jmc

We compute the first few terms: $$\begin{array}{cc}n& a_n\\ 0& 2\\ 1& 5\\ 2& 8\\ 3& 5\\ 4& 6\\ 5& 10\\ 6& 7\\ 7& 4\\ 8& 7\\ 9& 6\\ 10& 2\\ 11& 5\\ 12& 8\end{array}$$Since $a_{10}=a_0,$ $a_{11}=a_1,$ $a_{12}=a_2,$ and each term depends only on the previous three terms, the sequence becomes periodic at this point, with period 10. Therefore, $$a_{2018}{a}_{2020}{a}_{2022}=a_8{a}_0{a}_2=7\cdot 2\cdot 8=\boxed{112}.$$