Try the four possible combinations of values for $a_{0}$ and $a_{1}$. Since we can write $a_{n} \equiv$ $b_{n-1}-a_{n-2}-a_{n-1}$, these two numbers completely determine the solution $\left\{a_{i}\right\}$ beginning with them (if there is one). For $a_{0}=a_{1}=0$, we can check that the sequence beginning $0,0,0,0,1,1$ and repeating every 6 indices is a possible solution for $\left\{a_{i}\right\}$, so one possible value for $4 a_{0}+2 a_{1}+a_{2}$ is 0 . The other three combinations for $a_{0}$ and $a_{1}$ similarly lead to valid sequences (produced by repeating the sextuples $0,1,1,1,0,1 ; 1,0,1,1,1,0$; $1,1,0,1,0,1$, respectively); we thus obtain the values 3,5 , and 6.