Question 69
Let $p(n)$ denote the number of all $n$-digit positive integers containing only the digits $1$, $2$, $3$, $4$, $5$ and such that every two adjacent digits differ by at least $2$. Prove that for every positive integer $n$,
$$
5 \cdot 2.4^{n-1} \le p(n) \le 5 \cdot 2.5^{n-1}.
$$
$$
5 \cdot 2.4^{n-1} \le p(n) \le 5 \cdot 2.5^{n-1}.
$$