Computing a sequence in the Cantor set Let $\eta:2^\omega\to 2^\omega$ be such that:
If $x \neq$ all 1's and $n$ is least such that $x_n=0$, then $\eta(x)=y$ where $$y_i=\begin{cases}x_i &\text{for } i>n\\ 1-x_i&\text{for }i\leq n.\end{cases}$$
I am interested in starting with the particular $x$ defined below. I believe this point corresponds to the number $1/4$ in the middle-thirds Cantor set $C\subseteq [0,1]$. Because $\sum_{k=1}^\infty \frac{2}{9^k}=1/4$.
\begin{align}x=&0101010101...\\
\eta(x)=&1101010101...\\
\eta^2(x)=&0011010101...\\
\eta^3(x)=&1011010101...\\
\eta^4(x)=&0111010101...
\end{align}
QUESTION: Can you give a formula (closed form or recursive) for $\eta^n(x)$? 
It's obvious that all the numbers will correspond to fractions in $C$, which might make for a nice equation(?). It's also true that you will never get the same number twice.
Here are the first few numbers:
$\frac{1}{4}$
$\frac{11}{12}$
$\frac{11}{108}$
$\frac{83}{108}$
$\frac{35}{108}$
$\frac{107}{108}$
$\frac{11}{972}$
...
maybe a pattern?
 A: Applying $\eta$ to $y = 0000\dots$ instead would correspond to counting in binary, but backwards: 
\begin{align}
  y &= 0000\dots \\
 \eta(y) &= 1000\dots \\
 \eta^2(y) &= 0100\dots \\
 \eta^3(y) &= 1100\dots \\
 \eta^4(y) &= 0010\dots
\end{align}
This is closely related to sequence A030109 in the OEIS, and won't have a nice closed formula; you could write down the recurrence
\begin{align}
 \eta^{2n}(y) &= \frac13 \eta^n(y), \\
 \eta^{2n+1}(y) &= \frac13 \eta^n(y) + \frac23.
\end{align}
When we start with $x$ instead of $y$, there's a trailing infinite sequence of $\dots1010101\dots$ that periodically throws us off. So the sequence $\eta^n(x)$ can be naturally broken up into block of length $4^k$ for $k=0, 1, 2, \dots$: in each of these blocks, we are copying $\eta(y)$ on the first $2k$ bits, followed by $110101\dots$.
More precisely, when $n = 1 + 4 + 4^2 + \cdots + 4^{k-1} + j$, for $1 \le j \le 4^k$, we have
\begin{align}
 \eta^n(x) &= \eta^{j-1}(y) + \frac{2}{3^{2k+1}} + \sum_{i=k+1}^\infty \frac{2}{3^{2i}} \\
  &= \eta^{j-1}(y) + \frac{11}{4 \cdot 3^{2k+1}}.
\end{align}
For example:


*

*If $n=5 = 1 + 4$, then $k=1$ and $j=4$, so we have $\eta^{n}(x) = \eta^{3}(y) + \frac{11}{108} = \frac{8}{9} + \frac{11}{108} = \frac{107}{108}$. 

*If $n=6 = 1 + 4 + 1$, then $k=2$ and $j=1$, so we have $\eta^{n}(x) = \eta^{0}(y) + \frac{11}{972} = \frac{11}{972}$.

A: $\eta$ toggles all the bits up through the first zero in its argument.  As a result, the first bit of $x$ toggles with every application of $\eta$.  The second bit toggles unless the first bit is zero, so it goes through two $0$s and then two $1$s.  The third bit will toggle when the first two bits are $1$, so will go through four $0$s followed by four $1$s and so on.  The first four bits go through a cycle $$ \begin {align} &0000,1000,0100,1100,\\&0010,1010,0110,1110,\\&0001,1001,0101,1101,\\&0011,1011,0111,1111,\\ &0000,\ldots \end {align}$$  This cycle is just the reverse of the binary counting sequence.  For any $x$, to get the bit string of $\eta^k(x)$ take a long enough leading part of $x$ that we won't carry out, reverse that part, add $k$ to the reversed leading part in binary, reverse the result, and append the infinite tail that we ignored.
