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First appearance value conjecture for standard circulant matrices

From papers

Let CnC_n be the nn-th order standard circulant matrix. Define the triangular number

Tn−1=(n−1)n2.T_{n-1}=\frac{(n-1)n}{2}.

Let APD⁡m(Cn)\operatorname{APD}_m(C_n) denote its alternating power difference.

Standard-circulant first appearance value conjecture. For n≥2n \geq 2,

APD⁡n−1(Cn)=(−1)Tn−1nn−2APD⁡n−1(fix⁡),\operatorname{APD}_{n-1}(C_n)=(-1)^{T_{n-1}}n^{n-2}\operatorname{APD}_{n-1}(\operatorname{fix}),

where APD⁡n−1(fix⁡)=n!\operatorname{APD}_{n-1}(\operatorname{fix})=n!. This relationship has been verified for n≤10n \leq 10, but no general proof is supplied.

Progress summary

Open

The conjecture has only been checked in small cases, and no general proof or counterexample has been publicly reported.

The conjecture asserts that the first nonzero alternating power difference of the standard circulant matrix equals a signed power of its order times the corresponding fixed-point value. It was recorded as Conjecture 4 in a manuscript posted in December 2025.

Known results

  • Direct verification is reported for n≤10n \le 10; no general proof is supplied.

December 2025 manuscript

Kenichi Takemura’s arXiv manuscript presents the stated identity as a conjecture and explicitly identifies rigorous proofs for all parameters as future work. The scan found no later proof, counterexample, or independent verification.

Current status (as of August 2026): The identity is verified only for n≤10n \le 10; its validity for general n≥2n \ge 2 remains open.

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-differencecirculant-matricescombinatorics

Solutions 1

Proof

Write Tn−1=n(n−1)/2T_{n-1}=n(n-1)/2, and define

APD⁡m(A)=∑σ∈Snsgn⁡(σ)(∑j=1nAj,σ(j))m.\operatorname{APD}_m(A) =\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma) \left(\sum_{j=1}^nA_{j,\sigma(j)}\right)^m.

The determinant expansion gives the exponential generating function

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

For the standard circulant

(Cn)ij=((i+j−2) mod n)+1,(C_n)_{ij}=((i+j-2)\bmod n)+1,

put y=ezy=e^z. Reversing rows 2,…,n2,\ldots,n converts (y(Cn)ij)(y^{(C_n)_{ij}}) into the usual right-circulant with first row (y,y2,…,yn)(y,y^2,\ldots,y^n), contributing the sign (−1)(n−1)(n−2)/2(-1)^{(n-1)(n-2)/2}. If ω\omega is a primitive nn-th root of unity, its circulant eigenvalues are

∑j=0n−1yj+1ωkj=y(1−yn)1−yωk(0≤k<n).\sum_{j=0}^{n-1}y^{j+1}\omega^{kj} =\frac{y(1-y^n)}{1-y\omega^k} \qquad(0\le k<n).

Since

∏k=0n−1(1−yωk)=1−yn,\prod_{k=0}^{n-1}(1-y\omega^k)=1-y^n,

we obtain

∑m≥0APD⁡m(Cn)zmm!=(−1)(n−1)(n−2)/2enz(1−enz)n−1.\sum_{m\ge0}\operatorname{APD}_m(C_n)\frac{z^m}{m!} =(-1)^{(n-1)(n-2)/2}e^{nz}(1-e^{nz})^{n-1}.

This has a zero of exact order n−1n-1 at z=0z=0. Therefore

m1(Cn)=n−1,m_1(C_n)=n-1,

and its first nonzero coefficient is

APD⁡n−1(Cn)=(−1)n(n−1)/2nn−1(n−1)!=(−1)Tn−1nn−2n!.\operatorname{APD}_{n-1}(C_n) =(-1)^{n(n-1)/2}n^{n-1}(n-1)! =(-1)^{T_{n-1}}n^{n-2}n!.

For the fixed-point statistic, the same determinant identity gives

∑m≥0APD⁡m(fix⁡)zmm!=det⁡(Jn+(ez−1)In)=(ez−1)n−1(ez+n−1).\sum_{m\ge0}\operatorname{APD}_m(\operatorname{fix})\frac{z^m}{m!} =\det\big(J_n+(e^z-1)I_n\big) =(e^z-1)^{n-1}(e^z+n-1).

Consequently

m1(fix⁡)=n−1,APD⁡n−1(fix⁡)=n!.m_1(\operatorname{fix})=n-1, \qquad \operatorname{APD}_{n-1}(\operatorname{fix})=n!.

This proves both asserted comparisons. More generally, the complete closed formula is

APD⁡m(Cn)=(−1)Tn−1nm(n−1)! {m+1n},\operatorname{APD}_m(C_n) =(-1)^{T_{n-1}}n^m(n-1)!\,{m+1\brace n},

where {m+1n}{m+1\brace n} denotes a Stirling number of the second kind.

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