I am amateur and don't have very good understanding of mathematical proving. My proof is so simple i don't believe noone thought of this before. But I am so blinded by the hope it is correct, that I can't find any errors in it.

Here are all the conclusions I made in the process:

  1. You can always make smaller odd number from even, so we only need to prove that every odd number works.
  2. Out of the two ways (3n+1 and n/2) only the n/2 can be used to create n ≡ 0 (mod 3) from n-1, where n-1 ≡ 0 (mod 3) has to hold true
  3. n ≡ 0 (mod 3) always creates n ≡ 1 (mod 3) (EDIT: every odd n ≡ 0 (mod 3))
  4. therefore once we abandon n ≡ 0 (mod 3), we can never find any other n ≡ 0 (mod 3) later in the sequence
  5. in other words we can't compose any n ≡ 0 (mod 3) from smaller number
  6. But, we can "compose" any n !≡ 0 (mod 3) (sorry, if I invented the notation. !≡ means the remainder after division is not 0) from smaller number.
    • reverse Collatz Example: 5 => 10 => either 3 or 20
    • The only way to reverse n ≡ 2 (mod 3) is n*2, after which every n-1 ≡ 1 (mod 3)
    • n-1 ≡ 1 (mod 3) can be reversed in both ways, but the (n-1)/3 yielding n-2 will always be smaller then n
  7. If n !≡ 0 (mod 3) can be "composed" from smaller number, we need only to prove the Collatz conjecture works for the smaller number.
  8. which can be either n !≡ 0 (mod 3) and can be "compsed" from smaller number again or be n ≡ 0 (mod 3)
  9. So the only thing I have to prove now is that the Collatz conjecture works for every odd n ≡ 0 (mod 3)
  10. But since every odd n ≡ 0 (mod 3) will yield n+1 ≡ 1 (mod 3) and I already concluded that once we abandon n ≡ 0 (mod 3), no later number will be n ≡ 0 (mod 3), the Collatz conjecture works also for n ≡ 0 (mod 3)

I've already looked at this for two days and can't find any errors and it drives me crazy. Is it cyclic or something? Am I depending on my own proves that also depend on my own proves? Did I make algebraic mistake? Hopefully, some of you will be willing to help.

P. S. Please be mindful of my experience and don't use very mathematical language if not neccessary. I've tried to search for this and didn't understand many papers on this matter.


  • "be composed of" means "is preceded by"


  • don't overlook EDIT in point 3
  • made the points ordered so we can reference them more easily.
  • added explanations
  • 2
    $\begingroup$ Your proof is so simple i believe someone eliminated this before. $\endgroup$ – Yves Daoust Jun 12 at 9:37
  • $\begingroup$ I believe this too, but i did not find it anywhere. If anyone found it, please post it here. $\endgroup$ – Otakar Beinhauer Jun 12 at 9:39
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    $\begingroup$ I think the mistake is when you say "$n_{-1}\equiv 1\pmod 3$ can be reversed both ways". If you start from an odd number, your suggested predecessor $\frac{n_{-1}-1}{3}$ is even and thus the Collatz map does not map it to $n_{-1}$. $\endgroup$ – Arnaud D. Jun 12 at 9:41
  • $\begingroup$ For example $7\equiv 1\pmod 3$ but none of the smaller numbers go through $7$ on their way to $1$. $\endgroup$ – Arnaud D. Jun 12 at 9:44
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    $\begingroup$ @OtakarBeinhauer: as the proof is wrong, nobody posted that. You don't post non-proofs ! $\endgroup$ – Yves Daoust Jun 12 at 9:46

EDIT after OP edited the question.

Your reasoning fails after point $6$ (included). Indeed, you talk about a certain number in the chain being "smaller" or "bigger" than another, and then mistakenly apply this rule to multiple elements of the chain (for instance in your point $6c$).

This is not strong enough, as $n/2$ might be followed by $3(n/2)+1$, which is always bigger than your original $n$. And then the same behaviour might happen again, etc.

So you have not proved that every $n ≡ 1\ ( mod \ 3)$ eventually becomes smaller than $n$.

Nor have you proved that there does not exist a cycle other than $4\to2\to1\to4$, a problem that your reasoning did not address.

