Some determinants which are closely related to recurrences Let the sequence $(a(n,k))_{ n  \in \mathbb{Z}}$ satisfy  $$\sum_{j=0}^k c(k,j)a(n-j,k)=0$$ with $c(k,j)=c(k,k-j)$ and $c(k,0)=1$ and with initial values $a(-n,k)=0$ for $1\leq n\leq{k-1}$ and $a(0,k)=1.$
For  example for the binomial coefficients $c(k,j)=\binom{k}{j},$ we get $a(n,k)=\binom{-k}{n}.$
Let $A_k(n)$ be the matrix whose entries are $a(n+i+j-k+2,k)$ for $0 \leq i,j \leq {k-2}.$
Computer experiments suggest that $$\det{A_k(n)}=(-1)^{kn+1+\binom{k+1}{2}}a(n,k).$$
A similar result seems to hold if $c(k,j)=-c(k,k-j).$
Any idea how to prove this?  Is this a known result or a special case of a more general result?
 A: The question concerns the determinant of a Hankel matrix, or a fixed element of a Hankel matrix transform of a shifted sequence $a(n,k)$ for a fixed $k$, although I do not see how this fact alone can be useful. I give a standalone proof below.

Let $k$ be fixed. For the sake of simplicity, let's denote $c_j := c(k,j)$ and $a_n := a(n,k)$.
First, notice that the sequence $(a_n)$ has the characteristic polynomial
$$C(x) := \sum_{j=0}^k c_j x^{k-j},$$
while $C(-x)$ gives the characteristic polynomial for the sequence $((-1)^n a_n)$.
Now, let $b_t(n)$ for $t\in\{0,\dots,k-1\}$ be the determinant of the matrix obtained from the $(k-1)\times k$ matrix: $$\big(a_{n+i+j-k}\big)_{1\leq i\leq k-1\atop 0\leq j\leq k-1}$$
by removing the column with $j=t$.
Then $\det A_k(n) = b_0(n) = b_{k-1}(n+1)$.
From the recurrence for $a_n$, it follows that
$$
\begin{bmatrix} b_0(n) \\ b_1(n)\\ \vdots\\ b_{k-1}(n) \end{bmatrix} = M\cdot
\begin{bmatrix} b_0(n-1) \\ b_1(n-1)\\ \vdots\\ b_{k-1}(n-1) \end{bmatrix},$$
where
$$
M := \begin{bmatrix} 
(-1)^{k-1} c_{k-1} & (-1)^{k-1} c_k & 0 & \dots & 0\\
(-1)^{k-2} c_{k-2} & 0 & (-1)^{k-1} c_k & \dots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-c_1 & 0 & 0 & \dots & (-1)^{k-1} c_k\\
1 & 0 & 0 & \dots & 0
\end{bmatrix}.
$$
This is almost a companion matrix and its characteristic polynomial equals the reciprocal of $C((-1)^k x)$, which is same as $C((-1)^k x)$ thanks to the symmetry. In particular, $b_0(n)$ has a characteristic polynomial $C(x)$ when $k$ is even, $C(-x)$ when $k$ is odd.
It remains to notice that $b_0(n)=(-1)^{(k-1)(k-2)/2} a_n$ for $n=-(k-1),\dots, 0$, implying that $\det A_k(n) = b_0(n) = (-1)^{kn + (k-1)(k-2)/2} a_n$. QED

UPDATE. More generally, we can drop the symmetricity requirement $c_j = c_{k-j}$ and keep only $c_0 = c_k$. Then
$$\det A_k(n) = (-1)^{kn + (k-1)(k-2)/2} a_{-n-k},$$
where $a_n$ is extended to large negative indices by the same recurrence (see also this answer). This formula implies the original result, since for symmetric $c_j$, we have $a_{-n-k}=a_n$.
Similarly, if we have $c_0 = -c_k$, then
$$\det A_k(n) = (-1)^{(k+1)n + (k+1)k/2} a_{-n-k}.$$
