Let V_n be the standard Vandermonde matrix associated with (1,2,…,n), and let f_V be its associated permutation-sum function. Standard-Vandermonde value conjecture. For n ≥ 2, operatornameAPD_n-1(f_V)=(n-1)!∏_k=1^n-1k!. The formula completely matches exact calculation results for n ≤ 7, but no general proof is supplied.

First appearance value conjecture for standard Vandermonde matrices

From papers

Let $V_n$ be the standard Vandermonde matrix associated with $(1,2,\ldots,n)$ , and let $f_V$ be its associated permutation-sum function.

Standard-Vandermonde value conjecture. For $n \geq 2$ ,

\operatorname{APD}_{n-1}(f_V)=(n-1)!\prod_{k=1}^{n-1}k!.

The formula completely matches exact calculation results for $n \leq 7$ , but no general proof is supplied.

Progress summary

Open

The predicted formula matches all tested cases, but no general proof or counterexample has been found.

The conjecture predicts an exact first nonzero value for the permutation-sum function of the standard Vandermonde matrix. No proposer or original date is identified in the retrieved sources.

Known results

Exact computation verifies the value formula through dimension $n=10$ .
Exact computation verifies the first-appearance degree $m_1=n-1$ through dimension $n=7$ .

December 2025 paper

A paper on alternating power differences records both assertions as conjectures and explicitly lists rigorous general proofs as future work. It reports no proof, counterexample, withdrawal, or competing verification.

Current status (as of August 2026): The formula is established by exact computation for $n\leq 10$ , but its general validity and the corresponding first-appearance-degree claim remain open.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

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

1 week ago

For every $n\times n$ matrix $A$ , write $\mathbf1=(1,\ldots,1)^{\mathsf T}$ . The determinant expansion gives

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

As $z\to0$ ,

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

Since $J_n$ has rank one, determinant multilinearity implies

\det(e^{zA_{ij}}) =z^{n-1}\mathbf1^{\mathsf T}\operatorname{adj}(A)\mathbf1 +O(z^n).

The coefficient also follows directly from

\det(zA+\mathbf1\mathbf1^{\mathsf T}) =z^n\det A +z^{n-1}\mathbf1^{\mathsf T}\operatorname{adj}(A)\mathbf1.

Hence, universally,

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

For the standard Vandermonde matrix

(V_n)_{ij}=i^{j-1},

its first column equals $\mathbf1$ , so

V_n^{-1}\mathbf1=e_1.

Consequently

\mathbf1^{\mathsf T}\operatorname{adj}(V_n)\mathbf1 =\det(V_n)\mathbf1^{\mathsf T}V_n^{-1}\mathbf1 =\det(V_n).

The Vandermonde product gives

\det(V_n) =\prod_{1\le i<j\le n}(j-i) =\prod_{k=1}^{n-1}k!.

Substituting into (1) proves both equivalent formulas:

\operatorname{APD}_{n-1}(V_n) =(n-1)!\det(V_n) =(n-1)!\prod_{k=1}^{n-1}k!.

The positive leading coefficient also proves $m_1(V_n)=n-1$ .

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Shivam Patel · 1 day ago