4, 4, 4, 4s

We need to find $A \bmod 10$ .

$\begin{aligned} A & \equiv 4^0 + 4^1 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6 + \cdots + 4^{2017} + 4^{2018} \text{ (mod 10)} \\ & \equiv 1 + (4^1 + 4^2) + 4^2(4^1 + 4^2) + 4^4(4^1 + 4^2) + \cdots + 4^{2016}(4^1 + 4^2) \text{ (mod 10)} \\ & \equiv 1 + 20 + 4^2(20) + 4^4(20) + \cdots + 4^{2016}(20) \text{ (mod 10)} \\ & \equiv \boxed{1} \text{ (mod 10)} \end{aligned}$

$A = \frac{4^{2019} - 1}{3} = \frac{B}{3}$ and $B$ is divisible by 3. (Let $B = 4^{2019} - 1$ ).

$4^{2019}$ ends at 4. This implies that $B$ ends at 3 and hence $A$ finishes at 1, because $B$ without the last digit 3, is also divisible by 3 (the sum of its digits is divisible by 3).