Determinant of a Hankel matrix with sequence (1, 2, 3,..., n, 1, 2,..., n-1) I'm looking for a closed form of the determinant of matrices like 
$\begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\\ 3& 1 &2\end{bmatrix}$
or 
$\begin{bmatrix}1 & 2 & 3 &4\\2 & 3 &4 & 1\\ 3&4& 1 &2\\ 4& 1&2&3\end{bmatrix}$,
that means Henkel matrices of the size n with sequence (1, 2, 3,..., n, 1, 2,..., n-1)
 A: The magnitude of the determinant is $$\frac{n^{n-1}(n+1)}{2}.$$ The sign is given by $$f(n) = \begin{cases}+ & \mathrm{floor}(n/2) \text{ is even.} \\ - & \mathrm{floor}(n/2) \text{ is odd} .\end{cases}$$A useful formula is $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$.
A: These matrices are actually circulant matrices, whose determinant is well known.
For the matrices at hand, the determinant is
$
\Delta_n=\prod _{j=0}^{n-1}f(\omega _{j})
$,
where $f(z)=1+2z+3z^{2}+\dots +nz^{n-1}$ and $w_j$ is the $j$-th root of unity.
We have $f(\omega_0)=f(1)=1+2+3+\cdots +n = \dfrac{n(n+1)}{2}$.
We also have $f(z)=g'(z)$, where $g(z)=1+z+z^2+z^3+\cdots+z^n=\dfrac{z^{n+1}-1}{z-1}$. Therefore,
$$
g'(z)=\dfrac{n z^{n+1} - (n+1) z^n + 1}{(z-1)^2}
$$
and so, for $j\ne0$, 
$$
f(\omega_j)=g'(\omega_j)=\dfrac{n \omega_j  - n}{(\omega_j-1)^2}=\dfrac{n}{\omega_j-1}
$$
Thus,
$$
\Delta_n=\prod _{j=0}^{n-1}f(\omega _{j})
=\dfrac{n(n+1)}{2} \prod _{j=1}^{n-1}\dfrac{n}{\omega_j-1}
=\dfrac{n^n(n+1)}{2} \dfrac{1}{\prod _{j=1}^{n-1}(\omega_j-1)}
$$
Now $\prod _{j=1}^{n-1}(\omega_j-1)=h(1)$ for
$$
h(z)=\prod _{j=1}^{n-1}(\omega_j-z)
=(-1)^{n-1}\prod _{j=1}^{n-1}(z-\omega_j)
=(-1)^{n-1}\dfrac{z^{n}-1}{z-1}
=(-1)^{n-1}(1+z+z^2+z^3+\cdots+z^{n-1})
$$
and so $h(1)=(-1)^{n-1}n$. Thus
$$
\Delta_n
=\dfrac{n^n(n+1)}{2} \dfrac{1}{(-1)^{n-1}n}
=(-1)^{n-1}\dfrac{n^{n-1}(n+1)}{2}
$$
