11. In the Cartesian coordinate system $x O y$, there are infinitely many circles inside the parabola $y=a x^{2}$ (where $a$ is a positive constant), with centers $O_{1}, O_{2}, \cdots$ all on the $y$-axis. For each integer $n>1, \odot O_{n}$ is tangent to the parabola and externally tangent to $\odot O_{n-1}$, as shown in Figure 4. If the radius of $\odot O_{1}$ is $\frac{1}{2 a}$ and it passes through the vertex of the parabola, find the radius of $\odot O_{2016}$.
11. Let the radius of $\odot O_{n}$ be $r_{n}$, and denote
$$
S_{n}=r_{1}+r_{2}+\cdots+r_{n} \text {. }
$$

Let the center of $\odot O_{n+1}$ be $O_{n+1}\left(0,2 S_{n}+r_{n+1}\right)$, and the equation of the circle be
$$
x^{2}+\left(y-\left(2 S_{n}+r_{n+1}\right)\right)^{2}=r_{n+1}^{2} \text {. }
$$

By combining the above equation with the parabola equation $y=a x^{2}$ and eliminating $x$, we get
$$
\begin{array}{l}
y+a\left(y-\left(2 S_{n}+r_{n+1}\right)\right)^{2}=a r_{n+1}^{2} \\
\Rightarrow a y^{2}-\left(4 a S_{n}+2 a r_{n+1}-1\right) y+ \\
\quad a\left(4 S_{n}^{2}+4 S_{n} r_{n+1}\right)=0 .
\end{array}
$$

Since $\odot O_{n+1}$ is tangent to the parabola, the above quadratic equation in $y$ has a repeated root.
$$
\begin{array}{c}
\text { Hence } \Delta=\left(4 a S_{n}+2 a r_{n+1}-1\right)^{2}- \\
4 a^{2}\left(4 S_{n}^{2}+4 S_{n} r_{n+1}\right)=0 \\
\Rightarrow 4 a^{2} r_{n+1}^{2}+1-8 a S_{n}-4 a r_{n+1}=0 .
\end{array}
$$

Let $S_{0}=0$.
Also, $r_{1}=\frac{1}{2 a}$, so we have
$$
4 a^{2} r_{n}^{2}+1-8 a S_{n-1}-4 a r_{n}=0 \text {. }
$$
(1) - (2) gives
$$
\begin{array}{l}
a r_{n+1}^{2}-a r_{n}^{2}-r_{n+1}-r_{n}=0 \\
\Rightarrow\left(r_{n+1}+r_{n}\right)\left(a\left(r_{n+1}-r_{n}\right)-1\right)=0 .
\end{array}
$$

Since $r_{n}>0(n=1,2, \cdots)$, we have
$$
a\left(r_{n+1}-r_{n}\right)-1=0 \text {. }
$$

Thus, $r_{n+1}-r_{n}=\frac{1}{a}$.
Therefore, $\left\{r_{n}\right\}$ is an arithmetic sequence with a common difference of $\frac{1}{a}$.
Combining $r_{1}=\frac{1}{2 a}$, we get
$$
r_{n}=\frac{1}{2 a}+(n-1) \frac{1}{a}=\left(n-\frac{1}{2}\right) \frac{1}{a} \text {. }
$$

Thus, $r_{2016}=\frac{4031}{2 a}$.