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?

If you draw the graph of the orbit of the Collatz function through the odd subsequences then the numbers equivalent to $6k+3$ are the leaves of this graph.
This follows from the fact that no natural number of the form $2^{-p}(3x+1)$ with $x$ an odd number, is equivalent to $0\pmod3$
If all leaves converge and no leaves are fixed points (which they are not) then the entire graph converges.
