Five, divide a circle into $n(n \geqslant 2)$ sectors, sequentially denoted as $S_{1}, S_{2}, \cdots, S_{n}$. Each sector can be painted with any of the three different colors: red, white, and blue, with the requirement that adjacent sectors must have different colors. How many ways are there to color the sectors?
Let the total number of coloring methods be $a_{n}(n \geqslant 2)$.
When $n=2$, first color $S_{1}$, there are three ways to color it. After coloring $S_{1}$, continue to color $S_{2}$, there are only 2 ways, thus $a_{2}=2 \times 3=6$.
Now, let's determine the recursive relationship.
If we first color $S_{1}$, there are 3 ways to color it. Then color $S_{2}$, there are only 2 ways. Next, color $S_{3}$, if it is only required to be a different color from $S_{2}$, there are 2 ways, and so on, until finally coloring $S_{n}$, if it is only required to be a different color from $S_{n-1}$ (here, the color of $S_{n}$ and $S_{1}$ is not considered), there are still 2 ways. This results in a total of $3 \cdot 2^{n-1}$ ways to color, but these $3 \cdot 2^{n-1}$ ways can be divided into two categories: one category is where the color of $S_{n}$ is different from $S_{1}$, which meets the requirement, and the total number is $a_{n}$; the other category is where the color of $S_{n}$ is the same as $S_{1}$, which does not meet the requirement. If $S_{2}$ and $S_{1}$ are merged and considered as one sector, this type of coloring is equivalent to dividing into $n-1$ sectors, and the total number of coloring methods according to the problem's requirement is $a_{n-1}$. Thus, we get the recursive relationship:
$$
a_{n} + a_{n-1} = 3 \cdot 2^{n-1} (n=3,4, \cdots),
$$

which can be rewritten as $a_{n} = -a_{n-1} + 3 \cdot 2^{n-1} (n=3,4, \cdots)$.
To find $a_{n}$, let $b_{n} = \frac{a_{n}}{2^{n}}$, then the above equation becomes $b_{n} = -\frac{1}{2} b_{n-1} + \frac{3}{2}$, leading to $b_{n} - 1 = \frac{1}{2} (1 - b_{n-1})$ and $b_{2} - 1 = \frac{6}{2^{2}} - 1 = \frac{1}{2}$, thus $b_{n} - 1 = \frac{1}{2} \left(-\frac{1}{2}\right)^{n-2} = -\left(-\frac{1}{2}\right)^{n-1}$, so $b_{n} = -\left(-\frac{1}{2}\right)^{n-1} + 1$,
therefore $a_{n} = 2^{n} b_{n} = 2^{n} \left[-\left(-\frac{1}{2}\right)^{n-1} + 1\right] = 2 \left[2^{n-1} - (-1)^{n-1}\right] (n=2,3, \cdots)$,
hence, there are $2 \left[2^{n-1} - (-1)^{n-1}\right]$ ways to color.