Let C_n be the n-th order standard circulant matrix. Define the triangular number T_n-1=frac(n-1)n2. Let operatornameAPD_m(C_n) denote its alternating power difference. Standard-circulant first appearance value conjecture. For n ≥ 2, operatornameAPD_n-1(C_n)=(-1)^T_n-1n^n-2operatornameAPD_n-1(operatornamefix), where operatornameAPD_n-1(operatornamefix)=n!. This relationship has been verified for n ≤ 10, but no general proof is supplied.

First appearance value conjecture for standard circulant matrices

About 1 year old · traced to 2025

Let $C_n$ be the $n$ -th order standard circulant matrix. Define the triangular number

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

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

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

\operatorname{APD}_{n-1}(C_n)=(-1)^{T_{n-1}}n^{n-2}\operatorname{APD}_{n-1}(\operatorname{fix}),

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

An unverified posted argument claims a complete proof of the conjecture, while the author’s manuscript records only checks through ten.

Kenichi Takemura’s December 2025 manuscript states the identity as a conjecture for $n\geq 2$ , with $\operatorname{APD}_{n-1}(\operatorname{fix})=n!$ . It reports verification only for $n\leq 10$ .

Known results

The manuscript also conjectures $m_1(C_n)=m_1(\operatorname{fix})=n-1$ , verified computationally for $n\leq 10$ .

Posted attempt

A posted argument claims a complete proof, using the determinant generating function and obtaining $\operatorname{APD}_m(C_n)=(-1)^{T_{n-1}}n^m(n-1)!\,{m+1\brace n}$ . It therefore implies the conjectured first-appearance value and the fixed-point comparison, but the argument has not been independently verified.

Current status (as of August 2026): The published record establishes only verification for $n\leq 10$ ; a complete proof has been posted but remains unverified, so the general conjecture is not settled.

Sources

ProofThis solution needs a summarySee full solutionHide full solution

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

\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

\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

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

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

\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

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

we obtain

\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 $n-1$ at $z=0$ . Therefore

m_1(C_n)=n-1,

and its first nonzero coefficient is

\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

\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

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

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

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