Is the determinant of a Hessenberg matrix whose elements are the Bernoulli numbers positive? The Bernoulli numbers $B_r$ are generated by
\begin{equation}\label{Bernoullu=No-dfn-Eq}
\frac{z}{e^z-1}=\sum_{r=0}^\infty B_r\frac{z^r}{r!}
=1-\frac{z}2+\sum_{r=1}^\infty B_{2r}\frac{z^{2r}}{(2r)!}, \quad \vert z\vert<2\pi.
\end{equation}
For $r\in\mathbb{N}$, I proved that
\begin{equation*}
(-1)^r\begin{vmatrix}
B_2 & -B_0 & 0 & \dotsm & 0& 0 & 0\\
B_{4} & B_2 & -B_0 & \dotsm & 0 & 0& 0\\
B_6 & B_{4} & B_2 & \dotsm & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots &\vdots & \vdots\\
B_{2r-4} & B_{2r-6} & B_{2r-8} & \dotsm & B_2 & -B_0 & 0\\
B_{2r-2} & B_{2r-4} & B_{2r-6} & \dotsm & B_{4} & B_2 & -B_0\\
B_{2r} & B_{2r-2} & B_{2r-4} & \dotsm & B_{6} & B_{4} & B_2
\end{vmatrix}
<0.
\end{equation*}
For $r\in\mathbb{N}$, let
\begin{equation*}
\mathfrak{D}_{r}=
\begin{vmatrix}
B_2 &B_0& 0 & \dotsm & 0 & 0 & 0\\
B_4 & B_2 &B_0& \dotsm & 0 & 0 & 0\\
B_6& B_4 & B_2 & \dotsm & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
B_{2(r-2)} & B_{2(r-3)} & B_{2(r-4)} & \dotsm & B_2 & B_0& 0\\
B_{2(r-1)} & B_{2(r-2)} & B_{2(r-3)} & \dotsm & B_4 & B_2 &B_0\\
B_{2r} & B_{2(r-1)} & B_{2(r-2)} & \dotsm & B_6 & B_4 & B_2
\end{vmatrix}.
\end{equation*}
Straightforward computation gives
\begin{equation*}
\mathfrak{D}_1=\frac{1}{6}, \quad\mathfrak{D}_2=\frac{11}{180}, \quad\mathfrak{D}_3=\frac{299}{7560}, \quad\mathfrak{D}_4=\frac{10417}{226800}.
\end{equation*}
Therefore, we guess that, for $r\in\mathbb{N}$,
\begin{equation*}
\mathfrak{D}_r>0.
\end{equation*}
Prove or deny this guess.
Since $(-1)^{r+1}B_{2r}>0$ for $r\in\mathbb{N}$, it is trivial that
\begin{equation*}
\mathfrak{D}_1=B_2^2-B_0B_4>0
\end{equation*}
and
\begin{equation*}
\mathfrak{D}_2=B_2^3+B_0^2B_6-2B_0B_2B_4>0.
\end{equation*}
Does this imply that the guess $\mathfrak{D}_r>0$ is trivially true?
See also similar discussions on the ResearchGate.
 A: The so-called guess $\mathfrak{D}_r>0$ can be proved to be true via mathematical induction and by the following recursive relation.
Let $H_0=1$ and
\begin{equation*}
H_r=
\begin{vmatrix}
h_{1,1} & h_{1,2} & 0 & \dotsc & 0 & 0\\
h_{2,1} & h_{2,2} & h_{2,3} & \dotsc & 0 & 0\\
h_{3,1} & h_{3,2} & h_{3,3} & \dotsc & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
h_{r-2,1} & h_{r-2,2} & h_{r-2,3} & \dotsc & h_{r-2,r-1} & 0 \\
h_{r-1,1} & h_{r-1,2} & h_{r-1,3} & \dotsc & h_{r-1,r-1} & h_{r-1,r}\\
h_{r,1} & h_{r,2} & h_{r,3} & \dotsc & h_{r,r-1} & h_{r,r}
\end{vmatrix}
\end{equation*}
for $r\in\mathbb{N}$. Then the sequence $H_r$ for $r\ge0$, with $H_1=h_{1,1}$, satisfies
\begin{equation}
H_r=\sum_{\ell=1}^r(-1)^{r-\ell}h_{r,\ell} \Biggl(\prod_{j=\ell}^{r-1}h_{j,j+1}\Biggr) H_{\ell-1}
\end{equation}
for $r\ge2$, where the empty product is understood to be $1$.
This recursive relation can be found on page 222 of the paper:
N. D. Cahill, J. R. D'Errico, D. A. Narayan, and J. Y. Narayan, Fibonacci determinants, College Math. J. 33 (2002), no. 3, 221--225; available online at https://doi.org/10.2307/1559033.
