145 Let $a_{n}=4^{2 n-1}+3^{n-2}(n=1,2,3, \cdots), p$ be the smallest prime that can divide infinitely many terms of $a_{1}, a_{2}, a_{3}, \cdots$, and $q$ be the smallest prime that can divide every term of $a_{1}, a_{2}, a_{3}, \cdots$, then $p \cdot q$ is
A. $5 \times 7$
B. $7 \times 13$
C. $7 \times 11$
D. $5 \times 13$
145 D. Clearly, $p$ is not equal to 2 and 3. Let $n+2=4k (k \in \mathbf{N})$, then $a_{n}=4 \times 16^{n}+3^{n+2} \equiv 4+ 9^{2k} \equiv 4+(-1)^{2k} \equiv 0(\bmod 5)$, so $p=5$, which eliminates options B and C. Since $a_{1}= 7 \times 13, a_{2}=4^{5}+3^{4}=4 \times 16^{2}+9^{2} \equiv 4 \times 2^{2}+2^{2} \equiv 20 \not=0(\bmod 7)$, thus $q \neq 7$, eliminating A.
Since $a_{n}=4 \times 16^{n}+9 \times 3^{n} \equiv 4 \times 3^{n}+9 \times 3^{n} \equiv 13 \times 3^{n} \equiv 0(\bmod 13)$, so $q=13$.