# 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?

• There have been many papers on this topic, see wiki. – vadim123 Jun 5 '14 at 18:57
• Of course. However, I couldn't find exactly this result on many papers. – Hilario Fernandes Jun 5 '14 at 19:06
• What you have is normally called an "approach" rather than a "result". – vadim123 Jun 5 '14 at 19:22
• I handled this as a "sidenote" (though a medium interesting one) when I was fiddling with the collatz-problem and have some scribbles about it. But as long as I did not find any "explaining-power" in it I didn't include this, for instance, in my 2006-Collatz-treatize. However, it was somehow interesting for me in the comparing-approach with the 5x+1 variant, and a variant where I did a combined view in 3x+1 and 3x-1 , see an older discussion here in MSE math.stackexchange.com/questions/437261 – Gottfried Helms Jun 6 '14 at 10:34
• @GottfriedHelms The variant with $3x \pm 1$ is very interesting! But it seems to be more easy than Collatz conjecture... – Hilario Fernandes Jun 6 '14 at 14:57

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...

• But if one could prove that every number of the form $5+6x$ and $7+6x$ is reachable from a $3+6x$ number, a) becomes "that the table of iterates eventually contains all $3+6x$ numbers", right?. The second assertion really looks hard to prove, anyway. – Hilario Fernandes Jun 6 '14 at 15:04
• @Hilario: Well, you might do some examples of the trajectories in the $5x+1$-problem, forward and backward, to get familiar with some properties of (very likely) divergent trajectories, which do not cross the root-value $1$ – Gottfried Helms Jun 6 '14 at 15:34

Did you see this?

In a previous article, we reduced the unsolved problem of the convergence of Collatz sequences, to convergence of Collatz sequences of odd numbers, that are divisible by 3. In this article, we further reduce this set to odd numbers that are congruent to $21\bmod24$…