Harmonic APD relation between Hilbert and identity matrices

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

$$\sum_{m\ge0}\operatorname{APD}_m(A)\frac{z^m}{m!} =\det(e^{zA_{ij}}).$$

Because

$$(e^{zA_{ij}})=J_n+zA+O(z^2), \qquad J_n=\mathbf1\mathbf1^{\mathsf T},$$

and $J_n$ has rank one, determinant multilinearity yields the universal identity

$$\operatorname{APD}_{n-1}(A) =(n-1)!\mathbf1^{\mathsf T}\operatorname{adj}(A)\mathbf1. \tag{1}$$

Let

$$(H_n)_{ij}=\frac1{i+j-1}.$$

This is the Gram matrix of $1,x,\ldots,x^{n-1}$ in $L^2([0,1])$, since

$$(H_n)_{ij}=\int_0^1x^{i+j-2}\,dx.$$

The all-ones vector represents evaluation at $x=1$. Therefore $\mathbf1^{\mathsf T}H_n^{-1}\mathbf1$ is the value $K_n(1,1)$ of the reproducing kernel for polynomials of degree less than $n$. An orthonormal basis is

$$\phi_k(x)=\sqrt{2k+1}\,P_k(2x-1), \qquad 0\le k<n,$$

where $P_k$ denotes the $k$-th Legendre polynomial. Since $P_k(1)=1$,

$$\mathbf1^{\mathsf T}H_n^{-1}\mathbf1 =\sum_{k=0}^{n-1}\phi_k(1)^2 =\sum_{k=0}^{n-1}(2k+1) =n^2.$$

Equation (1) now gives

$$\operatorname{APD}_{n-1}(H_n) =(n-1)!\det(H_n)n^2 =n\,n!\det(H_n).$$

For the identity matrix,

$$\sum_{m\ge0}\operatorname{APD}_m(I_n)\frac{z^m}{m!} =(e^z-1)^{n-1}(e^z+n-1).$$

Hence

$$\operatorname{APD}_{n-1}(I_n)=n!,$$

and therefore

$$\operatorname{APD}_{n-1}(H_n) =n\det(H_n)\operatorname{APD}_{n-1}(I_n).$$

Both identities hold for all $n\ge2$. For the endpoint $n=1$, they also hold under the natural extension $\operatorname{APD}_0(A)=\sum_{\sigma\in S_1}\operatorname{sgn}(\sigma)=1$.