6. After removing all perfect squares from the sequence of positive integers $\{1,2, \cdots\}$, the remaining numbers form a sequence $\left\{a_{n}\right\}$ in their original order. Then $a_{2015}=$ $\qquad$ .
6.2060.

Let $k^{2}2014
\end{array}\right. \\
\Rightarrow k=45, a_{2015}=2060 .
\end{array}
$