First appearance degree conjecture for row-shifted second-power lattices

For any $n\times n$ matrix $A$, the determinant expansion gives

$$\sum_{m\ge0}\operatorname{APD}_m(A)\frac{z^m}{m!} =\det(e^{zA_{ij}})_{i,j=1}^n. \tag{1}$$

Set $T=\binom n2$. First consider the multiplication table $M_{ij}=ij$. With $y=e^z$, the Vandermonde determinant gives

$$\det(e^{zij})_{i,j=1}^n =\det(y^{ij})_{i,j=1}^n =y^{n(n+1)/2} \prod_{1\le i<j\le n}(y^j-y^i).$$

Each factor satisfies $y^j-y^i=(j-i)z+O(z^2)$. Since there are $T$ factors and

$$\prod_{1\le i<j\le n}(j-i)=\prod_{k=1}^{n-1}k!,$$

equation (1) yields

$$m_1(M)=T,\qquad \operatorname{APD}_T(M)=T!\prod_{k=1}^{n-1}k!.$$

Now take the row-shifted second-power lattice

$$A_{ij}=(j+(i-1)d)^2,\qquad d>0,$$

and put $q=e^{2dz}$. Since

$$A_{ij}=j^2+d^2(i-1)^2+2d(i-1)j,$$

extracting row and column factors gives

$$\det(e^{zA_{ij}}) = \exp\left( z\sum_{j=1}^n j^2+ zd^2\sum_{i=0}^{n-1}i^2 \right) q^T \prod_{0\le i<j\le n-1}(q^j-q^i).$$

Because

$$q^j-q^i=2d(j-i)z+O(z^2),$$

this determinant has exact vanishing order $T$ and leading coefficient

$$(2d)^T\prod_{k=1}^{n-1}k!.$$

Therefore

$$m_1(A)=\binom n2, \qquad \operatorname{APD}_{\binom n2}(A) =(2d)^{\binom n2} \left(\binom n2\right)! \prod_{k=1}^{n-1}k!.$$

Equivalently, if

$$V_{\mathrm{Core}}(n)=T!\prod_{k=1}^{n-1}k!,$$

then

$$\operatorname{APD}_T(A)=(2d)^TV_{\mathrm{Core}}(n).$$

These identities hold for every $n\ge2$ and every positive integer $d$.