We will double count the sum $a_1+\cdots+a_n$. Another way too see this quantity is to sum up the numbers $x_t$ where $x_t$ denotes number of elements in $\{1, 2, \cdots, n\}$ divisible by $p_t p_{t+1}$. Thus $E = a_1+\cdots+a_n = \sum_{t \geqslant 1} \left\lfloor\frac{n}{p_tp_{t+1}}\right\rfloor$. Then $2E < \frac{n}{3}+n\sum_{t \geqslant 2}\left( \frac{1}{p_t} - \frac{1}{p_{t+1}} \right) = 2n/3$. Thus $E < n/3$ as required.