$1 \cdot 40$ form "words" by taking $n$ numbers from the alphabet $\{0,1,2,3,4\}$, such that the difference between every two adjacent numbers is 1. How many words can be formed in total?
[Solution] Let the word to be counted be called a "qualified word". Let the number of qualified words with the last digit being $0,1,2$ be $a_{n}, b_{n}, c_{n}$, respectively. By symmetry, the number of qualified words with the last digit being 3 and 4 are $b_{n}$ and $a_{n}$, respectively. According to the requirements of the problem, we have
$$
a_{n+1}=b_{n}, b_{n+1}=a_{n}+c_{n}, c_{n+1}=2 b_{n} .
$$

From this, we can derive
$$
b_{n+2}=3 b_{n} .
$$

Since $b_{1}=1, b_{2}=2$, we get
$$
b_{n}=\left\{\begin{array}{l}
3^{k}, n=2 k+1, \\
2 \times 3^{k-1}, n=2 k .
\end{array}\right.
$$

Let the total number of qualified words be $m_{n}$, then $m_{1}=5$ and
$$
\begin{aligned}
m_{n} & =2 a_{n}+2 b_{n}+c_{n}=2 b_{n}+4 b_{n-1} \\
& =\left\{\begin{array}{l}
2 \times 3^{k}+4 \times 2 \times 3^{k-1}=14 \times 3^{k-1}, n=2 k+1, \\
2 \times 2 \times 3^{k-1}+4 \times 3^{k-1}=8 \times 3^{k-1}, n=2 k .
\end{array}\right.
\end{aligned}
$$