On a Collatz-like problem with two end cycles In an alternative version of Collatz problem, one iterates the function $f:\mathbb{N}\to\mathbb{N}$ defined by
$$f(n)=\begin{cases}n/3 &\mbox{if}\ n\equiv0\ (\mbox{mod}3)\\
2n+1&\mbox{if}\ n\equiv1\ (\mbox{mod}3)\\
2n+2&\mbox{if}\ n\equiv2\ (\mbox{mod}3)
\end{cases}$$
There are two cycles: $\{1,3\}$ and $\{2,6\}$. Starting from any $n$ in $\mathbb N$, one ends up in $\{1,3\}$ or in $\{2,6\}$.
To show this, use induction. The desired property holds for $n=1$ and $n=2$, obviously. If it holds for every $m<n$, then:


*

*if $n=3k$, then $f(n)=k$ and $k<3k=n$ hence, by the induction hypothesis, some number of iterations of $f$ brings $k$ to the cycle $\{1,3\}$ or to the cycle $\{2,6\}$; 

*if $n=3k+1$, then $f(n)=6k+3$, $f(f(n))=2k+1$ and $2k+1<3k+1=n$ hence, by the induction hypothesis, some number of iterations of $f$ brings $2k+1$ to the cycle $\{1,3\}$ or to the cycle $\{2,6\}$;

*if $n=3k+2$, then $f(n)=6k+6$, $f(f(n))=2k+2$ and $2k+2<3k+2=n$ hence, by the induction hypothesis, some number of iterations of $f$ brings $2k+2$ to the cycle $\{1,3\}$ or to the cycle $\{2,6\}$. 


Given any number $n$, let $c(n)=1$ if, starting at $n$ and iterating $f$, one ends in the $\{1,3\}$ cycle, and $c(n)=2$ if one ends in the $\{2,6\}$ cycle. The table of the first values of $c(n)$ is as follows:
\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}
n&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&20\\
\hline
c(n)&1&2&1&1&1&2&1&2&1&1&2&1&1&1&1&2&1&2&1&1\\
\end{array}

Is there any way to characterize numbers $n$ according to $c(n)$, the cycle they fall into? In other words, what can be proved about the set $\{n\in\mathbb N\mid c(n)=1\}$?

 A: Remark: this answer is not yet a complete answer, possibly a complete answer cannot be given

A good ansatz to define such a "character according to the cycle they fall into" is possibly to write down the reverse-iterates $C(a,A)$ beginning at root $r_1=1$ and at root $r_2=2$ and to create the two trees of numbers.     

Preliminaries: To make things a bit more compact, let's redefine our transformation-rule:     


*

*Assume $ a \in \Bbb N^+ \backslash 3 $ for example $a \in \{1,2,4,5,7,8,\ldots \} $        

*Define $\quad T(a) \overset{\text{def}}=   {2a+(a \% 3) \over 3^A} \quad$ where "$a\%b$" means the residue of integer division of $a$ by $b$ as in some programming languages       

*The inverse of $T(a)$ is then $\quad C(a,A)={3^Aa-2+(a \% 2) \over 2 } \quad $ where $A \in \Bbb N^+$ is now an additional free parameter


Tree generating from root $r_1=1$:
Let us consider that numbers, which, by one transformation fall down to $1$: of course, that are firstly all numbers $a_A(1) = 1 \cdot 3^A $ where $ A \in \Bbb N^+$ . But by our definition of the numbers $a$ that we want to look at we can omit that (trivial) considerations here.        
So let us look at that numbers $a_k$ which have the property that $T(a)=1$. This are all numbers $C(1,A)= { 1 \cdot3^A -2 + (1) \over 2} = { 3^A -1 \over 2} $ giving the vector of results $[a_{A}(1)]_{A>0}=[1,4,13,40,121,\ldots  ]$      
Next we use at each of those numbers and find all numbers $a_B(a_A(1)) = C(a_A(1),B)$. For $A=B=1$ this results in the value $1$ and is thus the "trivial cycle" on number $1$ and for $A=1, B>1$ we simply get back the same vector which we already have.
So we look at $a_B(a_A(1)) ; A \ge 2$ which means $a_B(4), a_B(13), a_B(40), \ldots $ and getting infinite vectors of numbers for each of them.
For $a_B(4)$ we get the vector $[a_B(4)]_{B\gt 0}=[5,17,53,161,\ldots]$
For $a_B(13)$ we get the vector $[a_B(13)]_{B\gt 0}=[19,58,175,\ldots]$
and so on.
I've made a little picture for the beginning of that tree, rooted on the first root $r_1=1$: 

