6. In the sequence $\left\{a_{n}\right\}$, each term $a_{n}$ represents the largest odd factor of the positive integer $n$. For example, the factors of 5 are $1, 5$, so $a_{5}=5$; the factors of 6 are $1, 2, 3, 6$, so $a_{6}=3$. Let the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ be $S_{n}$. Then $S_{2^{2000}-1}=$ $\qquad$
6. $\frac{1}{3}\left(4^{2020}-1\right)$.

From the definition of $a_{n}$, we know:
When $n$ is even, $a_{n}=a_{\frac{n}{2}}$;
When $n$ is odd, $a_{n}=n$.
Let $T_{n}=S_{2^{n}-1}$. Then
$$
\begin{array}{l}
T_{n+1}=S_{2^{n+1}-1} \\
=1+3+\cdots+\left(2^{n+1}-1\right)+a_{2}+a_{4}+\cdots+a_{2 n+1}-2 \\
=\frac{2^{n}\left(1+2^{n+1}-1\right)}{2}+S_{2^{n}-1} \\
\Rightarrow T_{n+1}=4^{n}+T_{n} .
\end{array}
$$

Notice that, $T_{1}=S_{1}=1$.
$$
\begin{array}{l} 
\text { Hence } T_{n}= T_{n}-T_{n-1}+T_{n-1}-T_{n-2}+\cdots+ \\
T_{2}-T_{1}+T_{1} \\
=4^{n-1}+4^{n-2}+\cdots+4^{1}+1
\end{array}
$$