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

From papers

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.

Progress summary

Open

No publicly verified progress appears to have been made on this determinant conjecture.

No public discussion or published progress was found for the factorial determinant conjecture.

Current status (as of August 2026): the conjecture remains open, with no recorded proof, counterexample, or verified progress.

Sources & referencesView supporting material

Primary source

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

1 week ago

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.

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Samuel Schlesinger · 5 days ago