In the orbit of the function $3x+2^{\nu_2(x)}$ through "accumulation points" of the Collatz graph I have:

$?\mapsto\dfrac{-\langle2\rangle\cdot\{5,7\}}{9}\mapsto\dfrac{-\langle2\rangle}{3}\mapsto \langle \Bbb 2 \rangle\cdot0$

What's the full set of predecessors that goes where the question mark is?

$2^{\nu_2(x)}$ is the highest power of $2$ that divides $x$.

The above stands alone as a question asking for the preimage of the given sequence by the given function with no mention of the Collatz graph, but a little background for any who may be interested:

I understand the Collatz function defined by $f:\Bbb N\to\Bbb N::f(x)=3x+2^{\nu_2(x)}$ is continuous in $\Bbb Z_2$

Note that this function commutes with $2x$ so we can ignore multiplication by $2$ throughout this question. The set $\langle2\rangle$ represents multiplication by any power of $2$ of your choosing.

For any odd $y\in2\Bbb N-1$, the set of all integers $x$ satisfying $f(x)\in\{2^my:m\in\Bbb Z\}$ accumulates to $\dfrac{-\langle2\rangle}{3}$

And the set of all integers $x$ satisfying $f^2(x)=2^my$ accumulates under the 2-adic metric to $\dfrac{-\langle2\rangle\cdot\{5,7\}}{9}$

We can immediately see that the continuity rule supports this analysis since $\bar f^{-2}(y)$ is in the preimage of $\bar f^{-1}(y)$

Analysis of the sequences themselves is a little tough going. (More on that here).

But is there some obvious induction over the boundary points alone which indicates the full sequence of predecessors?

Update: we now have the more complete question:

$?\mapsto\dfrac{-(2^n+3)}{9}\mapsto\dfrac{-1}{3}\mapsto 0$

  • 1
    $\begingroup$ The set of $x$ with $f(f(x)) \in \langle 2\rangle$ has many more accumulation points, even if (what I assume you want) you only allow odd natural numbers $x$. E.g. all $\dfrac{2^{6m+1} -11}{9}$ are in there, and they have accumulation point $-11/9$. (You are looking at $\lbrace \dfrac{2^{m+n} -2^n -3}{9}:m,n \in \Bbb Z\rbrace$ and maybe want to restrict to those $n$ for which this is an odd integer for an unbounded set of $m$'s; for those you get $(-2^n-3)/9$ as accumulation point.) $\endgroup$ – Torsten Schoeneberg May 9 at 20:42
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    $\begingroup$ Actually, all $n \in \Bbb N$ work: $\dfrac{2^{6m+5} -5}{9}$ (accumulates at $-5/9$), $\dfrac{2^{6m+4} -7}{9}$ (accumulates at $-7/9$), $\dfrac{2^{6m+1} -11}{9}$ (accumulates at $-11/9$), $\dfrac{2^{6m} -19}{9}$ (accumulates at $-19/9$), $\dfrac{2^{6m+3} -35}{9}$ (accumulates at $-35/9$), $\dfrac{2^{6m+2} -67}{9}$ (accumulates at $-67/9$), $\dfrac{2^{6m+5} -131}{9}$ (accumulates at $-131/9$), $\dfrac{2^{6m+4} -259}{9}$ (accumulates at $-259/9$), $\dfrac{2^{6m+1} -515}{9}$ (accumulates at $-515/9$), and the pattern in the exponents repeats. $\endgroup$ – Torsten Schoeneberg May 9 at 22:18
  • $\begingroup$ @TorstenSchoeneberg thank-you I'll try to understand those. My first thought is surprise to see any integers not $\equiv 1\pmod4$ as these sequences contain only those as all but their first element but I immediately see my mistake there. $\endgroup$ – user334732 May 10 at 3:07
  • $\begingroup$ @TorstenSchoeneberg if all $\Bbb N$ work, and the function is continuous, then does this not mean the graph of the function connects all $\Bbb N$? $\endgroup$ – user334732 May 10 at 7:44
  • $\begingroup$ @TorstenSchoeneberg (p.s. you were right to interpret that what I'm really talking about here is $\Bbb N/\langle2\rangle$) where the quotient's taken multiplicatively. $\endgroup$ – user334732 May 10 at 10:42

As alluded to in the comments, and not hard to see, the sequence of predecessor sets of $\langle 2 \rangle = \lbrace 2^n: n \in \Bbb Z\rbrace$ for $f$ is

$$\dots \mapsto\dfrac{2^{l+m+n}-2^{l+m}- 2^l \cdot 3-9}{27}\cdot\langle 2 \rangle\mapsto \dfrac{2^{m+n}-2^m-3}{9}\cdot\langle 2 \rangle\mapsto \dfrac{2^n-1}{3}\cdot\langle 2 \rangle \mapsto \langle 2 \rangle$$

so the set of $k$-th predecessors is

$$\lbrace\dfrac{2^{n_1+...+n_k} -2^{n_2+...+n_k} \cdot 3 - \dots -2^{n_{k-1}+n_k}\cdot 3^{k-3}-2^{n_k}\cdot3^{k-2}-3^{k-1}}{3^k} : n_1, ..., n_k \in \Bbb Z\rbrace, $$

let's call this $X_k$. (I.e. we have $X_0 = \langle 2 \rangle, X_{k+1}:= f^{-1}(X_k)$.) The Collatz conjecture is equivalent to: every natural number can be written this way for some $k$ with all $n_i \ge 0$.

What you seem to be asking for are the sets of $2$-adic accumulation points of each $X_k$ or, more interestingly, $X_k \cap \Bbb N$. For $k=2$ I pointed out some elements of $cl(X_2 \cap \Bbb N)$ in the comments, and I am relatively sure (although I have not proven) those are all elements of $cl (X_2 \cap \Bbb N) \setminus cl(X_1 \cap \Bbb N)$ ($cl$ meaning $2$-adic closure in $\Bbb Z_2$).

Obviously, $cl(X_3)$ contains all $\lbrace \dfrac{-2^{l+m}-2^l\cdot 3-9}{27}:l,m \in \Bbb Z\rbrace$, and a first guess would be that $cl(X_3 \cap \Bbb N) \setminus cl(X_2 \cap \Bbb N)$ is that set restricted to $l,m \ge 1$ (for such $(l,m)$, the set of $r$ with $2^r \equiv 2^{l+m}+2^l\cdot 3+9$ mod $27$ is unbounded, so these are all contained in $cl(X_3 \cap \Bbb N)$; the other inclusion looks likely, but I have no rigorous proof for it.) For higher $k$, the expressions get more complicated as one deals with more variables, but it's quite possible there is a general argument that

$$cl(X_k \cap \Bbb N) \setminus cl(X_k \cap \Bbb N) \\ \\ \stackrel{?!}= \lbrace\dfrac{-2^{n_2+...+n_k} \cdot 3 - \dots -2^{n_{k-1}+n_k}\cdot 3^{k-3}-2^{n_k}\cdot3^{k-2}-3^{k-1}}{3^k} : n_2, ..., n_k \ge 1\rbrace.$$

I want to emphasize that, even if we have this or something similar which describes all $cl(X_k\cap \Bbb N)$, I see no reason why the knowledge of those would immediately bring us closer to a knowledge of the $X_k$ or the Collatz conjecture. For example it is quite possible there is some easy argument that $\bigcup_{k \ge 1} cl(X_k \cap \Bbb N)$ contains $\Bbb N$, but as far as I see, that tells us next to nothing about whether $\bigcup_{k\ge 1} (X_k \cap \Bbb N) \stackrel{?}= \Bbb N$.

  • $\begingroup$ Thanks. I have three comments. The original question asks which Collatz sequences, seen as equivalence classes of 5-rough natural numbers the same number of steps from $0$, converge to $0$ (rather than converge to $1$ as the conjecture asks). The point of this is that the points converging to $0$ appear to be the limit points of the conventional graph so sequential continuity should mean if we have all limit points connected by the graph, we have all 5-rough naturals covered (a sufficient set to prove the conjecture). $\endgroup$ – user334732 May 21 at 19:26
  • $\begingroup$ Next, on the question of $\cap \Bbb N$. The given function $3x+2^{\nu_2(x)}$ can wlog be assumed to commute with $3x$ in the sense $f(3x)=3f(x)$ simply by considering only $\Bbb N\setminus3\Bbb N$, because $3\Bbb N$ are the leaves of the graph, and it is not necessary to prove for the leaves, provided all non-leaves are proven. So the problem of $\cap \Bbb N$ can essentially be ignored, provided we remain within $\Bbb Z[\frac16]\geq0$, because every element is simply a representative for its 5-rough positive integer. $\endgroup$ – user334732 May 21 at 19:31
  • $\begingroup$ Thirdly, at the bottom I acknowledge $\dfrac{-(2^n+3)}{9}\mapsto\dfrac{-1}{3}\mapsto 0$ is the full set of limit points "of order 2" and this is not too tricky to prove. The points of order $3$ are not that hard. (But harder). Just take $-\frac59$, find the lowest power of $2$ such $f(y)=-\frac{2^m5}9$ has a solution, then by induction find the rest for $m+1\ldots$. In this case we get $-\frac{19}{27},-\frac{29}{27},-\frac{49}{27},\ldots$, then it is necessary to repeat the induction for all of the (infinitely many) predecessors of order $2$... $\endgroup$ – user334732 May 21 at 19:56
  • $\begingroup$ But I don't see this as the challenge. The challenge is either to find the rule that chooses the "lowest" power of $2$ for each sequence (and show the rule gives a dense or complete set), or (and I see this as more likely) to characterise the induction in such a way that one can step back and have a look at what we've got, in order for the contraction mapping to become apparent. $\endgroup$ – user334732 May 21 at 19:58
  • $\begingroup$ ...for example, I could quickly throw a hypothesis out there already for what it's worth... the limit points (in the 2-adic metric) of the graph of the orbit of $f(x)=3x+2^{\nu_2(x)}$ through the $5$-rough numbers, contract under the 3-adic metric. Once we have this, we can check it for arbitrary values of the conjecture rather than ones known to converge to $1$, and see if it holds universally. $\endgroup$ – user334732 May 21 at 20:03

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