Unified APD formula for row-shifted second-power lattices

From papers

Let $f_{d,2}$ be the permutation-sum function of the $n$ -th order row-shifted second-power lattice with positive integer shift $d$ , and define $T_{n-1}=n(n-1)/2$ . Let $V_{\text{Core}}(n)$ be the core value

V_{\text{Core}}(n):=T_{n-1}!\prod_{k=1}^{n-1}k!.

Unified row-shifted second-power lattice conjecture. For $n \geq 2$ ,

\operatorname{APD}_{T_{n-1}}(f_{d,2})=(2d)^{T_{n-1}}V_{\text{Core}}(n).

The formula is presented as a reconstruction from the multiplication-table value and is not accompanied by a general proof.

Progress summary

Open

A December 2025 preprint reports numerical support for the formula, but no general proof or counterexample has appeared.

The conjecture asserts that the alternating power difference has the stated closed form for every $n \ge 2$ and positive integer shift $d$ . It appears in a December 2025 preprint as a conjecture, together with the associated claim that the first appearance degree is $T_{n-1}$ ; the source explicitly says rigorous proofs remain to be established.

December 2025 numerical confirmation

The preprint reports exact numerical checks, including the squared natural-square case $d=n$ for $n \le 7$ and an explicit check at $n=4$ . These computations support the formula but do not establish it for general $n$ and $d$ ; no retrieved source gives a proof, counterexample, refutation, or independent verification.

Current status (as of August 2026): Numerical cases are settled in the reported examples, but the unified formula and the associated first-appearance-degree claim remain unproved and open in general.

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 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