Hi everyone!
I found what seems to be a proof of the unicity of trajectory: what I mean by that is that if the collatz sequence of a given number x goes up-up-down-up-down… and the sequence of y goes the same, then x and y have to be equals.
A consequence of that is that there cannot be an exploding sequences that is ultimately periodic.
I would gladly explain my proof if someone is interested. It should be understandable by someone with a high school degree.
I’m very interested in that!
Great!
I’m not great with markdown, so I’ll try to keep the explanation not too calculatory while not hiding any tricky step…
I’m working with objects that I call tricots. A tricot is just a non-empty finite collection of functions from the set of integers to itself.
I note them as a two-columns matrix, each line corresponds to the affine factors of each function.
The height of the tricot is the number of those functions.
The tricot application is defined as follows: to get the image of x, do the euclidean division by the height. Take the function that corresponds to the module and apply it to the integer part of the quotient. I note it App(T).
More specifically I work on affine tricots, where all those functions also have to be affine, with height at least 2.
Under that formalism, the Collatz function would be noted as Tricot((1,0),(6,4)), and the reduced Collatz function would be Tricot((1,0),(3,2))
As an example, App(Tricot((1,0),(3,2))) (40) = 20
I also define the trajectory of a point on a Tricot by the (infinite) sequence of its modules (relative to the height). For example as 5 goes 5 → 16 → 8 → [4 → 2 → 1] under Tricot((1,0),(6,4)), its trajectory is 1,0,0,[0,0,1]. (here the brackets indicate that it repeats indefinitely).
Now my proof works for any affine tricot where all the linear coefficients are coprime with the height. It means it applies to 3n+1, but also 5n+1, and basically any kn+1 where k is odd.
So, let’s start with a lemma (that carries most of the job):
If
Proof: by induction on n.
From this lemma, to prove the theorem let’s consider x and y, two different integers.
Let’s get the difference, x-y, and decompose it in base h. Because of all the interesting properties of a decomposition in a base, this yields a sequence that contains a non-zero but finite amount of non-zero elements. If there is only one of those elements we are precisely in the case of the lemma, so we’re good. If there are more, we just need to apply it several times, once for each non-zero elements of the decomposition.
(I know that part seems fuzzy, but it is very hard to explain textually, while it is very clear on a picture. Just imagine you are decomposing the step between x and y into steps whose lengths are a power of h)
I guess that between my non-native english and my redaction some point may look unclear. I’d be glad to answer if you have any question.
If it makes sense can you apply this to cycles 3n-1 or 3n+c? They have cycles and you should be able to apply the same reasoning to them.
My theorem works for negative numbers, so 3n-1 has analog cycles to 3n+1.
As 3 is coprime with 2, it works for any 3n+c problem.
Anyway, it doesn’t refute the existence of cycles, it just says that you can’t have a periodic trajectory without having a cycle.
For example, is there in (reduced) 3n+15 a cycle that goes up-down repeating? To know that you just have to solve x = ((3x+15)/2)/2 and check.
x=((3x+15)/2)/2 => 4x=3x+15 => x=15
And indeed, 15 → 30 → 15
Is there one that goes up-down-down?
x=((3x+15)/2)/2/2 => 8x = 3x+15 => x = 3
indeed, 3 → 12 → 6 → 3
Is there one that goes up-down-down-down?
x=((3x+15)/2)/2/2/2 => 16x = 3x+15 => x = 15/13
There is a way to extend the definition to fraction to make this kind of cycle valid (it would go 15/13 → 120/13 → 60/13 → 30/13 → 15/13) but what matters is that it means there is no such cycle with integers.
it’s great. you have a self-contained proof that
(1) all numbers from 1 to 2^n have distinct initial parity sequences of length n.
and to reach the conclusion
(2) distinct start numbers p & q can’t share the same parity sequence,
i think you can just say, pick n such that 2^n is bigger than both p and q. then if p and q are distinct, then their parity sequences must diverge somewhere within the first n steps.
it’s also great observation that (2) implies:
(3) divergent trajectories have to be non-periodic.
i hadn’t realized that. it makes sense, because if n’s parity sequence is ab*, then n’s |a|th iterate will have the same parity sequence as its (|a|+|b|)th iterate … yet all iterates are distinct in a divergent trajectory.
Glad you understood me on the first try! I usually do maths in french and a lot of terms and notations have different meanings.
Once this is cleaned up, it’s a neat way to wrap some “standard” findings into one easy-to-read piece. It should work for any kn+1 map. Check out: https://math.colgate.edu/~integers/t8/t8.pdf
It doesn’t work for any kn+1 map. For example, let’s consider the 2n+1 map: f(2x)=x; f(2x+1)=4x+3.
Here all odd numbers have the same parity vector: [1111…]
I have it written in a cleaner way, but I cannot post pdfs here.
Really cool paper, thanks for sharing @stargazer07817 !
They extend the Collatz map from the regular integers to 2-adic integers (Z_2) and then define the function Q : Z_2 \to Z_2 which maps a 2-adic integer to it’s parity sequence (where the parity sequence is represented as a 2-adic integer (\dots a_2 a_1 a_0)_2). Q has a lot of cool properties: