Unified first appearance formula for row-shifted second-power lattices

Let $A$ be the $n$ -th order row-shifted second-power lattice with entries $A[i,j]=(j+(i-1)d)^2$ , where $d$ is a positive integer, and define

T_{n-1}=\frac{n(n-1)}{2}.

Let $f(\sigma)=\sum_{i=1}^n A[i,\sigma(i)]$ .

Row-shifted second-power lattice value conjecture. For $n \geq 2$ ,

\operatorname{APD}_{T_{n-1}}(f)=(2d)^{T_{n-1}}T_{n-1}!\prod_{k=1}^{n-1}k!.

This formula is presented as a closed form based on numerical evidence, with no general proof 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 complete proof has been posted, but it has not been independently checked, so the conjecture is not yet settled.

Kenichi Takemura proposed the formula in a December 2025 preprint, conjecturing both the first nonzero degree $T_{n-1}$ and its closed value for every $n\ge2$ and positive integer $d$ .

Known results

Takemura, 2025: exact numerical checks are reported through $n=7$ .
The preprint labels both statements conjectures and identifies rigorous proofs as future work.

Posted attempt (date unavailable)

A reader-written attempt claims a complete proof: it expresses the exponential generating function as a determinant, factors the row and column terms, and applies a Vandermonde factorization to obtain the exact vanishing order $T_{n-1}$ and the conjectured leading coefficient. The argument has not been independently verified.

Current status (as of August 2026): The conjecture has a posted but unverified complete-proof claim; the primary source still presents it as unproved, and no independent proof, counterexample, or verification was found.

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 $d$ .