Six. (20 points) As shown in Figure 3, let
$$
P_{1}(1, \sqrt{3}), P_{2}(4,2 \sqrt{3}), \cdots, P_{n}\left(x_{n}, y_{n}\right)
$$
$\left(0<y_{1}<y_{2}<\cdots<y_{n}\right)$ be $n$ points on the curve $C: y^{2}=3 x(y \geqslant 0)$. Points $A_{i}(i=1,2, \cdots, n)$ are on the positive $x$-axis, satisfying that $\triangle A_{i-1} A_{i} P_{i}$ is an equilateral triangle ($A_{0}$ is the origin).
(1) Find the expression for the $x$-coordinate $a_{n}$ of point $A_{n}\left(a_{n}, 0\right)$ in terms of $n$;
(2) Evaluate
$$
\left[\frac{4}{5}+\frac{1}{\sqrt[4]{a_{1}}}+\frac{1}{\sqrt[4]{a_{2}}}+\cdots+\frac{1}{\sqrt[4]{a_{2000}}}\right]
$$

where [a] denotes the integer part of the number. When $2005 < c < 2005.6$, take $\sqrt{c}=44.78$; when $2006 < c < 2006.5$, take $\sqrt{c}=44.785$. $\sqrt{2}=1.4$.
(1) Let point $A_{n}\left(a_{n}, 0\right)$. Using the graph, we can find that the x-coordinate of the point is the average of the x-coordinates of points $A_{n-1}$ and $A_{n}$, i.e., $x_{n}=\frac{a_{n-1}+a_{n}}{2}$. Therefore, the y-coordinate of point $P_{n}$ is $y_{n}=\sqrt{3\left(\frac{a_{n-1}+a_{n}}{2}\right)}$.
At the same time, $y_{n}=\frac{\sqrt{3}}{2}\left|A_{i-1} A_{i}\right|=\frac{\sqrt{3}}{2}\left(a_{n}-a_{n-1}\right)$.
Thus, $\sqrt{3\left(\frac{a_{n-1}+a_{n}}{2}\right)}=\frac{\sqrt{3}}{2}\left(a_{n}-a_{n-1}\right)$.
Hence, $a_{n}-a_{n-1}=\sqrt{2\left(a_{n}+a_{n-1}\right)}$.
Therefore, $a_{n}^{2}-2 a_{n-1} a_{n}+a_{n-1}^{2}=2\left(a_{n}+a_{n-1}\right)$.
Similarly, $a_{n+1}^{2}-2 a_{n} a_{n+1}+a_{n}^{2}=2\left(a_{n+1}+a_{n}\right)$.
Subtracting the two equations, we get
$\left(a_{n+1}-a_{n-1}\right)\left(a_{n+1}+a_{n-1}-2 a_{n}-2\right)=0$.
Thus, $a_{n+1}+a_{n-1}-2 a_{n}-2=0$.
Then, $\left(a_{n+1}-a_{n}\right)-\left(a_{n}-a_{n-1}\right)=2$.
Therefore, the sequence $\left\{a_{n+1}-a_{n}\right\}$ is an arithmetic sequence.
Hence, $a_{n}-a_{n-1}=\left(a_{1}-a_{0}\right)+(n-1) \times 2=2 n$.
Thus, by summing up, we can find $a_{n}=n(n+1)$.
(2) Since $\frac{1}{\sqrt[4]{a_{k}}}=\frac{1}{\sqrt[4]{k(k+1)}}\frac{1}{\sqrt{k+\frac{1}{2}}}$
$$
>\frac{2}{\sqrt{k+\frac{9}{8}}+\sqrt{k+\frac{1}{8}}} \\
=2\left(\sqrt{k+\frac{9}{8}}-\sqrt{k+\frac{1}{8}}\right),
$$

we have
$$
\begin{array}{l}
\frac{1}{\sqrt[4]{a_{1}}}+\frac{1}{\sqrt[4]{a_{2}}}+\cdots+\frac{1}{\sqrt[4]{a_{200}}} \\
2 \sqrt{2006 \frac{1}{8}}-\frac{3 \sqrt{2}}{2}>87.3 . \\
\text { Therefore, }\left[\frac{4}{5}+\frac{1}{\sqrt[4]{a_{1}}}+\frac{1}{\sqrt[4]{a_{2}}}+\cdots+\frac{1}{\sqrt[4]{a_{2000}}}\right]=88 .
\end{array}
$$