smc

In binary numbers, we have $$S_n=(x_{n-1}{x}_{n-2}{x}_{n-3}{x}_{n-4}\dots x_2{x}_1{x}_0{)}_2.$$ It follows that $$8S_n=(x_{n-1}{x}_{n-2}{x}_{n-3}{x}_{n-4}\dots x_2{x}_1{x}_{0}000{)}_2.$$ We obtain $7S_n$ by subtracting the equations: $$\begin{array}{clccrccccccr}& (x_{n-1}& x_{n-2}& x_{n-3}& x_{n-4}& \dots & x_2& x_1& x_0& 0& 0& 0{)}_2\\ -& & & & (x_{n-1}& \dots & x_5& x_4& x_3& x_2& x_1& x_0{)}_2\\ & & & & & & & & & & & \\ [-2.5ex]& (?& ?& ?& 0& \dots & 0& 0& 0& 0& 0& 1{)}_2\end{array}$$ We work from right to left: \begin{alignat}{6} x_0=x_1=x_2=1 \quad &\implies \quad &x_3 &= 0& \\ \quad &\implies \quad &x_4 &= 1& \\ \quad &\implies \quad &x_5 &= 1& \\ \quad &\implies \quad &x_6 &= 0& \\ \quad &\implies \quad &x_7 &= 1& \\ \quad &\implies \quad &x_8 &= 1& \\ \quad &\quad \vdots & & & \end{alignat} For all $n\ge 3,$ we conclude that $x_n=0$ if and only if $n\equiv 0(\mathrm{mod}3).$ $x_n=1$ if and only if $n\not\equiv 0(\mathrm{mod}3).$ Finally, we get $(x_{2019},x_{2020},x_{2021},x_{2022})=(0,1,1,0),$ from which $$x_{2019}+2x_{2020}+4x_{2021}+8x_{2022}=\boxed{6}.$$