In how many different ways can the set $\{1,2,\dots, 2006\}$ be divided into three non-empty sets such that no set contains two successive numbers?

$ 
\textbf{(A)}\ 3^{2006}-3\cdot 2^{2006}+1
\qquad\textbf{(B)}\ 2^{2005}-2
\qquad\textbf{(C)}\ 3^{2004}
\qquad\textbf{(D)}\ 3^{2005}-1
\qquad\textbf{(E)}\ \text{None of above}
$
1. **Initial Setup**: We need to divide the set $\{1, 2, \dots, 2006\}$ into three non-empty sets such that no set contains two successive numbers.

2. **First Observation**: The numbers $1$ and $2$ must be in different sets. This is because if they were in the same set, they would violate the condition that no set contains two successive numbers.

3. **Recursive Choice**: Given a division of the first $n$ numbers into three sets, we have only two choices for the $(n+1)$-th number: it can go into one of the two sets that do not contain the $n$-th number. This is because placing $(n+1)$ in the same set as $n$ would result in two successive numbers in the same set, which is not allowed.

4. **Counting

The final answer is $\boxed{2^{2004} - 1}$