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What's the closure of the equivalence classes of numbers the same distance from $\langle2\rangle$ by the Collatz graph?

Let the Collatz function be $f(x)=3x+2^{\nu_2(x)}$

The Collatz conjecture states that if $x_{n+1}=f(x_n)$ then for every integer $x_0$, there exists $n$ such that $x_n\in\langle2\rangle=\{1,2,4,8,\ldots\}$

Then letting $\Bbb N$ inherit the 2-adic topology, any given equivalence class $X$ of odd integers which is the same number of steps from $\langle2\rangle$, is a set closed under the relation $x\sim4x+1$. For example: $3, 13, 53, 213,\ldots$ are all two steps from $\langle2\rangle$.

Every such class shares the same accumulation point $-\frac13$ and therefore the closure of $X$ is $X\cup\{-\frac13\}$.

By this variant of the Collatz graph, each multiple of $2$ is the same number of steps from $\langle2\rangle$ as its largest odd factor. So $6,26,106,426\ldots$ are also two steps from $\langle2\rangle$. For these sets of even numbers the relation $x\sim4x+1$ can be generalised to $x\sim4x+2^{\nu_2(x)}$, and for any given class $X$ containing only numbers of the same 2-adic valuation $\nu_2(x)$, the accumulation point is $\dfrac{-2^{\nu_2(x)}}{3}$ so the closure of $X$ is $\overline X=X\cup\left\{\dfrac{-2^{\nu_2(x)}}{3}\right\}$

So the question is in two parts:

A. If $S$ is some sequence of odd integers that converges to $-\frac13$ by iteration of the relation $4x+1$ then does $S\cdot\langle2\rangle$ have further accumulation points over and above $\left\{-\frac{2^p}{3}:\leq\omega\right\}$?

and

B: If $S(2)$ is a sequence of sequences like $S$, all odd numbers, such that the sequence of least elements of every sequence itself converges to $-\frac13$, then does $S(2)\cdot\langle2\rangle$ have further accumulation points over and above $\left\{-\frac{2^p}{3}:\leq\omega\right\}$? (I'm writing $2^\omega=0$).


It's clear to me its closure is at least $\overline X=X\cup\left\{\dfrac{-2^p}{3}:p\in\Bbb N\right\}\cup\{0\}$. I don't think any further accumulation points are created but I'm not sure and I have no idea how to prove that.

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  • $\begingroup$ It's unclear what you mean by "some class X" in general, since your two examples before do not share a common definition, and your last paragraph seems to extend the definition further. If $X=\Bbb Z$ matches what you have in mind as "some class $X$", then $\bar X = \Bbb Z_p$ is much bigger. If it's not, please clarify what is "some class X [which] contains all the integers the same distance from $\langle 2 \rangle$, i.e. of every valuation" (if need be, by some examples for such a "class"). $\endgroup$ Oct 8, 2018 at 19:05
  • $\begingroup$ @TorstenSchoeneberg By "all integers 2 steps away", I mean "as opposed to only the odd integers 2 steps away" An example is perhaps the best clarification. The set $S=3,13,53,213,\ldots$ is the set of odd integers 1 iteration of $f$ from $5\cdot\langle2\rangle$. The set of all integers one iteration from $5\cdot\langle2\rangle$ is given by $S\cdot\langle2\rangle$. This has accumulation points I guess you might write it: $\left\{-\frac{2^p}{3}:p\leq\omega\right\}$ which I understand to contain $0$. $\endgroup$ Oct 8, 2018 at 19:29
  • $\begingroup$ @TorstenSchoeneberg What I'm not clear on is a) whether additional accumulation points can be constructed over and above $\left\{-\frac{2^p}{3}:p\leq\omega\right\}$, and then b) $3,13,53,213,\ldots$ is not the only sequence 2 steps from $\langle2\rangle$. In fact there are infinitely many sequences, the next-largest being $(113,453,...)\cdot\langle2\rangle$. But these continue (and in fact by induction for all such classes) a recurrence relation which again converges to $-\frac{2^p}{3}$ $\endgroup$ Oct 8, 2018 at 19:34
  • $\begingroup$ @TorstenSchoeneberg The conclusion I want determine whether I'm safe to arrive at, is that every finitely long subset of the Collatz graph (as defined above by iteration of $f(x)=3x+2^{\nu_2(x)}$), only has the accumulation points $\left\{-\frac{2^p}{3}:\leq\omega\right\}$ $\endgroup$ Oct 8, 2018 at 19:40
  • $\begingroup$ @TorstenSchoeneberg I've managed to substantially reduce the part of which I'm unclear - the last sentence of the partial answer I've posted expresses what's left to do. $\endgroup$ Oct 9, 2018 at 12:09

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This is a partial answer that identifies two further accumulation points...

Considering the sequences which are two steps before $5\cdot\langle2\rangle$ (although this generalises for any shared successor), then the sequence of first or immediate predecessors of $5\cdot\langle2\rangle$ is $S=3,13,53,213,\ldots$.

I will restrict to only considering the odd factors.

Assuming $S$'s $\Bbb N$-index begins with $0$ then every element $x$ of this sequence whose index is $\equiv1\pmod3$ has a least immediate pedecessor $(4x-1)/3$. The first such predecessor of $5$ is the number 13.

The general form for these every third elements $x$ of $S$ is $4^{3n}\cdot13+\frac{4^{3n}-1}{3}$

Then the sequence of their least predecessors $P$ is given by $\dfrac{3\cdot4^{3n+1}\cdot13+4^{3n+1}-7}{9}$

Therefore the set of sequences $2$ steps from any successor also has the accumulation point $-\frac{7}{9}$.

Then the elements $x$ whose indexes are $\equiv2\pmod3$ have leaast immediate predecessors $(2x-1)/3$. The first such predecessor of $5$ is $53$ and the general form for these is:

$4^{3n}\cdot53+\frac{4^{3n}-1}{3}$

Then the sequence of their predecessors is

$\dfrac{3\cdot2^{6n+1}\cdot53+2^{6n+1}-5}{9}$

So we have a further accumulation point of $-\frac59$

I am however still unclear whether and how the $\Bbb N$-indexed sequences of the form $4^n\cdot P+\frac{4^n-1}{3}$ where $P$ itself is the set $\left\{\dfrac{3\cdot4^{3m+1}\cdot13+4^{3m+1}-4}{9}:m\in\Bbb N\right\}$ may have further accumulation points over and above $-1/3$, $-5/9$ and $-7/9$.

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