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?

You can draw a graph for which your matrix is the matrix of probabilities of state change (some weights on the edges). 
Paint states A, D and E as white, and B and C as black. Then any possible move changes the color.
Since P^n is the probablility matrix of moving from X to Y in exacly n steps (or sum of products of weights spotted by any n-step path) then the oscillation occurs because your graph is bipartite. 
You can produce oscillation with other period if you wish this way.
