Let ${a_k}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201$, $a_2=(0.20101{)}^{a_1}$, $a_3=(0.201011{)}^{a_2}$, $a_4=(0.2010101{)}^{a_3}$, and in general, $a_k=\{\begin{array}{ll}(0.20101\dots 011{)}^{a_{k-1}}& \text{if }k\text{ is odd,}\\ (0.20101\dots 01{)}^{a_{k-1}}& \text{if }k\text{ is even.}\end{array}$ where the base has $k+2$ digits. 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 $a_k=b_k$?