15. The sequence $\left\{a_{n}\right\}$ satisfies: $a_{0}=1, a_{n}=\left[\sqrt{S_{n-1}}\right]$ $(n=1,2, \cdots,[x]$ represents the greatest integer not greater than $x$, $\left.S_{k}=\sum_{i=0}^{k} a_{i}(k=0,1, \cdots)\right)$. Find the value of $a_{2006}$.
15. Observe the initial terms of the sequence (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$a_{n}$ & 1 & 1 & 1 & 1 & 2 & 2 & 2 & 3 & 3 & 4 & 4 \\
\hline$S_{n}$ & 1 & 2 & 3 & 4 & 6 & 8 & 10 & 13 & 16 & 20 & 24 \\
\hline \hline$n$ & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & \\
\hline$a_{n}$ & 4 & 5 & 5 & 6 & 6 & 7 & 7 & 8 & 8 & 8 & \\
\hline$S_{n}$ & 28 & 33 & 38 & 44 & 50 & 57 & 64 & 72 & 80 & 88 & \\
\cline { 1 - 2 } & &
\end{tabular}

Notice that the sequence $\left\{a_{n}\right\}$ is strictly increasing, and each positive integer 1, 2, $\cdots$ appears in the sequence $\left\{a_{n}\right\}$ in order. Except for the first term $a_{0}=1$, each number of the form $2^{k} (k=0,1, \cdots)$ appears three times consecutively, and other numbers appear twice consecutively.
In general, the following properties of the sequence $\left\{a_{n}\right\}$ can be proven:
(1) For any $k \in \mathbf{N}$, if we denote $m=2^{k+1}+k+1$, then $a_{m-2}=a_{m-1}=a_{m}=2^{k}$.
(2) For any $k \in \mathbf{N}_{+}$, if we denote $m_{0}=2^{k}+k$, then when $1 \leqslant r \leqslant 2^{k-1}$, we have
$$
a_{m_{0}+2 r-1}=a_{m_{0}+2 r}=2^{k-1}+r \text {. }
$$

We use induction on $k$.
From the terms listed above, we know that the conclusion holds for $k \leqslant 2$. Assume that properties (1) and (2) hold for $k \leqslant n$, i.e., when $m=2^{n+1}+n+1$, $a_{m-2}=a_{m-1}=a_{m}=2^{n}$, then
$$
\begin{array}{l}
S_{m}=a_{0}+2\left(1+2+\cdots+2^{n}\right)+\left(2^{0}+2^{1}+\cdots+2^{n}\right) \\
=2^{2 n}+3 \times 2^{n} .
\end{array}
$$

We then use induction on $r$ for $1 \leqslant r \leqslant 2^{n}$.
When $r=1$, since $\left(2^{n}+1\right)^{2}0 \text {, }
$$

i.e., $\left(2^{n}+p+1\right)^{2} \leqslant S_{m+2 p}<\left(2^{n}+p+2\right)^{2}$.
Therefore, $a_{m+2 p+1}=\left[\sqrt{S_{m+2 p}}\right]=2^{n}+p+1$.
Since $S_{m+2 p}<S_{m+2 p+1}$
$$
\begin{array}{l}
=S_{m+2 p}+a_{m+2 p+1} \\
=2^{2 n}+2(p+2) \times 2^{n}+p^{2}+2 p+1 \\
<\left(2^{n}+p+2\right)^{2},
\end{array}
$$

Therefore, $a_{m+2 p+2}=\left[\sqrt{S_{m+2 p+1}}\right]=2^{n}+p+1$.
Hence, by induction, when $m=2^{n+1}+n+1, 1 \leqslant r \leqslant 2^{n}$, we have
$$
a_{m+2 r-1}=a_{m+2 r}=2^{n}+r \text {. }
$$

In particular, when $r=2^{n}$, the above formula becomes
$$
a_{2^{n+2}+n}=a_{2^{n+2}+n+1}=2^{n+1} \text {. }
$$

Also, from (1), we get
$$
\begin{array}{l}
S_{m+2 r}=2^{2 n}+(2 r+3) \times 2^{n}+r(r+1) . \\
\text { When } r=2^{n}, m=2^{n+1}+n+1, \text { we have } \\
S_{2^{n+2}+n+1}=2^{2 n}+\left(2^{n+1}+3\right) \times 2^{n}+2^{n}\left(2^{n}+1\right) \\
=2^{2 n+2}+2 \times 2^{n+1}<\left(2^{n+1}+1\right)^{2} .
\end{array}
$$

Therefore, $a_{2^{n+2}{ }_{n+2}}=\left[\sqrt{S_{2^{n+2}+n+1}}\right]=2^{n+1}$.
From (2) and (3), we know that for $m=2^{k+1}+k+1$, when $k=n+1$, we also have $a_{m-2}=a_{m-1}=a_{m}=2^{k}$.
Thus, properties (1) and (2) hold.
Since $2^{10}+10<2006<2^{11}+11$, let $m_{0}=2^{10}+10$, then $k-1=9, r=\frac{2006-m_{0}}{2}=486$.
Therefore, $a_{2006}=a_{m_{0}+2 r}=2^{9}+r=512+486=998$.