13. (i) (Grade 11) In the arithmetic sequence $\left\{a_{n}\right\}: a_{n}=4 n -1\left(n \in \mathbf{N}_{+}\right)$, after deleting all numbers that can be divided by 3 or 5, the remaining numbers are arranged in ascending order to form a sequence $\left\{b_{n}\right\}$. Find the value of $b_{2006}$.
(ii) (Grade 12) Given $\odot P: x^{2}+y^{2}=2 x$ and the parabola $S: y^{2}=4 x$, a line $l$ is drawn through the center $P$ intersecting the two curves at four points, which are sequentially labeled as $A, B, C, D$ from top to bottom. If the lengths of segments $A B, B C, C D$ form an arithmetic sequence in this order, find the equation of the line $l$.
(i) Since $a_{n+15}-a_{n}=60$, $a_{n}$ is a multiple of 3 or 5 if and only if $a_{n+15}$ is a multiple of 3 or 5.
Now, divide the positive direction of the number line into a series of intervals of length 60:
$$
(0,+\infty)=(0,60] \cup(60,120] \cup(120,180] \cup \cdots \text {. }
$$

Note that the first interval contains 15 terms of $\left\{a_{n}\right\}$, namely 3, $7,11,15,19,23,27,31,35,39,43,47,51,55,59$, of which 8 terms belong to $\left\{b_{n}\right\}$: $b_{1}=7, b_{2}=11, b_{3}=19, b_{4}=23$, $b_{5}=31, b_{6}=43, b_{7}=47, b_{8}=59$. Therefore, each interval contains 15 terms of $\left\{a_{n}\right\}$ and 8 terms of $\left\{b_{n}\right\}$, and $b_{8 k+r}-b_{r}=60 k(k \in \mathbf{N}, 1 \leqslant r \leqslant 8)$.
Since $2006=8 \times 250+6$, and $b_{6}=43$, we have
$$
b_{2006}=60 \times 250+b_{6}=60 \times 250+43=15043 \text {. }
$$
(ii) The equation of $\odot P$ is $(x-1)^{2}+y^{2}=1$, with a diameter of $|B C|=2$ and the center at $P(1,0)$.

Let $l: k y=x-1$, i.e., $x=k y+1$, substituting into the parabola equation gives $y^{2}=4 k y+4$.
Let $A\left(x_{1}, y_{1}\right)$ and $D\left(x_{2}, y_{2}\right)$, then
$$
\left\{\begin{array}{l}
y_{1}+y_{2}=4 k, \\
y_{1} y_{2}=-4 .
\end{array}\right.
$$

Thus, $\left(y_{1}-y_{2}\right)^{2}=\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}$.
Hence, $|A D|^{2}=\left(y_{1}-y_{2}\right)^{2}+\left(x_{1}-x_{2}\right)^{2}$
$=\left(y_{1}-y_{2}\right)^{2}+\left(\frac{y_{1}^{2}-y_{2}^{2}}{4}\right)^{2}$
$=\left(y_{1}-y_{2}\right)^{2}\left[1+\left(\frac{y_{1}+y_{2}}{4}\right)^{2}\right]=16\left(k^{2}+1\right)^{2}$.
Therefore, $|A D|=4\left(k^{2}+1\right)$.
According to the problem,
$2|B C|=|A B|+|C D|=|A D|-|B C|$,
so, $|A D|=3|B C|=6$, i.e.,
$4\left(k^{2}+1\right)=6$.
Solving for $k$ gives $k= \pm \frac{\sqrt{2}}{2}$.
Thus, the equation of $l$ is $x=\frac{\sqrt{2}}{2} y+1$ or $x=-\frac{\sqrt{2}}{2} y+1$.