What kind of matrix is this and why does this happen? So I was studying Markov chains and I came across this matrix  \begin{align*}P=\left( \begin{array}{ccccc}
        0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\
        \frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\
        \frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4}\\
        0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
        0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
    \end{array} \right).\end{align*}
I noticed (by brute force) that \begin{align*}P^2=\left( \begin{array}{ccccc}
        \frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
        0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
        0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
        \frac{3}{8} & 0 & 0 & \frac{1}{4}& \frac{1}{2}\\
        \frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
    \end{array} \right),\end{align*}
and \begin{align*}P^3=\left( \begin{array}{ccccc}
        0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
        \frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
        \frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
        0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
        0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
    \end{array} \right).\end{align*}
In fact; using a computer I found that every even power takes the form of the $P^2$ matrix and every odd power takes the form of the $P^3$ matrix.
I just wanted to know why that oscillation occurs? Is there a special name for the kind of matrix that $P$ is for it to exhibit that kind of behaviour?

Notice that your matrix $$\begin{align*}P=\left( \begin{array}{ccccc}
        0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\
        \frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\
        \frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4}\\
        0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
        0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
    \end{array} \right).\end{align*}$$
has a special block decomposition  as 
$$ P=\begin {pmatrix} 0&A &0\\B&0&C\\0&D&0\end{pmatrix}$$
Where A,B,C,D are non zero submatrices.
To find $P^2$ we use the new form to get
$$ P^2 =  \begin {pmatrix} 0&A &0\\B&0&C\\0&D&0\end{pmatrix}\begin {pmatrix} 0&A &0\\B&0&C\\0&D&0\end{pmatrix}=\begin {pmatrix} AB&0 &AC\\0&BA+CD&0\\DB&0&DC\end{pmatrix} $$
As you know  we can find powers of P by multiplying the new form of P as many times as we wish.
The alternating repeating form of powers of P is due to the block decomposition form of P.   
