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][1] for the (finite) additive group $\mathbb{Z}_n$. **EDIT.** Sorry, I was meant to write $[0]_q=0$. [1]: https://mathoverflow.net/questions/251971/converse-of-frobenius