First appearance value conjecture for multiplication-table lattices

From papers

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.

Progress summary

Open

The conjecture has been checked in small cases but no general proof or counterexample has appeared.

The conjecture asserts a factorial product formula for the first-appearance value of the multiplication-table lattice, namely $\operatorname{APD}_{T_{n-1}}(f_1)=T_{n-1}!\prod_{k=1}^{n-1}k!$ . The relevant paper records it as Conjecture 11 and explicitly says that a rigorous proof is still needed.

Known results

Exact integer arithmetic verifies the formula for $n\le 7$ .
The associated first-appearance degree has been numerically verified as $m_1(f_1)=T_{n-1}$ in those cases.
No general proof, counterexample, correction, or independent verification was found.

December 2025 arXiv record

The paper linked on December 23, 2025, presents the formula as a conjecture rather than a theorem and identifies rigorous proofs of the first-appearance degree and value conjectures as a priority. No later source in the scan claims progress.

Current status (as of August 2026): The formula is verified for $n\le 7$ , but its validity for general $n\ge 2$ remains open, with no public proof or counterexample recorded.

Sources

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 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 $d$ .

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Shivam Patel · 1 day ago