The repeating sequence

Suppose the sequence contains $a$ copies of $2$ , $b$ copies of $3$ and $c$ copies of $6$ . Then we want to solve $2a + 3b + 6c \; = \; 633 \hspace{2cm} 9a + 4b + c \; = \; 2017$ for nonnegative integers $a,b,c$ . Eliminating $b$ from these equations gives $\begin{aligned} 19a - 21c & = \; 3519 \; = \; 3519(10 \times 19 - 9 \times 21) \\ 19(a - 35190) & = \; 21(c - 31671) \end{aligned}$ and hence $a \,=\, 35190 - 21u$ , $c = 31671 - 19u$ for some integer $u$ . This tells us that $b = 52u - 86591$ . Since $a,b,c$ all need to be nonnegative, we deduce that $u \le 1675$ , $u \ge 1666$ and $u \le 1666$ respectively, and hence $u = 1666$ .

Thus $a = 204$ , $b = 41$ and $c = 17$ , so that $n = a + b + c = \boxed{262}$ .