Some properties are immediately visible:         


*

*The yellow boxes.
  a. They have an odd "parent"
("parent" means: all numbers of the box transform to that "parent" by the $T()$-transformation).
  Example: the box with first element $19$ has "parent" $13$ because $T(19)=13$ (and of course $T(13)=1=r_1$ so $19$ falls down to the first root).     
  b. The progression in the box is $a_{k+1}=3a_k+1$.
  c. The numbers are alternatively odd and even.
  d. The numbers are $1 \pmod 3$.          

*The green boxes.
  a. They have an even "parent".
Example: the box with first element $5$ has "parent" $4$ because $T(5)=4$ (and $T(4)=1=r_1$ ). 
  b. The progression in the box is $a_{k+1}=3a_k+2$.
  c. The numbers have all the same parity; they are either all odd or all even (possibly depending on the factor power-of-2 in their parent, didn't check this).
  d. The numbers are $2 \pmod 3$.          

Provisorical resume: 
This should be enough for the beginning. The analoguous tree for the root $r_2=2$ can be constructed analoguously. I've not yet a simpler formula for the accordances of the numbers in the trees with their roots.          

The trees of $a$ sorted as increasing sequences, according to their roots:
 for $r_1$ oeis A183213
 for $r_2$ oeis A178931
(note, that the OEIS-sequences have the numbers from the trees and all multiples by powers of 3) 
We can relate that tree to the explanation in the OEIS-sequences, which deal with the floor-function and the relation is intransparent at first glance. But we can note, that we can generate our tree using the floor-function as well.
Let $a_i$ denote the elements from the vector $[1,4,13,40,...]$ (which fall down to root $r_1=1$ by one step) such that we have $$a_i \underset{i \gt 0}= \left \lfloor {r_1 3^i-1\over 2} \right \rfloor $$.
Then we can recursively denote the elements of their "children" by a second index $j>0$ writing $$a_{i,j} \underset{i,j \gt 0} =\left \lfloor { a_i \cdot 3^j-1\over 2} \right \rfloor $$ and can immediately recurse arbitrarily deep by adding more indices: 
$$a_{i,j,k} \underset{i,j,k \gt 0} =\left \lfloor { a_{i,j} \cdot 3^k-1\over 2} \right \rfloor $$
$$a_{i,j,k,l} \underset{i,j,k,l \gt 0} =\left \lfloor { a_{i,j,k} \cdot 3^l-1\over 2} \right \rfloor $$
and so on.
For instance, the entry $22$ in the lower box in the above picture would have the index $a_{2,1,2}$ and this should be computed by
$$\begin{array}{} a_2 &= \left \lfloor {r_1 \cdot 3^2 -1 \over 2} \right \rfloor &= 4 \\ 
   a_{2,1} &= \left \lfloor  {4 \cdot 3^1-1 \over 2} \right \rfloor &= 5 \\
   a_{2,1,2} &= \left \lfloor  {5 \cdot 3^2-1 \over 2} \right \rfloor &= 22 \\
\end{array}$$
This scheme using the floor function seems to agree with some of the given formulae in the comments at the OEIS-sequences and their linked "eigen-sequences".

The tree from root $r_2$


Context-remark: it might be instructive to compare similar trees for the collatz-problem (only the graphic is a bit more primitive, especially the representation of the numbers in the digitsystem to base 2 (there) resp base 3 (here) . See here and click to "About numerical and graphical trees" (subpage without frame)
