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File: 1598070039574.png (1.52 MB, 3696x3653, 3696:3653, All_Collatz_sequences_of_a….png) ImgOps Google

Do you believe the Collatz Conjecture is true?



I don't have a strong enough background in mathematics to have an opinion


Uh, I might be reading this wrong, but if the only operations performed in the sequence are halving the previous integer and tripling it while adding 1, then for it to reach 1 the sequence would also have to reach 0?  But you definitely can't reach 0 by halving a positive integer, and even more can't reach 0 by tripling and adding 1 to a positive integer.

So my take is no, it's clearly, immediately, demonstrably false.



I completely overlooked that you could also reach 0 if the current integer was ever 2, which is much easier.  Clearly I need more sleep.


2 -> 1 -> 4 -> 2 -> ...


It appears to be empirically true.  I would have to think on it awhile.


File: 1600156566228.png (243.38 KB, 828x442, 414:221, Sleep.png) ImgOps Google

I've been working on trying to prove it for a few years now. Though I'm only really making progress on half of the proof so far, that half being proving that cycles are impossible in a hailstone sequence unless the only odd number in the sequence is 1.

The other half that will definitely be a much greater challenge to prove is the idea that an infinite sequence that just climbs up and up forever is impossible.

The difference between the two is that cycles can be defined in a compact manner as a functional relationship between a number in the sequence and itself. Infinite sequences can't be defined compactly so easily, making it more difficult to find any sort of rigid statement about such a sequence to grapple with.

However, despite all of this, my hunch is that the Collatz Conjecture is absolutely true. Just to give a rough probabilistic "proof", we know that 3n+1 always results in an n that is coprime to the previous n. Assuming that the prime factorization is basically random after every 3n+1, there is always a probability that the prime factorization will be composed entirely of 2's, at which point it's over.

I could go on and on about all kinds of features and patterns in the hailstone sequence.

For example, a fun little relationship between the definitions of different odd numbers and how many divisions by two come before it:

(Where N is a whole number)
- Any n that can be represented by 6N+1 is always preceded by an even number of divisions by 2. (except at the beginning)
- Any n that can be represented by 6N+5 is always preceded by an odd number of divisions by 2. (except at the beginning)

(6N+3 numbers don't matter because they will always be the first and only number divisible by 3 in the sequence.)

It's reasonably simple to find out why this is true, but I think it's still a super fun feature.

Anyway, I think I've rambled enough about this.

TLDR: I think it's definitely true. I hope to prove it, too.

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