Four, (50 points) A password lock's password setting involves assigning one of the two numbers, 0 or 1, to each vertex of a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, and coloring each vertex one of two colors, red or blue, such that for any two adjacent vertices, at least one of the number or color is the same. How many different password settings are there for this type of password lock?
For this type of password lock setting, if the numbers assigned to two adjacent vertices are different, mark the edge between them with $a$; if the colors are different, mark it with $b$; if both the number and color are the same, mark it with $c$. Thus, for a given setting at point $A_{1}$ (with 4 possible settings), the settings at points $A_{2}, A_{3}, \cdots, A_{n}$ can be sequentially determined based on the letters on the edges. To ensure that the setting returns to $A_{1}$ and is the same as the initial setting, the number of edges marked with $a$ and $b$ must both be even.

Therefore, the number of different password settings for this lock is equal to 4 times the number of ways to mark the edges with $a$, $b$, and $c$ such that the number of edges marked with $a$ and $b$ are both even.

Let the number of edges marked with $a$ be $2 i\left(0 \leqslant i \leqslant\left[\frac{n}{2}\right]\right)$, and the number of edges marked with $b$ be $2 j\left(0 \leqslant j \leqslant\left[\frac{n-2 i}{2}\right]\right)$.

The number of ways to choose $2 i$ edges to mark with $a$ is $\mathrm{C}_{n}^{2 i}$, and the number of ways to choose $2 j$ edges from the remaining edges to mark with $b$ is $\mathrm{C}_{n-2 i}^{2 j}$. The remaining edges are marked with $c$.
By the multiplication principle, there are $\mathrm{C}_{n}^{2 i} \mathrm{C}_{n-2 i}^{2 j}$ ways to mark the edges.
Summing over $i$ and $j$, the number of different password settings for the lock is
$$
f(n)=4 \sum_{i=0}^{\left[\frac{n}{2}\right]}\left(\mathrm{C}_{n}^{2 i} \sum_{j=0}^{\left[\frac{n-2 i}{2}\right]} \mathrm{C}_{n-2 i}^{2 j}\right) .
$$

Here, we agree that $\mathrm{C}_{0}^{0}=1$.
When $n$ is odd, $n-2 i>0$, in this case,
$$
\sum_{j=0}^{\left.\frac{n-2 i}{2}\right]} C_{n-2 i}^{2 j}=2^{n-2 i-1} \text {. }
$$

Substituting into equation (1) we get
$$
\begin{array}{l}
f(n)=4 \sum_{i=0}^{\left[\frac{n}{2}\right]} 2^{n-2 i-1} \mathrm{C}_{n}^{2 i}=2 \sum_{i=0}^{\left[\frac{n}{2}\right]} 2^{n-2 i} \mathrm{C}_{n}^{2 i} \\
=\sum_{k=0}^{n} 2^{n-k} \mathrm{C}_{n}^{k}+\sum_{k=0}^{n} 2^{n-k}(-1)^{k} \mathrm{C}_{n}^{k} \\
=(2+1)^{n}+(2-1)^{n}=3^{n}+1 .
\end{array}
$$

When $n$ is even, if $i<\frac{n}{2}$, equation (2) still holds; if $i=\frac{n}{2}$, then all edges of the $n$-sided polygon are marked with $a$, in which case there is only one marking method. Therefore, the number of different password settings is
$$
\begin{array}{l}
f(n)=4\left(1+\sum_{i=0}^{\left[\frac{n}{2}\right]-1} 2^{n-2 i-1} \mathrm{C}_{n}^{2 i}\right) \\
=2+4 \sum_{i=0}^{\left[\frac{n}{2}\right]} 2^{n-2 i-1} \mathrm{C}_{n}^{2 i}=3^{n}+3 .
\end{array}
$$

In summary, the number of different password settings for this lock is: when $n$ is odd, there are $3^{n}+1$ ways; when $n$ is even, there are $3^{n}+3$ ways.
(Ding Longyun provided)