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We know that there are two possibilities to disprove the Collatz Conjecture.

  • We find a nontrivial cycle.

  • We find a sequence that diverges to $\infty$

A non-constructive disproof is imaginable as well. I am particular interested in the cycles that have been ruled out.

I read the questions and answers about the Collatz conjecture in MSE. I would like to learn.

What is the longest cycle that has been ruled out until now? For example, is it possible to prove that there is no cycle of $10 ^ {1000} $ (or otherwise)?

I present an example for negative integer number that best describes the definition of the length of the cycle.

$$17 → −50 → −25 → −74 → −37 → −110 → −55 → −164 → −82 → −41 → −122 → −61 → −182 → −91 → −272 → −136 → −68 → −34 → −17$$

So, we have $\large 7$ odd-value cycle length.

But here, Collatz Conjecture doesn't include negative number.

Gottfried Helms
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Learner
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    You may want to look at this:https://en.wikipedia.org/wiki/Collatz_conjecture – Vinyl_cape_jawa Feb 23 '19 at 16:25
  • @Vinyl_coat_jawa I know . $k=68$? – Learner Feb 23 '19 at 16:26
  • What is the problem with the question? – Learner Feb 23 '19 at 16:40
  • @Peter wiki is the oldest source. – Learner Feb 23 '19 at 17:15
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    @Learner I highly doubt that there is a better result. Wikipedia updates such articles usually quite soon after a new discovery. – Peter Feb 23 '19 at 17:16
  • @Peter hmmm..Then this is an empty question. I have to delete..? – Learner Feb 23 '19 at 17:24
  • I edited the question. I would wait with the deletion. – Peter Feb 23 '19 at 17:38
  • @Peter Thank you so much.. – Learner Feb 23 '19 at 18:02
  • @Peter Because, I thought maybe a mathematician has disproved a new cycle that isn't on Wikipedia.. – Learner Feb 23 '19 at 18:06
  • See here: the minimum cycle length is at least 338,466,909. However, I think the limit should be recomputed and lifted, since according to the same site in the meantime convergence has been computer checked up to $10^{20}$ by the yoyo@home project. – Fabius Wiesner Feb 23 '19 at 21:01
  • @mbjoe - the Steiner/Simons/deWeger-key is here, that the number 338,466,909 is only relevant if that cycle is assumed to have more than, say, 68 local minima/maxima. For cycles with less than such number of local minima/maxima it is known that no overall length, even 10^10^10^10^10..., allows a cycle. – Gottfried Helms Feb 23 '19 at 23:23
  • Is my answer accepted? – rukhin Mar 04 '19 at 23:35
  • I asked in meta for redaction of the current title of the question due to honoring of answers which answer a different focus of the question. See my meta-question here: https://math.meta.stackexchange.com/q/29911 – Gottfried Helms Mar 06 '19 at 07:48
  • I've adapted the title of the question to match the given answers which are most welcomed/honored. I'm doing a new question with the original title and my answers (which I'm still goind to expand) and shall move my answers towards that question. See my discussion of this on meta (link in previous comment) – Gottfried Helms Mar 07 '19 at 05:42
  • Here is my new own question which re-asks what I understood of your initial question: https://math.stackexchange.com/q/3138566/1714 My answer here I shall delete soon. – Gottfried Helms Mar 07 '19 at 20:19

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The following link asserts an improvement on Eliahou’s lower bound on cycle lengths: p. 13 in the slides of T. Ian Martiny’s talk The author gives a lower bound for a cycle length to be 10, 439, 860, 591.

