Inverse of a circulant matrix with a specific pattern I'm trying to invert the following circulant matrix:
$$\begin{bmatrix}1 & -1/4 & 0&0  &0&\cdots&0&-1/4\\ -1/4 & 1 & -1/4 & 0&0&\cdots&0&0\\0 & -1/4 & 1 & -1/4&0&\cdots&0&0\\\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots
\\-1/4 & 0 & 0 & 0 & 0&\cdots&-1/4 & 1 \end{bmatrix}$$
As it turns out, Fuyong (2011) "The inverse of circulant matrix" proposes the following method:
1-) Find the roots of the polynomials $g(z)=g^{\prime}(z)=1 - \frac{1}{4}z- \frac{1}{4}z^{-1}$ that are inside the unit circle:
$z_1 = z_1^{\prime}= 2-\sqrt{3}$ and $z_2 = z_2^{\prime}= 2+\sqrt{3}$
Only $z_1$ and $z_1^{\prime}$ are inside the unit circle.
2-) Compute $g_1(z_1)=-\frac{1}{4}z_1^{-1}(z_1-z_2)$ and $g_1^{\prime}(z^{\prime}_1)=-\frac{1}{4}z_1^{\prime-1}(z_1^{\prime}-z_2^{\prime})$:
$g_1(z_1)=g_1^{\prime}(z^{\prime}_1)= \frac{\sqrt{3}}{2(2-\sqrt{3})}$
3-) The elements of the inverse are given by:
$b_k= \dfrac{z_1^{k_1}}{g_1(z_1)(1-z_1^n)} +
\dfrac{z_1^{\prime k_2}}{z_1^{\prime s}g_1^{\prime}(z^{\prime}_1)(1-z_1^{\prime n})} $ , $k=0,\ldots,n-1$
where:
$k_1 = \mathrm{mod}\,e (k-1,n) $
$k_2 = \mathrm{mod}\,e (n-k-1+s,n) $
I don't know how $s$ is determined and what the operator $\mathrm{mod}\,e$ does. Can anyone help?

The polynomial $g$ is defined as $g(z) = a_0 + a_1 z + \cdots + a_s x^s$. Thus, $s=-1$.
Let $B=A^{-1}$:
\begin{align*}
&B= \begin{bmatrix}b_0 & b_1 & b_2&b_3  &b_4&\cdots &b_{n-2}&b_{n-1}\\ b_{n-1} & b_0 & b_1 & b_2&b_3&\cdots & b_{n-3}&b_{n-2}\\b_{n-2} & b_{n-1} & b_0 & b_1&b_2&\cdots&b_{n-4}&b_{n-3}\\\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots
\\b_1 & b_2 & b_3 & b_4 & b_5 &\cdots&b_{n-1} & b_0 \end{bmatrix}&
\end{align*}
\begin{align*}
b_k = \dfrac{2\bigl(2-\sqrt{3}\bigr)^{k}}{\sqrt{3}\bigl[1-\bigl(2-\sqrt{3}\bigr)^n\bigr]} + \dfrac{2\bigl(2-\sqrt{3}\bigr)^{n-k}}{\sqrt{3}\bigl[1-\bigl(2-\sqrt{3}\bigr)^n\bigr]}\qquad k = 0,\ldots, n-1 
\end{align*}
