Let $[0]\_q:=0$ and $[n]\_q:=\frac{1-q^n}{1-q}=1+q+\cdots+q^{n-1}$, for $n\geq1$.



> 
> **Question.** Is there a closed formula (with proof) for the determinant of the matrix of $(i,j)$-entries
>  $$[i+j\bmod n]\_q, \qquad i,j=1,2,\dots,n.$$
> 
> 
> 


**Remark.** To bring in some context to the problem, this determinant is the specialization 
$$x\_{i+j\bmod n}\rightarrow [i+j\bmod n]\_q$$ 
in the [group determinant of Frobenius](https://mathoverflow.net/questions/251971/converse-of-frobenius) for the (finite) additive group $\mathbb{Z}\_n$.


**EDIT.** Sorry, I was meant to write $[0]\_q=0$ not $[0]\_q=1$.



Determinant of the "quantum" version of the group $\mathbb{Z}_n$
Let me give a few details on Fedor's calculation. If we swap the columns with index $j$ and $n+1-j$ we get the matrix $M\_n(q)$ with $(i,j)$ entry equal to $[1+i-j \mod n]\_q$. Since we have swapped $\lfloor \frac{n}{2}\rfloor$ columns, the determinant of your original matrix is equal to
$$(-1)^{\lfloor \frac{n}{2}\rfloor}\det M\_n(q)=(-1)^{\frac{n(n-1)}{2}}\det M\_n(q).$$
Now, $M\_n(q)$ is a circulant matrix so it has eigenvectors $(1,\omega^k,\dots,\omega^{k(n-1)})$, with corresponding eigenvalue $[1]\_q+[0]\_q\omega^k+[n-1]\_q\omega^{2k}+\cdots+[2]\_q\omega^{(n-1)k}$, for all $0\le k\le n-1$ where $\omega=\exp(\frac{2\pi i}{n})$.


Next we observe that
$$[1]\_q+[0]\_q\omega^k+[n-1]\_q\omega^{2k}+\cdots+[2]\_q\omega^{(n-1)k}=\omega^{k}\left([0]\_q+\frac{1+\frac{q}{\omega^k}+\cdots \frac{q^{n-1}}{\omega^{k(n-1)}}}{q-1}\right)=\omega^k\left([0]\_q+\frac{q^n-1}{(q-1)(\frac{q}{\omega^k}-1)}\right)$$
when $k\neq 0$. So we can calculate
$$\det M\_n(q)=\omega^{\frac{n(n-1)}{2}}\left([0]\_q+[1]\_q+\cdots+[n-1]\_q\right)\prod\_{k=1}^{n-1}\frac{[0]\_q(q-\omega^k)+\omega^k [n]\_q}{q-\omega^k}$$
if you take $[0]\_q=0$ this gives 
$$\det M\_n(q)=[n]\_q^{n-2}\frac{[n]\_q-n}{q-1}$$
and if you take $[0]\_q=1$ it gives
$$\det M\_n(q)=(-q)^{n-1}\left(1+\frac{[n]\_q-n}{q-1}\right)\frac{1+(-1)^{n-1}[n-1]\_q^n}{[n]\_q(1+[n-1]\_q)}.$$


