First appearance value conjecture for multiplication-table lattices

About 1 year old · traced to 2025

Let $M_n$ be the $n$ -th order multiplication-table first-power lattice and let $f_1$ be its associated permutation-sum function. Define $T_{n-1}=n(n-1)/2$ .

Multiplication-table lattice value conjecture. For $n \geq 2$ ,

\operatorname{APD}_{T_{n-1}}(f_1)=T_{n-1}!\prod_{k=1}^{n-1}k!.

The relationship has been verified for $n \leq 7$ by exact integer arithmetic, 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 posted argument claims a complete proof of the factorial formula, but it has not been independently checked and the published source still presents the result as a conjecture.

Kenichi Takemura’s paper, posted in December 2025, conjectures that the first nonzero value for the first-power multiplication table is $T_{n-1}!\prod_{k=1}^{n-1}k!$ .

Known results

Exact integer arithmetic verifies the formula for $n\le 7$ ; the paper supplies no proof for general $n\ge 2$ .

Posted attempt

A posted argument claims a complete proof: it derives the first-appearance coefficient from the determinant generating function and a Vandermonde factorization for $M_{ij}=ij$ . The argument has not been independently verified.

Current status (as of August 2026): The formula is verified for $n\le 7$ and has an unverified claimed proof, while general validity for $n\ge 2$ remains unsettled.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

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

\sum_{m\ge0}\operatorname{APD}_m(A)\frac{z^m}{m!} =\det(e^{zA_{ij}})_{i,j=1}^n. \tag{1}

Set $T=\binom n2$ . First consider the multiplication table $M_{ij}=ij$ . With $y=e^z$ , the Vandermonde determinant gives

\det(e^{zij})_{i,j=1}^n =\det(y^{ij})_{i,j=1}^n =y^{n(n+1)/2} \prod_{1\le i<j\le n}(y^j-y^i).

Each factor satisfies $y^j-y^i=(j-i)z+O(z^2)$ . Since there are $T$ factors and

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

equation (1) yields

m_1(M)=T,\qquad \operatorname{APD}_T(M)=T!\prod_{k=1}^{n-1}k!.

Now take the row-shifted second-power lattice

A_{ij}=(j+(i-1)d)^2,\qquad d>0,

and put $q=e^{2dz}$ . Since

A_{ij}=j^2+d^2(i-1)^2+2d(i-1)j,

extracting row and column factors gives

\det(e^{zA_{ij}}) = \exp\left( z\sum_{j=1}^n j^2+ zd^2\sum_{i=0}^{n-1}i^2 \right) q^T \prod_{0\le i<j\le n-1}(q^j-q^i).

Because

q^j-q^i=2d(j-i)z+O(z^2),

this determinant has exact vanishing order $T$ and leading coefficient

(2d)^T\prod_{k=1}^{n-1}k!.

Therefore

m_1(A)=\binom n2, \qquad \operatorname{APD}_{\binom n2}(A) =(2d)^{\binom n2} \left(\binom n2\right)! \prod_{k=1}^{n-1}k!.

Equivalently, if

V_{\mathrm{Core}}(n)=T!\prod_{k=1}^{n-1}k!,

then

\operatorname{APD}_T(A)=(2d)^TV_{\mathrm{Core}}(n).

These identities hold for every $n\ge2$ and every positive integer $d$ .