Unified APD formula for row-shifted second-power lattices

About 1 year old · traced to 2025

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.

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

The original paper reports only numerical evidence, but a reader-posted calculation now claims a complete proof for every dimension and shift; that calculation has not been independently verified.

Kenichi Takemura's December 2025 preprint formulates the unified identity as Conjecture 12 for the row-shifted square lattice, alongside the claim that the first nonzero degree is $T_{n-1}$ . The paper explicitly presents both as conjectural and says rigorous proofs remain to be established.

Known results

The preprint reports exact numerical checks for the formula, including the specialization $d=n$ through $n=7$ and a detailed case at $n=4$ .
The associated first-appearance claim $m_1(f_{d,2})=T_{n-1}$ is likewise reported as numerically confirmed, not proved.

Posted attempt

A reader-posted determinant-generating-function and Vandermonde calculation claims a complete proof for all $n\ge2$ and positive integer $d$ , deriving both the first-appearance degree and the stated value. The attempt has not been independently verified.

Current status (as of August 2026): The preprint establishes only numerical instances, while a complete-proof claim exists in the discussion but remains unverified; consequently the general conjecture is not settled.

Sources

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 $d$ .