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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 for the (finite) additive group $\mathbb{Z}_n$.

EDIT. Sorry, I was meant to write $[0]_q=0$.

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 for the (finite) additive group $\mathbb{Z}_n$.

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

Let $[0]_q:=1$ 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 for the (finite) additive group $\mathbb{Z}_n$.

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 for the (finite) additive group $\mathbb{Z}_n$.

EDIT. Sorry, I was meant to write $[0]_q=0$.

Let $[0]_q:=1$ 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 $$[i+j\mod n]_q, \qquad i,j=1,2.\dots,n.$$

Remark. To bring in some context to the problem, this determinant is the specialization $$x_{i+j\mod n}\rightarrow [i+j\mod n]_q$$ in the group determinant of Frobenius for the (finite) additive group $\mathbb{Z}_n$.

Let $[0]_q:=1$ 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 for the (finite) additive group $\mathbb{Z}_n$.