AMC 12 2012a Problem 24

Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$, and in general,

$$a_k=\begin{cases} (0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}} & \text{if }k\text{ is odd,}\\ (0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}& \text{if }k\text{ is even.} \end{cases}$$

Rearranging the numbers in the sequence $\{a_k\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\{b_k\}_{k=1}^{2011}$. What is the sum of all integers $k$, $1\le k \le 2011$, such that $a_k=b_k?$

\textbf{(A)}\ 671 \textbf{(B)}\ 1006 \textbf{(C)}\ 1341 \textbf{(D)}\ 2011 \textbf{(E)}\ 2012