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:









maybe a pattern?


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 $\eta^{n}(x) = \eta^{0}(y) + \frac{11}{972} = \frac{11}{972}$.

$\eta$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.