Let $x_0 = 3;\ x_{n+1}=3x_n\ $ if $\ \frac{x_n}{2}<1;\ x_{n+1}=\frac{x_n}{2}\ $ if $\ \frac{x_n}{2}>1.\ $ Is $\ \liminf_{n\to\infty} x_n=1?$

Question

This is a natural follow-up question of this previous question of mine.

Let $x_0 = 3.$ Let $\ x_{n+1} = 3x_n\ $ if $\ \frac{x_n}{2}<1;\quad x_{n+1} = \frac{x_n}{2}\ $ if $\ \frac{x_n}{2}>1.\quad $ Is $\ \displaystyle\liminf_{n\to\infty} x_n = 1?$

I don't think the proposition follows from either of the answers to my previous question, but maybe I am missing something.

I tried testing this algorithm with a calculator, but couldn't conclude either way after $40$ or so iterations, at which point the calculator has probably lost some accuracy of the value of $x_n.$

Answer 1

Let $ y_n = \log x_n$. Then the problem transforms into:

$ y_0 = \log 3 $.
$y_{n+1} = y_n + \log 3 $ if $ y_n < \log 2 $.
$ y_{n+1} = y_n - \log 2 $ if $ y_n > \log 2 $.
Is $ \liminf y_n = 0 $?

Note that $y_n \geq 0$.

Claim: For each positive integer $k$, there exists an $m_k$ such that $y_{m_k} = \quad k \log 3 \pmod{ \log 2 }$. (Note that I don't just want $y_{m_k} {\color{red} \equiv} k \log 3 \pmod {\log 2}$.)

We're adding $ \log 3 $ one at a time, so there is some $y_m \equiv k \log 3 \pmod{ \log 2 } $.
In the next 1 or 2 steps, we then reduce it to the desired value of $ y_{m_k} = k \log 3 \pmod{\log 2 } $.

Corollary: From the irrationality of $ \frac{ \log 3 } { \log 2 } $ and the density of $ k \log 3 \pmod{\log 2 } $, it follows that $\lim \inf y_n = 0$.