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

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.

• 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"). Oct 8, 2018 at 19:05
• @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 $0$. Oct 8, 2018 at 19:29
• @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 $-\frac{2^p}{3}$ Oct 8, 2018 at 19:34
• @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 $\left\{-\frac{2^p}{3}:\leq\omega\right\}$ Oct 8, 2018 at 19:40
• @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. Oct 9, 2018 at 12:09

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 $$-7/9$$.