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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?

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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 $$\det A_k(n) = (-1)^{(k+1)n + (k+1)k/2} a_{-n-k}.$$

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  • $\begingroup$ @ Max Alekseyev: Thank you for this very nice proof! $\endgroup$ – Johann Cigler Jan 8 at 7:40

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