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?



Some determinants which are closely related to recurrences
The question concerns the determinant of a [Hankel matrix](https://en.wikipedia.org/wiki/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.




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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](https://en.wikipedia.org/wiki/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




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**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](https://mathoverflow.net/q/372663)). 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
$$\det A\_k(n) = (-1)^{(k+1)n + (k+1)k/2} a\_{-n-k}.$$


