Let us define the two functions $F(n)$ and $G(n)$ as given:
$$ F(n) = \begin{cases} 3n+1 & \text{ and then dividing through by any powers of $2$} \end{cases} \\ G(n) = \begin{cases} 3n-1 & \text{ and then dividing through by any powers of $2$} \end{cases} $$
Assume for all $n$ we work with that $$G^{k}(n)=1 \qquad \text{ and } \qquad F^{k}(n)=1$$ for some $k$.
For example, $F^4(11) = 1$. Let us consider the set $\mathcal{O}(F) = \{n \in \mathbb{N} : F^k(n) = 1 \text{ for some k }\}$ and $\mathcal{O}(G)$ similarly.
Let's try to organize the sets as follows:
$$\mathcal{O}(F) \dashleftarrow\dashrightarrow \mathcal{O}(G)$$
For $$k=1$$ $1 \longleftrightarrow 1$
$5 \longleftrightarrow 11$
$21 \longleftrightarrow 43$
$\dots$ Notice these are all numbers which on their first iteration will go to $1$. For $$k=2$$
$3 \longleftrightarrow 15$
$227 \longleftrightarrow 911$
$909 \longleftrightarrow 3643$
$\dots$
For any $k$ we can write
$a_1 \longleftrightarrow b_1$
$a_2 \longleftrightarrow b_2$
$a_3 \longleftrightarrow b_3$
$\dots$
$a_i \longleftrightarrow b_i$
Question: If the set of $\mathcal{O}(F)$ is in one-to-one correspondence with the set $\mathcal{O}(G)$, can we say there is a counterexample for $F(n)$ ? Because, There is a counterexample for $G(n)$ which that, they are $(5,7,17)$ and $$\mathcal{O}(F) \dashleftarrow\dashrightarrow \mathcal{O}(G)$$
P.S: Please help me for improve my question.
Thank you!!
No; any two infinite sets of natural numbers are in bijection with each other. In particular, any infinite set of natural numbers is in bijection with $\mathbb{N}$, even though most (all but one, in fact) infinite sets of natural numbers don't contain every natural number. Just knowing that $\mathcal{O}(F)$ and $\mathcal{O}(G)$ are both infinite, and that $\mathcal{O}(G)\not=\mathbb{N}$, doesn't tell you anything about what $\mathcal{O}(F)$ could be.