4. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Calculate: $\sum_{k=0}^{2019}\left[\frac{4^{k}}{5}\right]=$ $\qquad$ .
4. $\frac{4^{2020}-1}{15}-1010$.

Notice, $\sum_{k=0}^{2019} \frac{4^{k}}{5}=\frac{4^{2020}-1}{15}$.
$$
\begin{array}{l}
\text { By } \frac{4^{k}}{5}+\frac{4^{k+1}}{5}=4^{k} \Rightarrow\left[\frac{4^{k}}{5}\right]+\left[\frac{4^{k+1}}{5}\right]=4^{k}-1 . \\
\text { Therefore } \sum_{k=0}^{2019}\left[\frac{4^{k}}{5}\right]=\frac{4^{2020}-1}{15}-1010 .
\end{array}
$$