Group with 2 orbits on the nonnegative integers -- description of the orbits Definition: Let $r(m)$ denote the residue class $r+m\mathbb{Z}$,
where $0 \leq r < m$. Given disjoint residue classes $r_1(m_1)$ and $r_2(m_2)$,
let the class transposition $\tau_{r_1(m_1),r_2(m_2)}$ be the permutation
of $\mathbb{Z}$ which interchanges $r_1+km_1$ and $r_2+km_2$ for every
$k \in \mathbb{Z}$ and which fixes everything else.
The group $G := \langle \tau_{0(2),1(2)}, \tau_{1(2),4(6)}, \tau_{0(3),4(6)}
\rangle$ has 3 orbits on $\mathbb{Z}$ -- one of them consists of the negative
integers, and the other two form a partition of the nonnegative integers.
The latter partition does look somewhat complicated.
There is computational evidence that the orbits have natural densities,
where $0^G$ seems to have density about 0.685, and $2^G$ correspondingly
seems to have density about 0.315. Also, both orbits seem to be approximately
uniformly distributed on the residue classes (mod $m$), for every positive
integer $m$. It seems that differences of consecutive members of an orbit
are always odd.

Questions:
  
  
*
  
*Is there an alternative description of the 2 orbits of the group $G$
  on $\mathbb{N}_0$ 
  (i.e. not just as "the orbits of $G$ on $\mathbb{N}_0$")?
  
*What are the natural densities of the orbits, if they exist?
  Are they rational, algebraic or transcendental?

Background
The Collatz conjecture is equivalent to the assertion that the group
$H := \langle \tau_{0(2),1(2)}, \tau_{1(2),2(4)}, \tau_{1(4),2(6)} \rangle$
acts transitively on $\mathbb{N}_0$.
Clearly the action of $G$ on $\mathbb{Z}$ is much easier to understand --
e.g. one can relatively easily count orbits. Nevertheless it is not
a trivial case, thus one might hope that its investigation might provide
some insights into the action of $H$ as well.
Data
There are tables of the numbers less than 10000 in any of the 2 orbits of $G$
on $\mathbb{N}_0$: orbit1_10000.txt, orbit2_10000.txt.
Larger versions with bound 1000000 rather than 10000 are available as well:
orbit1_1000000.txt, orbit2_1000000.txt.
 A: These are just an attempt to reformulate everything in a more convenient way.


*

*Let us define the following function on nonnegative integers (the definition can be simplified a bit, but I prefer to present it in this form):
$$
  f(n)=\begin{cases}
    4n, & x=6n\; \text{ or }\;x=6n+1;\cr
    4n+2, & x=6n+2\; \text{ or }\;x=6n+3;\cr
    2n, & x=6n+4\; \text{ or }\;x=6n+5.
  \end{cases}
$$
Then $f(x) < x$ for all $x>2$ and $x=1$, and $f(x)$ lies in the orbit of $x$. Moreover, it can be checked straightforwardly that for every two numbers interchanged by one of the generators, they have a common image under some iterations of $f$. This means that one orbit consists of all the numbers coming to $0$ after a sufficient number of iterations, and the other orbit consists of those coming to $2$. 

*Now let us check what happens in this process. Surely one may concentrate only on the even numbers, since $f(2n)=f(2n+1)$ is always even. Thus we change the variable by $y=x/2$ and introduce the function 
$$
  g(y)=\frac{f(2y)}2=\begin{cases}
    2n, & y=3n;\cr
    2n+1, & y=3n+1;\cr
    n, & y=3n+2\cr
  \end{cases} =
  \begin{cases}
    \lfloor y/3\rfloor, & 3\;\big|\; (y+1);\cr
    \lceil 2y/3\rceil, & \text{otherwise.}
  \end{cases}
$$
Considering the preimages under $g$, we see that each $y$ generates the numbers $3y+2$ and $\lfloor 3y/2\rfloor$ lying in the $y$'s orbit, and all the numbers are generated by such process from $0$ and $1$ (or from $1$ and $2$), thus partitioning into two orbits.

*Now, if we denote by $\mu(n)$ the density of, say, elements of the first orbit on $[2n,3n)$, then we obtain
$$
  \mu(3n)=\frac{\mu(n)+2\mu(2n)}3.
$$
It seems that this, together with the observation that $\mu(n)$ and $\mu(n+1)$ are close to each other, should suffice to see that the limiting density exists. 
