MathDB

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

From papers

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

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

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

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

APD⁡Tn−1(f)=(2d)Tn−1Tn−1!∏k=1n−1k!.\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.

Progress summary

Open

A December 2025 preprint reports numerical evidence for the proposed formula, but no proof or counterexample has been found.

The conjecture predicts both the first nonzero degree, Tn−1T_{n-1}, and the stated closed value of APD⁡Tn−1(f)\operatorname{APD}_{T_{n-1}}(f) for every n≥2n \ge 2 and positive integer dd. Kenichi Takemura’s preprint, posted in December 2025, presents the claim as numerically supported rather than proved.

December 2025 preprint

Takemura reports exact numerical checks for small cases, including n=2,…,7n=2,\ldots,7, and records the first-appearance and value statements as conjectures. The preprint explicitly identifies rigorous proofs as future work; no counterexample or proof is reported.

Current status (as of August 2026): The formula remains an unproved conjecture, with numerical checks reported for small cases and no public proof or counterexample located.

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).

alternating-power-differencecombinatoricslattice-matrices

Solutions 1

Proof

For any n×nn\times n matrix AA, the determinant expansion gives

∑m≥0APD⁡m(A)zmm!=det⁡(ezAij)i,j=1n.(1)\sum_{m\ge0}\operatorname{APD}_m(A)\frac{z^m}{m!} =\det(e^{zA_{ij}})_{i,j=1}^n. \tag{1}

Set T=(n2)T=\binom n2. First consider the multiplication table Mij=ijM_{ij}=ij. With y=ezy=e^z, the Vandermonde determinant gives

det⁡(ezij)i,j=1n=det⁡(yij)i,j=1n=yn(n+1)/2∏1≤i<j≤n(yj−yi).\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 yj−yi=(j−i)z+O(z2)y^j-y^i=(j-i)z+O(z^2). Since there are TT factors and

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

equation (1) yields

m1(M)=T,APD⁡T(M)=T!∏k=1n−1k!.m_1(M)=T,\qquad \operatorname{APD}_T(M)=T!\prod_{k=1}^{n-1}k!.

Now take the row-shifted second-power lattice

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

and put q=e2dzq=e^{2dz}. Since

Aij=j2+d2(i−1)2+2d(i−1)j,A_{ij}=j^2+d^2(i-1)^2+2d(i-1)j,

extracting row and column factors gives

det⁡(ezAij)=exp⁡(z∑j=1nj2+zd2∑i=0n−1i2)qT∏0≤i<j≤n−1(qj−qi).\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

qj−qi=2d(j−i)z+O(z2),q^j-q^i=2d(j-i)z+O(z^2),

this determinant has exact vanishing order TT and leading coefficient

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

Therefore

m1(A)=(n2),APD⁡(n2)(A)=(2d)(n2)((n2))!∏k=1n−1k!.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

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

then

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

These identities hold for every n≥2n\ge2 and every positive integer dd.

0 endorsements
Shivam Patel ·