510 For a regular $n$-prism $A_{1} A_{2} \cdots A_{n}-A_{1}^{\prime} A_{2}^{\prime} \cdots A_{n}^{\prime}$, each vertex is colored with one of four colors. If adjacent vertices must have different colors, how many different coloring methods are there?
Consider unfolding a regular $n$-prism into $n$ columns, as shown in Figure 2. The point at the top of the $i$-th column is denoted as $i_{1}$, and the point at the bottom of the $i$-th column is denoted as $i_{2}$.

First, consider Figure 3. The points $A$ and $B$ in the first column can be colored in $4 \times 3 = 12$ ways.

If point $C$ is the same color as $A$, then point $D$ has 3 ways to be colored;
If point $C$ is a different color from $A$, then point $C$ has 2 ways to be colored and point $D$ has 2 ways to be colored.
Thus, points $C$ and $D$ have a total of $3 + 2 \times 2 = 7$ ways to be colored.
Similarly, the total number of ways to color the $n$ columns such that no two adjacent points have the same color is $12 \times 7^{n-1}$.

Next, consider the various scenarios when the last column and the first column form the original regular $n$-prism.
For $m \geqslant 2 (m \in \mathbf{Z})$:
Let $a_{m}$ be the number of ways to color such that $m_{1}$ is the same color as $1_{1}$ and only up to the $m$-th column is colored; $b_{m}$ be the number of ways to color such that $m_{2}$ is the same color as $1_{1}$ and only up to the $m$-th column is colored; $c_{m}$ be the number of ways to color such that $m_{1}$ is the same color as $1_{1}$ and $m_{2}$ is the same color as $1_{2}$, and only up to the $m$-th column is colored.

When $(m-1)_{1}$ is a different color from $1_{1}$ and $(m-1)_{2}$ is also a different color from $1_{1}$, coloring $m_{1}$ the same color as $1_{1}$, $m_{2}$ has 2 ways to be colored; when $(m-1)_{1}$ is a different color from $1_{1}$ but $(m-1)_{2}$ is the same color as $1_{1}$, coloring $m_{1}$ the same color as $1_{1}$, $m_{2}$ has 3 ways to be colored.
Thus, the recurrence relation is
$$
a_{m} = \left(12 \times 7^{m-2} - a_{m-1} - b_{m-1}\right) \times 2 + b_{m-1} \times 3.
$$

Similarly,
$$
b_{m} = \left(12 \times 7^{m-2} - a_{m-1} - b_{m-1}\right) \times 2 + a_{m-1} \times 3.
$$

If $m_{1}$ is the same color as $1_{1}$ and $m_{2}$ is the same color as $1_{2}$, then $(m-1)_{1}$ and $1_{1}$, $(m-1)_{2}$ and $1_{2}$ are respectively different colors.

By symmetry, the number of ways to color $(m-1)_{2}$ the same color as $1_{2}$ is the same as the number of ways to color $(m-1)_{1}$ the same color as $1_{1}$, both being $a_{m-1}$.
By the principle of inclusion-exclusion, we get
$$
\begin{array}{l}
c_{m} = 12 \times 7^{m-2} - \mathrm{C}_{2}^{1} a_{m-1} + c_{m-1}. \\
\text{Thus, } a_{n} = \left(12 \times 7^{n-2} - a_{n-1} - b_{n-1}\right) \times 2 + b_{n-1} \times 3, \\
b_{n} = \left(12 \times 7^{n-2} - a_{n-1} - b_{n-1}\right) \times 2 + a_{n-1} \times 3, \\
c_{n} = 12 \times 7^{n-2} - \mathrm{C}_{2}^{1} a_{n-1} + c_{n-1},
\end{array}
$$

where $a_{2} = 0, b_{2} = 36, c_{3} = 84$.
$$
\begin{array}{l}
\text{(1) + (2) gives} \\
a_{n} + b_{n} = 48 \times 7^{n-2} - (a_{n-1} + b_{n-1}). \\
\text{Let } x_{n} = \frac{a_{n} + b_{n}}{7^{n}}. \\
\text{Then } x_{n} = \frac{48}{49} - \frac{x_{n-1}}{7} \\
\Rightarrow x_{n} - \frac{6}{7} = -\frac{1}{7}\left(x_{n-1} - \frac{6}{7}\right) \\
\Rightarrow x_{n} = \frac{6}{7^{n}}(-1)^{n-1} + \frac{6}{7} \\
\Rightarrow a_{n} + b_{n} = 6 \times 7^{n-1} + 6(-1)^{n-1} \\
\Rightarrow a_{n-1} + b_{n-1} = 6 \times 7^{n-2} + 6(-1)^{n-2}.
\end{array}
$$

Substituting into equation (1) gives
$$
\begin{aligned}
a_{n} & = 24 \times 7^{n-2} - 3 a_{n-1} + 6 \times 7^{n-2} + 6(-1)^{n-2} \\
& = 30 \times 7^{n-2} - 3 a_{n-1} + 6(-1)^{n} \\
\Rightarrow & a_{n} - 3 \times 7^{n-1} + 3(-1)^{n} \\
& = (-3)\left(a_{n-1} - 3 \times 7^{n-2} + 3(-1)^{n-1}\right) \\
\Rightarrow & a_{n} = 3 \times 7^{n-1} - 2(-3)^{n} - 3(-1)^{n}.
\end{aligned}
$$

Substituting into equation (3) gives
$$
c_{n} = 6 \times 7^{n-2} + 4(-3)^{n-1} + 3(-1)^{n-1} + c_{n-1}.
$$

Summing and combining with $c_{3} = 84$ gives
$$
c_{n} = 7^{n-1} - (-3)^{n} - 3(-1)^{n} + 5.
$$

Finally, the unfolded diagram can be reassembled into the original regular $n$-prism, and the conditions of the problem are satisfied, then $n_{1}$ and $1_{1}$, $n_{2}$ and $1_{2}$ are all different colors.
Thus, the total number of coloring methods is
$$
c_{n+1} = 7^{n} - (-3)^{n+1} - 3(-1)^{n+1} + 5 \left(n \in \mathbf{Z}_{+}\right).
$$