6. Given $S_{n}$ as the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, $\boldsymbol{a}=\left(S_{n}, 1\right), \boldsymbol{b}=\left(-1,2 a_{n}+2^{n+1}\right), \boldsymbol{a} \perp \boldsymbol{b}$. If $b_{n}=\frac{n-2011}{n+1} a_{n}$, and there exists $n_{0}$ such that for any $k\left(k \in \mathbf{N}_{+}\right)$, the inequality $b_{k} \leqslant b_{n_{0}}$ holds. Then the value of $n_{0}$ is $\qquad$.
6.2009 or 2010.

From $\boldsymbol{a} \perp \boldsymbol{b} \Rightarrow-S_{n}+2 a_{n}+2^{n+1}=0$
$$
\begin{array}{l}
\Rightarrow-S_{n+1}+2 a_{n+1}+2^{n+2}=0 \\
\Rightarrow a_{n+1}=2 a_{n}-2^{n+1} \Rightarrow \frac{a_{n+1}}{2^{n+1}}=\frac{a_{n}}{2^{n}}-1
\end{array}
$$
$\Rightarrow\left\{\frac{a_{n}}{2^{n}}\right\}$ is an arithmetic sequence
$$
\begin{array}{l}
\Rightarrow \frac{a_{n}}{2^{n}}=-2-(n-1)=-(n+1) \\
\Rightarrow b_{n}=(2011-n) 2^{n} .
\end{array}
$$

Let $b_{n+1} \geqslant b_{n}$. Then
$$
(2010-n) 2^{n+1} \geqslant(2011-n) 2^{n} \text {. }
$$

Solving gives $n \leqslant 2009$.
Thus, the maximum value of $b_{n}$ is $b_{2010}=b_{209}$.
Therefore, $n_{0}=2009$ or 2010.