Update (redux): I read the slides too hastily. The result rules out ranges of cycle lengths: a cycle length must admit the representation $$ 630 138 877a + 10 439 860 591b + 103 768 467 013c $$ where $a,b,c$ are non-negative integers, $b>0$, and $ac=0$. (Thanks--again--for the clarification and keeping me honest, G. Helms)

rukhin
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    Unfortunately, the question of the OP is "what is the longest cycle that is ruled out?" due to the title of the question. I've recently found that the cycle with $N=127940101513462006853$ odd steps can be ruled out based on $a_{min}>87 \times 2^{60}$ (using a very simple method). This is so far the longest (except for the $m$-cycles with small $m$ which are disproven for any length). I write this although I think the OP means actually that what you've in fact answered... – Gottfried Helms Feb 27 '19 at 07:58
  • @rukhin then the numerical value $10,439,860,591$ should be recomputed and increased using the better yoyo@home convergence test maximum at $1.003 \times 10^{20} \approx 87 \times 2^{60}$. – Fabius Wiesner Feb 27 '19 at 15:53
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    @rukhin - just a minor doubt: you say "can include arbitrarily large integers". Do you really mean arbitrarily ? And with "integers" do you mean elements of a cycle (what I call $a_k$) or lengthes-of-nontrivial-cycles (what I call $N$)? The reason why I ask is that I think for both parameters there are upper bounds existent. – Gottfried Helms Mar 08 '19 at 08:05
  • Using the knowledge, that all elements $a_k$ of a cycle must be larger than $\chi = 87 \times 2^{60}$, a cycle with $N=208'576'659'753'891'832'997$ odd steps cannot have integer elements. (This is a refining of my previous comment and the largest $N$ for which the non-integrality can be proved based on that $\chi$ with that simple method (see at https://math.stackexchange.com/a/3138585/1714)) – Gottfried Helms Jan 03 '23 at 12:00
  • Recently, the user @DaBler has announced (https://math.stackexchange.com/questions/3314430) $\chi_{2022}=645 \cdot 2^{60}$. I've not seen a confirmation of this, But if this is true, then a cycle of length $N=1'546'343'459'710'263'321'923$ odd steps is disproved (with my simple method) and after a series of heuristics, it seems, this is also the longest disprovable cycle using the new $\chi_{2022}$ (this assumes that the structure is a "k-cycle" with $k>90$, because "k-cycles" with smaller $k$ are already disproved for any length) – Gottfried Helms Jan 08 '23 at 14:22
  • *(Correction of my previous comment)* There has been an error in the given $N$ in my previous comment with $\large {\chi }_{2020} $ $ =645 \cdot 2^{60}$ (source of error unknown at moment). Recalculation with my procedures moreover gave now an even larger $N$ where the general cycle (not a small m-cycle with $m<90$) can be disproved by my simple method $N=1'546'344'187'744'718'534'528$ . I leave this as a challenge now. – Gottfried Helms Jan 09 '23 at 13:18
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According to this and, since according to the same site convergence has been computer tested up to $1.003 \times 10^{20}$ by a yoyo@home project (see also this and click "Start" to order from highest to lower number checked), the minimum cycle length should be by now 9,283,867,937. This length is computed counting all sequence steps for both odd and even values, and with one step only from odd $x$ to the following $(3x+1)/2$.

The yoyo guys claim it is 17 billion (search here for "Collatz: Search finished"), however I think Eric Roosendaal is more trustworthy. Maybe this value refers to a double step for odd $x$: $x \to 3x+1 \to (3x+1)/2$.

Fabius Wiesner
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  • mbjoe - please see my extended answer giving some much larger numbers $N$ of odd steps for which the existence of general cycles is disproved. – Gottfried Helms Feb 25 '19 at 02:59
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    I think the OP should clarify the question. My answer would be fitted for a question like "what is the minimum length of a possible cycle?", because it excludes all cycles lengths below 9,283,867,937. But if one makes some assumption on properties of the cycle, or specific cycle length, there are bigger cycle lengths which have been ruled out, as @Gottfried Helms explained. – Fabius Wiesner Feb 25 '19 at 09:31
  • @mbjoe Example: We have $\large 7$ odd value cycle length for $−17 → −50 → −25 → −74 → −37 → −110 → −55 → −164 → −82 → −41 → −122 → −61 → −182 → −91 → −272 → −136 → −68 → −34 → −17$ – Learner Feb 25 '19 at 10:18
  • yes, that reformulation of the focus towards the smallest would be helpful... and I think is most likely also intended. – Gottfried Helms Feb 25 '19 at 11:25