# Do all Collatz orbits begin with a 3 mod 6? [closed]

It can be shown that the output of the Collatz algorithm (3n + 1)/2^k where n is 1 mod 2 and k is a natural number is 1 or 5 mod 6. Does this mean that all Collatz orbits begin with 3 mod 6?

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• An orbit (or trajectory) starts with a given number which can be any positive integer. It is possible that we can reduce Collatz to a proof for all numbers of the form $6k+3$. I do not know which such reductions have been established. Sep 22 at 18:16
• And obviously, there are trajectories having no number of the form $6k+3$ , for example if we begin with a power of $2$. Sep 22 at 18:39
• Only the first element of a trajectory can be divisible by 3. In your style of transformation-rule (often called "syracuse style") indeed each iterate (except the initial value) is moreover of the form $6x \pm 1$. Sep 22 at 22:37
• @Peter - reading your first comment: just tried to put together the reduction-argument, just composed as draft, but I think it's working. Pls see my answer-box. Sep 22 at 23:12

This is to @Peter's hint towards the possibility of proof on a reduced set of numbers, namely $$6k+3$$. I think, this is a simple proof for such a reducibility. We want to show, that not only all numbers divisible by $$3$$ are in the branches of the tree of trajectories, but even that no number of the form $$6k \pm 1$$ exists which has no $$6j+3$$-precedessor on its branch of the tree.
The reverse operation of the Syracuse-transformation is $$a_1 = {2^A a_2 -1 \over 3}$$ where $$A$$ must be of the form such that the numerator becomes divisible by $$3$$. The following table shall show that all numbers $$6k \pm 1$$ must have a $$6j+3$$-precedessor applying some $$A \in \{1,2,3,4,...\}$$: $$\small \begin{array} {rl|rl} (6k+1) \cdot 2^1 - 1 &= 12k+1 & (6k+1) \cdot 2^2 - 1 &= 24k+3 \\ (6k+1) \cdot 2^3 - 1 &= 36k+7 & (6k+1) \cdot 2^4 - 1 &= 64k+15 \\ \vdots && \vdots \\ (6k-1) \cdot 2^1 - 1 &= 12k-3 &(6k-1) \cdot 2^2 - 1 &= 24k-5 \\ (6k-1) \cdot 2^3 - 1 &= 36k-9 &(6k-1) \cdot 2^4 - 1 &= 36k-17 \\ \vdots && \vdots \\ \end{array}$$ So every number of the form $$a_2=6k \pm1$$ has a predecessor of the form $$3j$$, so
every odd number not of the form $$6k+3$$ has a predecessor of the form $$6j+3$$ and this implies that there are no odd numbers $$6k+3$$ which are not in a (branch of a) trajectory with a leading number $$6j+3$$ and so proving Collatz only for all odd numbers divisible by $$3$$ suffices to prove Collatz for all odd numbers.
Comment: this does not interfere with the question of divergent trajectories: whether a trajectory is divergent or not does not affect the question of existence of branches having heads divisible by $$3$$.