Let H_n be the n-th order Hilbert matrix, and let det(H_n) denote its determinant. Write operatornameAPD_m(H_n) for its alternating power difference. Hilbert-matrix first appearance value conjecture. For n ≥ 2, operatornameAPD_n-1(H_n)=det(H_n),n,n!. The supplied status evidence says that the relation was confirmed by exact rational calculation for n=1 through n=7; it is therefore recorded as solved.

First appearance value conjecture for Hilbert matrices

About 1 year old · traced to 2025

Let $H_n$ be the $n$ -th order Hilbert matrix, and let $\det(H_n)$ denote its determinant. Write $\operatorname{APD}_m(H_n)$ for its alternating power difference.

Hilbert-matrix first appearance value conjecture. For $n \geq 2$ ,

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

The supplied status evidence says that the relation was confirmed by exact rational calculation for $n=1$ through $n=7$ ; it is therefore recorded as solved.

References

Primary source

Kenichi Takemura, “Alternating Power Difference and Matrix Symmetry: Closed-Form Formulas for the First Appearance Degree m_1”, arXiv:2512.18169 (2025).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A December 2025 paper verified the formula only in small dimensions, while a reader-posted argument now claims a complete general proof that has not been independently checked.

Takemura’s December 2025 paper states the Hilbert-matrix first-appearance formula as Conjecture 8, predicting $\operatorname{APD}_{n-1}(H_n)=\det(H_n)n\,n!$ for $n\geq2$ . The paper explicitly treats the all-dimensional statement as unproved.

Known results

Exact rational computation verifies $m_1(H_n)=n-1$ and the conjectured value for $2\leq n\leq7$ .
The related identity-matrix formula was checked for $1\leq n\leq7$ .

Posted attempt

A reader-posted argument claims a complete proof for every $n$ : it derives a general adjugate identity for the first APD and evaluates $\mathbf{1}^{\mathsf T}H_n^{-1}\mathbf{1}=n^2$ using the reproducing kernel of polynomials on $[0,1]$ . This attempt has not been independently verified.

Current status (as of August 2026): The formula is verified computationally through $n=7$ , and a complete proof has been posted but remains unverified; the all-dimensional conjecture is therefore not settled.

Sources

ProofThis solution needs a summarySee full solutionHide full solution

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 $\operatorname{APD}_0(A)=\sum_{\sigma\in S_1}\operatorname{sgn}(\sigma)=1$ .