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:
$?\mapsto\dfrac{-(2^n+3)}{9}\mapsto\dfrac{-1}{3}\mapsto 0$
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
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 $\bigcup_{k\ge 1} (X_k \cap \Bbb N) \stackrel{?}= \Bbb N$.