13. (2018 湖南湘潭四模) 已知数列 $\left\{\sqrt{a_{n+1}}-\sqrt{a_{n}}\right\}$ 是公差为 2 的等差数列, 且 $a_{1}=1, a_{3}=9$, 则 $a_{n}=$ \$ \qquad \$ .
依题意, $\left(\sqrt{a_{3}}-\sqrt{a_{2}}\right)-\left(\sqrt{a_{2}}-\sqrt{a_{1}}\right)=2$, 则 $3-2 \sqrt{a_{2}}+1=2$,

所以 $a_{2}=1$, 所以 $\sqrt{a_{n+1}}-\sqrt{a_{n}}=\sqrt{a_{2}}-\sqrt{a_{1}}+2(n-1)=2 n-2$,

所以 $\sqrt{a_{n}}-\sqrt{a_{1}}=\left(\sqrt{a_{1}}-\sqrt{a_{1}}\right)+\left(\sqrt{a_{3}}-\sqrt{a_{2}}\right)+\cdots+\left(\sqrt{a_{n}}-\sqrt{a_{n-1}}\right)=0+2+\cdots+(2 n-4)$

$=\frac{(2 n-4)(n-1)}{2}$, 又 $a_{1}=1$, 所以 $\sqrt{a_{n}}=n^{2}-3 n+3$, 故 $a_{n}=\left(n^{2}-3 n+3\right)^{2}$.