Group with 2 orbits on the nonnegative integers -- description of the orbits

Question

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:

1. 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 $\mathbb{N}_0$")?

2. 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.

Answer 1

These are just an attempt to reformulate everything in a more convenient way.

1. 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 $2$.

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.

3. 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.