1. Define the sequence $\left\{a_{n}\right\}$ :
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, a_{3}=5, \\
a_{n}=a_{n-1}-a_{n-2}+a_{n-3}(n=4,5, \cdots) .
\end{array}
$$

Then the sum of the first 2015 terms of this sequence $S_{2015}=$
-、1.6045.
From the given information, we have $a_{4}=3$, and the sequence $\left\{a_{n}\right\}$ has a period of 4. Therefore, $S_{2015}$
$$
\begin{array}{l}
=503\left(a_{1}+a_{2}+a_{3}+a_{4}\right)+a_{1}+a_{2}+a_{3} \\
=503(1+3+5+3)+1+3+5=6045 .
\end{array}
$$