4. Let $a_{n}=6^{n}+8^{n}$. Then the remainder when $a_{2018}$ is divided by 49 is $\qquad$ .
\begin{array}{l}\text { 4.2. } \\ a_{2018}=(7-1)^{2018}+(7+1)^{2018} \\ =2 \sum_{k=0}^{1009} \mathrm{C}_{2018}^{2 k} 7^{2 k} \equiv 2\left(\bmod 7^{2}\right)\end{array}