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Question

The dynamical system:

Let $f(x)=\begin{cases}(x+1)/2&&x\textrm{ odd}\\x/2&&x\textrm{ even}\end{cases}$

is the the bit shift map with binary strings reversed. It terminates for all $x\in\Bbb N$ because $f$ is monotonic decreasing and the natural numbers are well-founded.

But for any given odd $x>1$ which is not divisble by $3$, let $y_n(x)=3^n(x):n\in\Bbb N$

Strong conjecture: For any given $x$, is there always some sufficiently large $n$ such that the orbit of $f$ through $y_n$ never encounters a second odd multiple of $3$?

Weak conjecture: Is there always some sufficiently large $n$ such that the orbit of $f$ through $y_n$ never encounters a larger odd multiple of $3$ than the first?

EDIT: Alternative conjecture: Is there always some sufficiently large $n$ such that the orbit of $f$ through $y_n$ has constant 3-adic value? (I believe this case could occur if the conjugation by $h$ goes the opposite way, i.e. if it were the case that we seek a sequence of ternary rationals of the form $3^{-n}x$, terminating in $\frac13$)

Motivation

I think the above is within reach of modular arithmetic and it looks like it may prove the Collatz conjecture because the appropriate orbit conjugates to the Collatz orbit of the underlying 5-rough (divisble by neither two nor three) number.

More details

BIll asked for more details on the motivation so I have added here:

  • Termination of the reverse bit shift dynamical system above can be rewritten in such a way that the powers of $2$ accumulate instead of being divided out, as follows: $x\mapsto x+2^{\nu_2(x)}$ goes to $2^p:p\in\Bbb N$
  • Suppose $x\mapsto x+2^{\nu_2(x)}$ goes to $2^p:p\in\Bbb N$ and all but the first term are indivisible by $2, 3$
  • Then $x\mapsto x+2^{\nu_2(x)}\cdot3^{\nu_3(x)}$ goes to $2^p3^q:p,q\in\Bbb N$ for all but the first term.
  • Then $x\mapsto 3(x+2^{\nu_2(x)}\cdot3^{\nu_3(x)})$ also goes to $2^p3^q:p,q\in\Bbb N$
  • And since conjugation by $h(x)=x/3$ conjugates that to $x\mapsto 3x+2^{\nu_2(x)}\cdot3^{\nu_3(x)}$, this also goes to $2^p3^q:p,q\in\Bbb N$
  • This final statement is equivalent to the Collatz conjecture.

Example

Example for $x=19$ (showing odd numbers only):

$19,5,3\implies \textrm{ hits a multiple of }3$

$57,29,15\implies \textrm{ hits a multiple of }3$

$171,43,11,3\implies \textrm{ hits a multiple of }3$

$513,257,129\implies \textrm{ hits a multiple of }3$

$1539,385,193,97,49,25,13,7,1\implies \textrm{19 is not a counterexample}$

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    $\begingroup$ What is $3^n(x)$ please? $\endgroup$ Commented Nov 16 at 18:00
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    $\begingroup$ For this to be a bit shift, you'd need the +1 to be a -1, right? Your function has $f(1111_2)=1000_2$. $\endgroup$ Commented Nov 16 at 20:49
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    $\begingroup$ You have not supplied any strong evidence to support the conjecture, nor any justification for your claims that it it "within reach" or "may prove Collatz". It is very easy to make all sorts of conjectures - even beginners can do so. It is much more difficult to make conjectures that may be amenable to current methods, and may also prove useful. Generally only those of the latter type have some hope of being on-topic here. $\endgroup$ Commented Nov 16 at 20:59
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    $\begingroup$ @RobertFrost Can you please add a couple of examples "testing" your conjectures for specific values of $x$ ($x>1$, $x\equiv 1\pmod{2}$, $x\not\equiv 0\pmod{3}$), to make sure the interpretation is clear? In particular, what happens for these values? $$x\in\left\{ 19, 35, 47, 53, 59, 67, 71, 77, 83, 89 \right\}$$ $\endgroup$ Commented Nov 17 at 14:16
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    $\begingroup$ @mezzoctane I added one example for your first value $x=19$, apologies it's all I had time to add. $\endgroup$ Commented Nov 17 at 14:56

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