Toeplitz matrices are matrices where each descending diagonal is constant. A $n\times n$ Toeplitz matrix can generated by a sequence $(c_{1}\ldots c_{2n-1})$. For example, here is a $5\times5$ matrix:
$$ \left(\begin{array}{ccccc} c_{5} & c_{6} & c_{7} & c_{8} & c_{9}\\ c_{4} & c_{5} & c_{6} & c_{7} & c_{8}\\ c_{3} & c_{4} & c_{5} & c_{6} & c_{7}\\ c_{2} & c_{3} & c_{4} & c_{5} & c_{6}\\ c_{1} & c_{2} & c_{3} & c_{4} & c_{5} \end{array}\right) $$
Let us define k-periodic Toeplitz matrices as Toeplitz matrices generated by a sequence $(c_{1}\ldots c_{2n-1})$ such that $c_{i+k}=c_{i}$ for all $i<2n-k$. For example, a $6$-periodic $5\times5$ Toeplitz matrix would look like this:
$$\left(\begin{array}{ccccc} c_{5} & c_{6} & c_{1} & c_{2} & c_{3}\\ c_{4} & c_{5} & c_{6} & c_{1} & c_{2}\\ c_{3} & c_{4} & c_{5} & c_{6} & c_{1}\\ c_{2} & c_{3} & c_{4} & c_{5} & c_{6}\\ c_{1} & c_{2} & c_{3} & c_{4} & c_{5} \end{array}\right)$$
My question is: for any $n\in\mathbb{N}$ and any $k\in\{n+1\ldots2n-1\}$, does there exist invertible k-periodic Toeplitz matrices and can we proove it ?
I have done a lot of numerical experiments (I tried all values of $n$ up to 60), and up to now I have found ways to build these invertible matrices, but with no guarantee.
For all $n\in\{3\ldots60\}$ and all periods $k\in\{n+1\ldots2n-1\}$, choosing $c_{i}=i^{th}$ prime number induces invertible k-periodic Toeplitz matrices
For the same values of $n$ and $k$, choosing $c_{i}\sim\mathcal{N}(0,1)$ also induces invertible k-periodic Toeplitz matrices
For $n>7$ choosing $c_{i}\sim bernoulli(0.5)$ induces invertible k-periodic Toeplitz matrices with high probability
My best guess would be choosing normally distributed $c_{i}$ would be the way to go to get a theorem, but I don't really know. There are lots of results on invertible random circulant matrices, but I don't see how to translate them in my case. Does anyone have a clue ?