Let \left\{a_{k}\right\}_{k=1}^{2011} be the sequence of real numbers defined by
a_{1}=0.201, \quad a_{2}=(0.2011)^{a_{1}}, \quad a_{3}=(0.20101)^{a_{2}}, \quad a_{4}=(0.201011)^{a_{3}}
and more generally
a_{k}= \begin{cases}(0 . \underbrace{20101 \ldots 0101}_{k+2 \text { digits }})^{a_{k-1}}, \text { if } k \text { is odd, } \\ (0 . \underbrace{20101 \ldots 01011}_{k+2 \text { digits }})^{a_{k-1}}, \text { if } k \text { is even. }\end{cases}
Rearranging the numbers in the sequence \left\{a_{k}\right\}_{k=1}^{2011} in decreasing order produces a new sequence \left\{b_{k}\right\}_{k=1}^{2011}. What is the sum of all the integers k, 1 \leq k \leq 2011, such that a_{k}=b_{k} ?
Answer Choices
A. 671
B. 1006
C. 1341
D. 2011
E. 2012