1. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=1, a_{2}=4, a_{3}=9, \\
a_{n}=a_{n-1}+a_{n-2}-a_{n-3}(n=4,5, \cdots) .
\end{array}
$$

Then $a_{2011}=$
$-、 1.8041$.
From the problem, we have
$$
a_{2}-a_{1}=3, a_{3}-a_{2}=5 \text {, }
$$

and $a_{n}-a_{n-1}=a_{n-2}-a_{n-3}(n \geqslant 4)$.
Thus, $a_{2 n}-a_{2 n-1}=3, a_{2 n+1}-a_{2 n}=5\left(n \in \mathbf{N}_{+}\right)$.
Therefore, $a_{2 n+1}-a_{2 n-1}=8$.
Hence, $a_{2011}=\sum_{k=1}^{1005}\left(a_{2 k+1}-a_{2 k-1}\right)+a_{1}$
$$
=1005 \times 8+1=8041 \text {. }
$$