 * 2^3
sage: factor(pasc(3).determinant())
-1 * 2 * 3^7 * 7
sage: factor(pasc(4).determinant())
2^23 * 3 * 13^2 * 17^2
sage: factor(pasc(5).determinant())
2^2 * 3^2 * 5^18 * 11^2 * 13^2 * 17 * 19 * 23^3
sage: factor(pasc(6).determinant())
-1 * 2^27 * 3^24 * 5^2 * 13^2 * 17^2 * 19^3 * 29^2 * 31^4 * 37^4
sage: factor(pasc(7).determinant())
-1 * 2^4 * 5^4 * 7^33 * 17^2 * 23^4 * 29 * 31 * 37^3 * 41^3 * 43^5 * 47^5 * 53^2
sage: factor(pasc(8).determinant())
2^122 * 3^2 * 5^2 * 7^4 * 11 * 13^2 * 17^3 * 19^2 * 29^4 * 31^4 * 41^2 * 43^2 * 47^2 * 53^4 * 59^6 * 61^6 * 67^4
sage: factor(pasc(9).determinant())
2^10 * 3^102 * 5^3 * 7 * 19^4 * 23 * 29^2 * 37^6 * 41^6 * 43^2 * 47 * 53 * 59^3 * 61^3 * 67^5 * 71^5 * 73^7 * 79^7 * 83^6
sage: factor(pasc(10).determinant())
-1 * 2^64 * 3^7 * 5^60 * 7^7 * 13 * 17^3 * 23^2 * 31^6 * 41^4 * 43^4 * 47^6 * 53^3 * 61^2 * 67^2 * 71^4 * 73^4 * 79^4 * 83^6 * 89^6 * 97^8 * 101^8 * 103^6 * 107^2