Five. (20 points) The sequence $\left\{a_{n}\right\}$ is
$$
1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots \text {. }
$$

Its construction method is:
First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, thus, we get $a_{2}=1, a_{3}=2$;

Then copy all the previous items $1,1,2$, and add the successor number 3 of 2, thus, we get $a_{4}=1, a_{5}=1, a_{6}=$ $2, a_{7}=3$;

Next, copy all the previous items $1,1,2,1$, $1,2,3$, and add the successor number 4 of 3, thus, the first 15 items are $1,1,2,1,1,2,3,1,1,2,1,1,2,3,4$.
Continue in this manner.
Try to find $a_{2006}$ and the sum of the first 2006 terms $S_{2006}$.
Five, according to the construction method of $\left\{a_{n}\right\}$, it is easy to know
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \text {. }
$$

Generally, there is $a_{2^{n}-1}=n$, that is, the number $n$ first appears at the $2^{n}-1$ term, and if
$m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$, then $a_{m}=a_{i}$.
Since $2006=2^{10}-1+983,983=2^{9}-1+472$,
$$
\begin{array}{l}
472=2^{8}-1+217,217=2^{7}-1+90, \\
90=2^{6}-1+27,27=2^{4}-1+12, \\
12=2^{3}-1+5,
\end{array}
$$

thus $a_{2006}=a_{963}=a_{42}=a_{217}=a_{90}=a_{27}=a_{12}=a_{5}=1$.
To find $S_{20 \pi}$, first calculate $S_{2^{\pi}-1}$.
According to the construction method of $\left\{a_{n}\right\}$, in the first $2^{n}-1$ terms of the sequence, there are exactly 1 $n$, 2 $n-1$, $2^{2}$ $n-2$, ..., $2^{k}$ $n-k$, ..., $2^{n-1}$ 1. Therefore,
$$
\begin{aligned}
S_{2^{n-1}}= & n+2(n-1)+2^{2}(n-2)+\cdots+ \\
& 2^{n-2} \times 2+2^{n-1} \times 1 .
\end{aligned}
$$

Thus, $2 S_{2^{n}-1}=2 n+2^{2}(n-1)+2^{3}(n-2)+\cdots+$
$$
2^{n-1} \times 2+2^{n} \text {. }
$$

According to equations (1) and (2),
$$
\begin{array}{l}
S_{2^{n}-1}=-n+\left(2+2^{2}+\cdots+2^{n-1}+2^{n}\right) \\
=2^{n+1}-(n+2) .
\end{array}
$$

Secondly, when $m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$,
$$
\begin{array}{l}
S_{m n}=S_{2^{n}-1}+\left(a_{\left(2^{n}-1\right)+1}+a_{\left(2^{n}-1\right)+2}+\cdots+a_{\left(2^{n}-1\right)+k}\right) \\
=S_{2^{n}-1}+\left(a_{1}+a_{2}+\cdots+a_{k}\right)=S_{2^{n}-1}+S_{k} .
\end{array}
$$

Therefore, $S_{2 \alpha 66}=S_{2^{10}-1}+S_{\text {कe3 }}, S_{\text {*43 }}=S_{2}{ }^{9}-1+S_{472}$,
$$
\begin{array}{l}
S_{42}=S_{2^{8}-1}+S_{217}, S_{217}=S_{2^{7}-1}+S_{90}, \\
S_{90}=S_{2^{6}-1}+S_{z 7}, S_{77}=S_{2^{4}-1}+S_{12}, S_{12}=17 .
\end{array}
$$

Thus, by equation (3),
$$
\begin{array}{l} 
S_{2006}=\left(2^{11}-12\right)+\left(2^{10}-11\right)+\left(2^{9}-10\right)+ \\
\left(2^{8}-9\right)+\left(2^{7}-8\right)+\left(2^{5}-6\right)+17 \\
=3961 .
\end{array}
$$