8. [2] Define the sequence $\left\{x_{i}\right\}_{i \geq 0}$ by $x_{0}=x_{1}=x_{2}=1$ and $x_{k}=\frac{x_{k-1}+x_{k-2}+1}{x_{k-3}}$ for $k>2$. Find $x_{2013}$.
Answer: 9 We have $x_{3}=\frac{1+1+1}{1}=3, x_{4}=\frac{3+1+1}{1}=5, x_{5}=\frac{5+3+1}{1}=9, x_{6}=\frac{9+5+1}{3}=5$. By the symmetry of our recurrence (or just further computation - it doesn't matter much), $x_{7}=3$ and $x_{8}=x_{9}=x_{10}=1$, so our sequence has period 8. Thus $x_{2013}=x_{13}=x_{5}=9$.