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

About 1 year old · traced to 2025

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.

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 reader-written argument claims a complete proof of the conjecture, but it has not been independently checked.

Kenichi Takemura’s 2025 paper conjectures that the first nonzero alternating power difference of the standard Vandermonde matrix has the stated factorial value for every dimension. The paper explicitly leaves a rigorous general proof open.

Known results

Exact calculations verify the value formula for $n\leq 7$ (Takemura, 2025).
The same paper rewrites the conjectured value as $(n-1)!\det(V_n)$ , using $\det(V_n)=\prod_{k=1}^{n-1}k!$ .

Posted attempt

A reader-written argument claims a complete proof: it derives the universal identity $\operatorname{APD}_{n-1}(A)=(n-1)!\mathbf{1}^{\mathsf T}\operatorname{adj}(A)\mathbf{1}$ and applies $V_n^{-1}\mathbf{1}=e_1$ . The argument has not been independently verified.

Current status (as of August 2026): The conjecture has a complete but unverified posted proof claim; the published source still treats it as open, so independent verification remains outstanding.

Sources

ProofThis solution needs a summarySee full solutionHide full solution

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 $m_1(V_n)=n-1$ .