Determinant formula for the first appearance value of standard Vandermonde matrices

About 16 years old · traced to 2010

Let $V_n$ be the standard Vandermonde matrix associated with $(1,2,\ldots,n)$ , let $f_V$ be its associated permutation-sum function, and let $\det(V_n)$ denote its determinant. The first appearance degree is $m_1=n-1$ .

Standard-Vandermonde determinant formula. The first appearance value is

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

The relationship was verified by exact calculation for $n=2$ through $n=10$ ; the supplied status evidence therefore records this candidate 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).

Additional references

5 papers in this index state this conjecture (2010–2025). The statement above is taken from the most recent of them; the others are arXiv:2410.02286, arXiv:2309.01123, arXiv:2207.11200, arXiv:1012.2719.

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A new posted argument claims a complete proof, but no independent verification has confirmed it, so the formula is not settled.

Kenichi Takemura proposed the identity in a December 2025 preprint, together with the claim that the first nonzero degree is $m_1=n-1$ .

December 2025 conjecture

The preprint records exact agreement for $2\le n\le10$ and gives the equivalent product formula $\operatorname{APD}_{n-1}(f_V)=(n-1)!\prod_{k=1}^{n-1}k!$ , but labels the general statements conjectural and calls for rigorous proofs.

Posted attempt

A posted argument claims a complete proof: it derives $\operatorname{APD}_{n-1}(A)=(n-1)!\mathbf1^{\mathsf T}\operatorname{adj}(A)\mathbf1$ for every square matrix, then uses the first column of $V_n$ to obtain the proposed formula and the Vandermonde product. The argument has not been independently verified.

Current status (as of August 2026): the formula and $m_1=n-1$ have computational support and a complete proof has been posted, but neither the proof nor the general claim has independent verification.

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 $m_1(V_n)=n-1$ .