  • 1
    $\begingroup$ thanks for the answer, I said "every ODD n ≡ 0 (mod 3) will yield n+1 ≡ 1 (mod 3)" $\endgroup$ – Otakar Beinhauer Jun 12 at 9:36
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    $\begingroup$ Then perhaps you might want to rephrase your question, as your 3rd point clearly says "n ≡ 0 (mod 3) always creates n ≡ 1 (mod 3)" $\endgroup$ – Klangen Jun 12 at 9:43
  • $\begingroup$ Oh, thanks. Edited. $\endgroup$ – Otakar Beinhauer Jun 12 at 9:56
  • $\begingroup$ Yes, point 6c is wrong. As for other cycle than 4→2→1→4, try imagine a cycle x->y->z where either x, y or z is the smallest of the cycle. The smallest has to be odd. If we could prove, that every odd number can be preceded by smaller number and the smaller number is proven to work, it would effectively break the other cycle. I think. $\endgroup$ – Otakar Beinhauer Jun 12 at 11:40
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    $\begingroup$ @OtakarBeinhauer If you like my answer feel free to accept/upvote it :) As for your question, that would work but a proof is currently out of reach with today's tools. As Lagarias (a specialist of the Collatz conjecture) put it, "Spending time on the Collatz conjecture is a fantastic time sink". $\endgroup$ – Klangen Jun 12 at 11:50

But, we can "compose" any $n \not≡ 0\,(\text{mod}\space 3)$ from smaller number.

Not true. How do we get $11$ from a smaller number? You've only proven that you don't need to consider odd cases of the form $3k+1$, and you've also proven that even $3k+2$ can be 'composed' from a smaller term. You still have $n ≡ 2\,(\text{mod}\space 3)$ (where $n$ is odd) to think about.

  • $\begingroup$ You get 11 from 7. It doesn't have to be immediate. Though Arnauld D. gave similar answer in my question comments. For example, you can't get 7 from smaller number (at least not that I know of). $\endgroup$ – Otakar Beinhauer Jun 12 at 9:54
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    $\begingroup$ Okay, but what do you mean by "it doesn't have to be immediate"? This opens up a whole new can of worms -- your proof does literally nothing to attempt to prove that you can make anything non-immediately. You can't just state a result without a proof. $\endgroup$ – auscrypt Jun 12 at 9:59
  • $\begingroup$ I meant, that for every 𝑛≡2(mod 3) you need exactly two steps to come to the smaller number. because when you doble n, you get 𝑛<sub>-1</sub>≡2(mod 3) you can always subtract one and divide by 3 (which is always smaller). Though the can of worms still exists - every odd 𝑛≡1(mod 3) (for example 7, 13, 19) as Arnaud D. stated above. $\endgroup$ – Otakar Beinhauer Jun 12 at 10:12

Alright, some people post answers to the comments and I'd like them to be seen. Also to have all the errors of my proof at one place. I will edit this as someone finds new errors.

  1. (point 6) Not every n !≡ 0 (mod 3) can be preceded by smaller elements (credit to Arnaud D.). More precisely every odd n !≡ 1 (mod 3) DOESN'T HAVE TO. Example (7, 13, 19 (19 will be eventually preceded number 5, but not after just two steps as i thought before)).

Change in my phrasing, because it might have caused confusion: - "composed of" -> "be preceded by"

  • 2
    $\begingroup$ In fact the numbers that can't be preceded by a smaller one are recorded in OEIS A061641. They are apparently called pure numbers and studied in this paper. $\endgroup$ – Arnaud D. Jun 12 at 12:52
  • $\begingroup$ Among other things, the paper shows that there is an infinite number of them that are congruent to $1$ modulo $3$ (they form the sequence OEIS A127633). $\endgroup$ – Arnaud D. Jun 12 at 12:59
  • $\begingroup$ I'm surprised, 13 is not among them. It has rather long chain preceding it. $\endgroup$ – Otakar Beinhauer Jun 12 at 13:08
  • $\begingroup$ If you start from $7$ (or $9$) you eventually get $11$, $17$ and $13$ (in that order). $\endgroup$ – Arnaud D. Jun 12 at 13:40

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