1. Given the sequence $\left\{a_{n}\right\}$ :
$$
a_{0}=1, a_{n}=2 \sum_{i=0}^{n-1} a_{i}(n \geqslant 1) \text {. }
$$

Then the maximum positive integer $n$ that satisfies $a_{n} \leqslant 2018$ is . $\qquad$
-,1.7.
When $n \geqslant 1$,
$S_{n}-S_{n-1}=a_{n}=2 S_{n-1} \Rightarrow S_{n}=3 S_{n-1}$.
Also, $S_{0}=a_{0}=1$, so $S_{n}=3^{n}$.
Thus, $a_{n}=S_{n}-S_{n-1}=2 \times 3^{n-1}$.
From $2 \times 3^{n-1}<2018 \Rightarrow n<8$.