Is it sufficient to prove collatz conjecture doing it for $3+6k, k \geq 0$? Thinking about this problem, I saw two interesting properties of Collatz graph. Firstly, if we consider that every even number $e$ can be represented (on a single way) as $e = o 2^n$, where $o$ is an odd number and $n$ is an integer ($n \geq 1$), then after $n$ steps our $e$ will generate $o$. It means that we "only" have to prove that every odd number is in Collatz graph.
The second point is: multiples of 3 are never generated by other odd numbers ($3n+1=3k$?). Also one can realize that every odd number that isn't multiple of 3 is generated (when we skip steps that generates even numbers) by infinitely many odd multiples of 3. So, if there are some number of the form $3+6k$ (an odd number multiple of 3) that isn't on Collatz graph, then his "function" ($3(3+6k)+1 \over 2^M$, where $M$ is the greatest integer for which previous division is integer) also isn't on the same graph. So we can treat odd numbers multiples of 3 as "odd endpoints" of our graph.
If all those propositions are correct, then Collatz conjecture can be, in some way, simplified. My question is: what is wrong with my propositions? If nothing is wrong, why I didn't find that information anywhere?

Since a cycle in your notation (using odd numbers only) cannot have a number divisible by 3, your proposal would suffice, if you could prove, that     
a) that the table of iterates eventually contains all odd integers       
b) and none of the trajectories diverges.               
Discussing divergence instead of cycles might have some special flavour, because we observe apparently divergent trajectories in all problem-modifications like ${mx+1\over 2^A}$ instead of ${3x+1\over 2^A}$. However, I've not seen any proof concerning the divergence of trajectories either, so...
