Let $C$ be the Collatz map on the natural numbers, defined by:
$$C(n) :=
\begin{cases}
n/2 & \text{if} \;n \;\text{even} \\
(3n+1)/2 & \text{if} \;n \;\text{odd}
\end{cases}$$

The inverse of $C$ is:
$$C^{-1}(\{n\}) =
\begin{cases}
\{2n\} & \text{if} \;n \not \equiv 2 \pmod 3 \\
\{2n, (2n-1)/3\} & \text{if} \;n \equiv 2 \pmod 3
\end{cases}$$
Let $n$ be a natural number with $n \equiv 2 \pmod 3$. Let $C^{-k}$ be $C^{-1} \circ \cdots \circ C^{-1}$ ($k$ times).

Consider the cardinals $c_1(n,k):= \vert C^{-k}(\{2n\}) \vert $ and $c_2(n,k):= \vert C^{-k}(\{(2n-1)/3\}) \vert$.

**Question**: Is it true that $\forall n \equiv 2 \pmod 3$, $\exists k>0$ such that $c_1(n,k) \neq c_2(n,k)$?

*Remark*: It is checked for $n \le 10^8$.

Assuming the answer is no, let $n$ be a counter-example.

*Motivation-Question*: Is $n$ also a counter-example of the Collatz conjecture?

Assuming the answer is yes, let $\alpha$ be the map defined by:

$$\alpha (n) :=
\begin{cases}
1 & \text{if} \;n \not \equiv 2 \pmod 3 \\
1+\min\{k>0 \ \vert \ c_1(n,k) \neq c_2(n,k) \} & \text{if} \;n \equiv 2 \pmod 3
\end{cases}$$

By looking to the graph below, we get for example that $\alpha(8)=2$ and $\alpha(20)>3$.

A natural number $N$ is called a *champion* if $\forall n<N$ we have $\alpha(n)<\alpha(N)$.

Below is the list of the first champions (read as $[N,\alpha(N)]$):

```
[2, 4]
[20, 6]
[182, 8]
[1640, 10]
[14762, 12]
[132860, 14]
[1195742, 16]
[10761680, 18]
[96855122, 20]
```

*Observation*: Two consecutive champions $N$ and $N'$ in the finite list above satisfy the relations: $N'=9N+2$ and $\alpha(N')=\alpha(N)+2$.

*Bonus question*: Is the list of champions exactly $(N_r)_{r \ge 1}$ with $N_r = \frac{9^r-1}{4}$ and $\alpha(N_r) = 2r+2$?

*Remark*: I don't know whether $N_r = \frac{9^r-1}{4}$ is champion $ \forall r \ge 10$; but I have checked that $\alpha(N_r) = 2r+2$, $\forall r \le 26$, and we can confidently conjecture that it is true in general (the proof should be workable using the fact that the base-$9$ representation of $N_r$ is $222 \cdots 2$).