4. Let $a_{n}=4^{2 n+1}$
$$
+3^{n+2}(n=1,2,3 \text {. }
$$
$\cdots)$, $p$ is the smallest prime that can divide infinitely many terms of
$a_{1}, a_{2}, a_{3}, \cdots$, $q$ is the smallest prime that can divide every term of $a_{1}, a_{2}, a_{3}, \cdots$, then $p$ - $q$ is ( ).
(A) $5 \times 7$
(B) $7 \times 13$
(C) $13 \times 13$
(D) None of the above answers

直接输出翻译结果，保留了源文本的换行和格式。
4. (D).

Obviously, $p$ is not equal to 2 and 3. Let $n+2=4k (k \in \mathbb{N})$, then
$$
\begin{aligned}
a_{n} & =4 \times 16^{n}+3^{n+2}=4+3^{4 k} \\
& =4+9^{2 k} \equiv 4+(-1)^{2 k} \equiv 0(\bmod 5) .
\end{aligned}
$$

Therefore, $p=5$. This eliminates (B) and (C). Since $a_{1}=7 \times$
13, $a_{2}=4^{5}+3^{4}=4 \times 16^{2}+9^{2} \equiv 4 \times 2^{2}+2^{2} \equiv 20 \not=0$ $(\bmod 7)$, so $q \neq 7$, eliminating $(\mathrm{A})$.