Let rho be the evaluation map used in the paper. Assuming the divisibility conjecture, for every integer n define the square matrix mathbfM_n by mathbfM_n(i,j)=rho(x^i(1+x)^j)2^-lfloor j/2 rfloor,qquad 0≤ i,j≤ n. Let d_n=det(mathbfM_n). Factorial determinant conjecture. For all n≥ 0, d_n=(n-1)!^varepsilon(1)(n-2)!^varepsilon(2)(n-3)!^varepsilon(3)… 1!^varepsilon(n-1), where varepsilon(k)= 2&if k is odd,4&if k is even. This is a conjectural determinant evaluation built on the preceding divisibility assertion; no proof or resolution is supplied in the given text.

Factorial determinant conjecture for the normalized evaluation matrices

About 6 years old · traced to 2020

Let $\rho$ be the evaluation map used in the paper. Assuming the divisibility conjecture, for every integer $n$ define the square matrix $\mathbf{M}_n$ by

\mathbf{M}_n(i,j)=\rho(x^i(1+x)^j)2^{-\lfloor j/2\rfloor},\qquad 0\leq i,j\leq n.

Let $d_n=\det(\mathbf{M}_n)$ . Factorial determinant conjecture. For all $n\geq 0$ ,

d_n=(n-1)!^{\varepsilon(1)}(n-2)!^{\varepsilon(2)}(n-3)!^{\varepsilon(3)}\dots 1!^{\varepsilon(n-1)},

where

\varepsilon(k)=\begin{cases}2&\text{if }k\text{ is odd},\\4&\text{if }k\text{ is even.}\end{cases}

This is a conjectural determinant evaluation built on the preceding divisibility assertion; no proof or resolution is supplied in the given text.

References

Primary source

Frédéric Chapoton and Guo-Niu Han, “On the roots of the Poupard and Kreweras polynomials”, arXiv:2001.01449 (2020).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Open

A reader-submitted correction says the recorded statement is false as written, while the intended version has reportedly been proved, but that claim has not been independently verified.

The conjecture asserts a factorial formula for the determinant of normalized evaluation matrices, conditional on the preceding divisibility conjecture. With the stated inclusive indexing, the submission identifies a direct contradiction at $n=2$ : the determinant is $4$ , whereas the displayed product is $1$ .

Community submission (unverified), August 15, 2026

A submitted correction argues that the issue is an indexing error: the intended matrix should be $N\times N$ with $0\le i,j\le N-1$ . It claims that Guo-Niu Han's 2026 preprint Dilated Hankel determinants, Theorem 26.1, proves the corresponding factorial determinant formula for every $N\ge 1$ . The submission is unverified here.

Current status (as of September 2026): the literal inclusive-index statement is claimed false, while the corrected $N\times N$ formulation is claimed proved but remains unverified.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

Indexing correction and resolution

The statement is false under its literal inclusive indexing.

It defines $M_n$ using $0\le i,j\le n$ , making $M_n$ an $(n+1)\times(n+1)$ matrix. At $n=2$ , this is the leading $3\times3$ block of the matrix displayed in the source:

M_2= \begin{pmatrix} 1&1&1\\ 1&2&3\\ 2&8&18 \end{pmatrix}.

Its determinant is

\det(M_2)=4.

The product printed for $n=2$ , however, is only

\prod_{k=1}^{1}((2-k)!)^{\varepsilon(k)} =(1!)^2=1.

Thus the literal MathDB statement gives $4=1$ and is false.

The original source itself indicates the intended correction: immediately after specifying $0\le i,j\le n$ , it displays a $6\times6$ matrix as $M_6$ , rather than the $7\times7$ matrix required by the inclusive bounds.

This ambiguity is resolved in Guo-Niu Han's 2026 preprint Dilated Hankel determinants. Section 26 explicitly defines $M_N$ as the $N\times N$ matrix

M_N(i,j) = 2^{-\lfloor j/2\rfloor}\rho\!\left(x^i(1+x)^j\right), \qquad 0\le i,j\le N-1.

Theorem 26.1 then proves, for every $N\ge1$ ,

\det M_N = \prod_{k=1}^{N-1}((N-k)!)^{\varepsilon(k)}, \qquad \varepsilon(k)= \begin{cases} 2,&k\text{ odd},\\ 4,&k\text{ even}. \end{cases}

Therefore the intended $N\times N$ conjecture is proved, while the inclusive-index version currently recorded by MathDB is false. This reply is an indexing/status correction, not a claim of a new proof of Han's theorem.