First appearance degree conjecture for standard circulant matrices

About 1 year old · traced to 2025

Let $C_n$ be the $n$ -th order standard circulant matrix, and let $m_1(C_n)$ denote its first appearance degree. Let $\operatorname{fix}$ be the fixed-point function associated with the identity matrix of order $n$ .

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

m_1(C_n)=m_1(\operatorname{fix})=n-1.

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

The conjecture was checked through ten dimensions, and a reader-written complete proof has now been posted but has not been independently verified.

Kenichi Takemura’s December 2025 paper formulates the conjecture that the standard circulant and fixed-point statistic first become nonzero at degree $n-1$ for every $n \geq 2$ . It reports numerical verification through $n=10$ but supplies no general proof.

Known results

Takemura, 2025: $m_1(\operatorname{fix})=n-1$ and $\operatorname{APD}_{n-1}(\operatorname{fix})=n!$ for general $n$ .
Takemura, 2025: $m_1(C_n)=n-1$ was verified numerically for $n \leq 10$ ; the general circulant assertion remained conjectural in the paper.

Posted attempt

A reader-written argument claims a complete proof by evaluating the exponential generating determinants for both $C_n$ and $\operatorname{fix}$ , and derives explicit formulas including $m_1(C_n)=m_1(\operatorname{fix})=n-1$ . The attempt has not been independently verified.

Current status (as of August 2026): The fixed-point case and the circulant case through $n \leq 10$ are established, while the posted general proof claim is unverified and the conjecture is not independently 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.