Is it sufficient to prove collatz conjecture doing it for $3+6k, k \geq 0$?

Question

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?

Answer 1

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...

Answer 2

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 $21\bmod24$…

Answer 3

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.

Answer 4

Perhaps a better expression of the argument pro your question:

If we look at the backwards-trajectory then we can say:

• any odd number $a_k$ (not divisible by $3$) has (even infinitely) many precedessors of the form $3m$ by $ a_{k-1} = {2^A \cdot a_k-1 \over 3}$ with some appropriate exponent $A$ such that $2^A a_k \equiv 1 \pmod 9 $. For instance $a_k = 7$, $a_{k-1} = {2^A \cdot 7-1 \over 3}$ then for $A=\{2,8,14,...\}$ gives precedessors $a_{k-1}$ which all are divisible by $3$.
• Even if any number $a_k$ is member of a cycle or of a divergent sequence, it has such precedessors.
• From this you can conclude in the reverse order: if you can adress (somehow) all odd numbers divisible by $3$ by some analytic argument, and look at their forward trajectories, then all other odd numbers shall occur in the infinite set of all possible trajectories.

Now if you are able to prove, that all numbers divisible by $3$ fall down towards $1$ then you have proved the same thing for all odd numbers (and then as well for all even numbers)

Answer 5

This paper proves your idea to be correct:
Kenneth M. Monks, The sufficiency of arithmetic progressions for the 3x+1 conjecture (2006)

The result of this paper is more general and directly proves your idea. Here is its abstract:

Abstract. Define T: Z⁺ → Z⁺ by T(x) = (3x+1)/2 if x is odd and T(x) = x/2 if x is even. The 3x+1 Conjecture states that the T-orbit of every positive integer contains 1. A set of positive integers is said to be sufficient if the T-orbit of every positive integer intersects the T-orbit of an element of that set. Thus to prove the 3x+1 Conjecture it suffices to prove it on some sufficient set. Andaloro proved that the sets 1+2ⁿℕ are sufficient for n ≤ 4 and asked if 1+2ⁿℕ is also sufficient for larger values of n. We answer this question in the affirmative by proving the stronger result that A+Bℕ is sufficient for any nonnegative integers A and B with B ≠ 0, i.e. every nonconstant arithmetic sequence forms a sufficient set. We then prove analagous results for the Divergent Orbits Conjecture and Nontrivial Cycles Conjecture.

"sufficiency sets" is the term to use when searching for this kind of result. Probably you were simply unlucky with the keywords when you tried to find it.