Four. (50 points) $n$ points $P_{1}, P_{2}, \cdots, P_{n}$ are sequentially on a straight line, each point is colored one of white, red, green, blue, or purple. If for any two adjacent points $P_{i}, P_{i+1}$ $(i=1,2, \cdots, n-1)$, either these two points are the same color, or at least one of them is white, it is called a good coloring. Find the total number of good colorings.
Four, Solution: Let $a_{n}$ denote the number of good colorings of $P_{n}$ where $P_{n}$ is colored white, and $b_{n}$ denote the number of good colorings of $P_{n}$ where $P_{n}$ is not colored white. Then, $c_{n}=a_{n}+b_{n}$ represents the total number of good colorings. By the definitions of $a_{n}, b_{n}, c_{n}$, we have: $a_{n}=c_{n-1}, b_{n}=4 a_{n-1}+b_{n-1}$, so $c_{n}=a_{n}+b_{n}=c_{n-1}+b_{n-1}+4 a_{n-1}=c_{n-1}+b_{n-1}+4 c_{n-2}$. Noting that $b_{n}=c_{n}-a_{n}=c_{n}-c_{n-1}$, substituting this into the above equation gives $c_{n}=2 c_{n-1}+3 c_{_{n-2}}$. It is easy to see that $c_{1}=5, c_{2}=13$. Using the characteristic root method, we get $c_{n}=\frac{1}{2}\left(3^{n+1}+(-1)^{n+1}\right)$.
Therefore, the total number of good colorings is $\frac{1}{2}\left(3^{n+1}+(-1)^{n+1}\right)$.