Example 18 (2008 Southeast Mathematical Olympiad Problem) Let $n$ be a positive integer, and $f(n)$ denote the number of $n$-digit numbers (called wave numbers) $\overline{a_{1} a_{2} \cdots a_{n}}$ that satisfy the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and $a_{i} \neq a_{i+1}, i=1,2, \cdots$;
(ii) When $n \geqslant 3$, the signs of $a_{i}-a_{i+1}$ and $a_{i+1}-a_{i+2}$ are opposite, $i=1,2, \cdots$.
(1) Find the value of $f(10)$;
(2) Determine the remainder when $f(2008)$ is divided by 13.
Let $n \geqslant 2$, and call an $n$-digit waveform number $a_{1} a_{2} \cdots a_{n}$ that satisfies $a_{1}a_{2}$ a $B$-type number. By symmetry, when $n \geqslant 2$, the number of such numbers is also $g(n)$, so $f(n)=2 g(n)$.
Now we find $g(n)$: Let $m_{k}(i)$ represent the number of $k$-digit $A$-type waveform numbers ending in $i$ $(i=1,2,3,4)$, then $g(n)=\sum_{i=1}^{4} m_{n}(i)$.
Since $a_{2 k-1}a_{2 k+1}$, then
(1) When $k$ is even,
$$
\begin{array}{l}
m_{k+1}(4)=0, m_{k+1}(3)=m_{k}(4), m_{k+1}(2)=m_{k}(4)+m_{k}(3), \\
m_{k+1}(1)=m_{k}(4)+m_{k}(3)+m_{k}(2) ;
\end{array}
$$
(2) When $k$ is odd,
$$
\begin{array}{l}
m_{k+1}(1)=0, m_{k+1}(2)=m_{k}(1), m_{k+1}(3)=m_{k}(1)+m_{k}(2), \\
m_{k+1}(4)=m_{k}(1)+m_{k}(2)+m_{k}(3),
\end{array}
$$
(2)
It is easy to see that $m_{2}(1)=0, m_{2}(2)=1, m_{2}(3)=2, m_{2}(4)=3$, so $g(2)=6$.
From this, we can get,
$$
\begin{array}{l}
m_{3}(1)=m_{2}(2)+m_{2}(3)+m_{2}(4)=6, \\
m_{3}(2)=m_{2}(3)+m_{2}(4)=5, \\
m_{3}(3)=m_{2}(4)=3, m_{3}(4)=0,
\end{array}
$$

So $g(3)=\sum_{i=1}^{4} m_{3}(i)=14$.
And by
$$
\begin{array}{l}
m_{4}(1)=0, m_{4}(2)=m_{3}(1)=6, \\
m_{4}(3)=m_{3}(1)+m_{3}(2)=11, \\
m_{4}(4)=m_{3}(1)+m_{3}(2)+m_{3}(3)=14,
\end{array}
$$
In the problem of numbers in mathematics,
So $g(4)=\sum_{i=1}^{4} m_{4}(i)=31$.
Similarly, we can find that $g(5)=70, g(6)=157, g(7)=353, g(8)=793$.
In general, when $n \geqslant 5$,
$$
g(n)=2 g(n-1)+g(n-2)-g(n-3) \text {. }
$$

Now we prove (3) as follows:
By induction on $n$, for $n=5,6,7,8$ it has been verified, assume (3) holds up to $n$, consider the case for $n+1$. When $n$ is even, according to (1) and (2) we get
$$
\begin{array}{l}
m_{n+1}(4)=0, m_{n+1}(3)=m_{n}(4), m_{n+1}(2)=m_{n}(4)+m_{n}(3), \\
m_{n+1}(1)=m_{n}(4)+m_{n}(3)+m_{n}(2),
\end{array}
$$

And $m_{n}(1)=0$, then
$$
\begin{aligned}
g(n+1) & =\sum_{i=1}^{4} m_{n+1}(i)=2\left(\sum_{i=1}^{4} m_{n}(i)\right)+m_{n}(4)-m_{n}(2) \\
& =2 g(n)+m_{n}(4)-m_{n}(2) .
\end{aligned}
$$

Because
$$
\begin{aligned}
m_{n}(4) & =m_{n-1}(1)+m_{n-1}(2)+m_{n-1}(3)+0 \\
& =\sum_{i=1}^{4} m_{n-1}(i)=g(n-1), \\
m_{n}(2) & =m_{n-1}(1)=m_{n-2}(4)+m_{n-2}(3)+m_{n-2}(2)+0 \\
& =g(n-2),
\end{aligned}
$$

At this point, we have $g(n+1)=2 g(n)+g(n-1)-g(n-2)$.
When $n$ is odd, $g(n+1)=\sum_{i=1}^{4} m_{n+1}(i)$, and
$$
\begin{array}{l}
m_{n+1}(1)=0, m_{n+1}(2)=m_{n}(1), m_{n}(4)=0, \\
m_{n+1}(3)=m_{n}(1)+m_{n}(2), m_{n+1}(4)=m_{n}(1)+m_{n}(2)+m_{n}(3),
\end{array}
$$

Then $g(n+1)=\sum_{i=1}^{4} m_{n+1}(i)=2 \sum_{i=1}^{4} m_{n}(i)+m_{n}(1)-m_{n}(3)$
$$
=2 g(n)+m_{n}(1)-m_{n}(3) \text {. }
$$

Because
$$
\begin{array}{l}
m_{n}(1)=m_{n-1}(4)+m_{n-1}(3)+m_{n-1}(2)+0=g(n-1), \\
m_{n}(3)=m_{n-1}(4)=m_{n-2}(1)+m_{n-2}(2)+m_{n-2}(3)+0=g(n-2),
\end{array}
$$

At this point, we also have
$$
g(n+1)=2 g(n)+g(n-1)-g(n-2),
$$

Thus (3) holds for $n+1$, and by induction, (3) holds for all $n \geqslant 5$.
According to (3),
$$
\begin{array}{l}
g(9)=2 g(8)+g(7)-g(6)=1782, \\
g(10)=2 g(9)+g(8)-g(7)=4004,
\end{array}
$$

So $f(10)=2 g(10)=8008$.
Now consider the sequence of $\{g(n)\}$ modulo 13:
Using (3), it is easy to calculate that when $n=2,3,4, \cdots, 14,15,16,17, \cdots$, the remainders of $g(n)$ when divided by 13 are:
$$
6,1,5,5,1,2,0,1,0,1,1,3 ; 6,1,5,5, \cdots,
$$

Therefore, when $n \geqslant 2$, the sequence of remainders of $\{g(n)\}$ when divided by 13 is a periodic sequence, with the smallest period length being 12. And $2008=12 \times 167+4$, so
$$
g(2008) \equiv 5(\bmod 13),
$$

Therefore, $f(2008) \equiv 10(\bmod 13)$.