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This page from 2017 by Eric Roosendaal says that the yoyo@home project checked for convergence all numbers up to approx. 266. Is it still a valid record? I am aware of the ongoing BOINC project, but I cannot find how far they are.

The same question from 2014: For how many consecutive numbers Collatz conjecture was checked?

DaBler
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    This lists the record as of last month: http://www.ericr.nl/wondrous/progress.html – David G. Stork Aug 05 '19 at 17:51
  • https://boinc.thesonntags.com/collatz/highest_steps.php this was three clicks away from the BOINC homepage – Bananach Aug 05 '19 at 17:51
  • @DavidG.Stork If I understand this page correctly, then the project verified all numbers up to $509,040 \times 20 \times 10^{12}$, which corresponds to approx. $2^{63}$. This is less than $2^{66}$. How does this project relate to the original page by Eric Roosendaal claiming that checked all numbers below $2^{66}$? – DaBler Aug 06 '19 at 11:47
  • @Bananach This page just lists the numbers having the highest number of steps so far. It says nothing about how much the problem was verified. Or did I misunderstand something? – DaBler Aug 06 '19 at 11:54
  • This (in German) seems to be an article explaining the Collatz sub-project. Would anyone be able to translate it for me? – DaBler Aug 06 '19 at 16:15
  • Just open in Chrome browser and click translate... it gives a great translation. – David G. Stork Aug 06 '19 at 19:55
  • I believe this answer needs to be updated – klutt Oct 17 '22 at 22:16
  • @klutt Updated! – DaBler Oct 21 '22 at 08:32

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Since nobody provided an answer to my question, I will answer myself.

  • As of August 2019, I am aware of ongoing BOINC project [1](https://boinc.thesonntags.com/collatz/). By personal correspondence with Eric Roosendaal I found that this ongoing BOINC project is meant to disprove the Collatz conjecture by trying to find a counter-example. The project started off in the middle of nowhere, at $2^{71}$ apparently, without specifying any arguments why this was chosen or why this would be a sensible point to use. It looks like they have reached roughly $2^{72.3}$ or so. No info is given as to whether all numbers up to that limit have indeed be checked.

  • As of August 2019, I am also aware of another ongoing project [2] by Eric Roosendaal. All numbers up to $2^{60} \approx 10^{18}$ have been checked for convergence.

  • In 2017, the yoyo@home project [3] [4] checked for convergence all numbers up to $10^{20} \approx 2^{66.4}$.

  • The paper by Tomás Oliveira e Silva [5] from 2010 claims that the author verified the conjecture up to $2^{62.3} \approx 5.76 \times 10^{18}$. Source: Tomás Oliveira e Silva, "Empirical Verification of the 3x+1 and Related Conjectures." In "The Ultimate Challenge: The 3x+1 Problem," (edited by Jeffrey C. Lagarias), pp. 189-207, American Mathematical Society, 2010.

  • The page [6] by Tomás Oliveira e Silva states that, in 2009, they verified the conjecture up to $2^{62.3}$.

  • Earlier, in 2008, Tomás Oliveira e Silva [6] tested all numbers below $19\times 2^{58}$.

  • Much earlier, in 1992, Leavens and Vermeulen verified the convergence for all numbers below $5.6 \times 10^{13} \approx 2^{45.67}$. Source: Leavens, G. T. and Vermeulen, M. "3x+1 Search Programs." Comput. Math. Appl. 24, 79-99, 1992.

  • By the way, the paper [7] from 2019 confirms to me that the largest integer being (consecutively) verified is about $2^{60}$, referring to above sources.

When I put it all together, I get the upper bound $2^{66.4}$.

UPDATE:

From September 2019 to May 2020, my project [8] managed to verify the Collatz conjecture for all numbers below $2^{68}$. So the current upper bound is $2^{68}$.

UPDATE 2:

On December 10, 2021, my ongoing project managed to verify the convergence of all numbers below $2^{69}$. As of October 19, 2022, the current bound is $645 × 2^{60}$ (≈ $2^{69.33}$).

UPDATE 3:

As of July 9, 2023, my project was finally able to verify the validity of the Collatz conjecture for all numbers less than $2^{70}$.

UPDATE 4:

On November 3, 2023, my project verified the validity of the Collatz conjecture for all numbers less than $1.5 \times 2^{70}$ ($= 1536 \times 2^{60}$). This is the moment when the length of a non-trivial cycle raises to 355 504 839 929. See the article from Hercher, C. (2023). "There are no Collatz m-cycles with m <= 91" (PDF). Journal of Integer Sequences. 26 (3): Article 23.3.5.

DaBler
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  • Applause! Please do also a notification on mathoverflow.net – Gottfried Helms Jul 03 '20 at 12:36
  • @GottfriedHelms The notification is here. However, it does not seem to be welcome there. – DaBler Jul 03 '20 at 19:19
  • Verry sorry, @DaBler. Didn't consider my proposal well. Hope, the closing of your msg does not hurt you... Perhaps it would have been wiser to make a question "what is the current status of the numerical verification" and then give a self answer. But really - I didn't think about this enough. – Gottfried Helms Jul 03 '20 at 22:05
  • @DaBler That's a very clever way to aggregate steps, congrats! I might have an idea to hopefully add 1 to the exponent with only a slight change in the formula and without increasing the computations and I'm curious for an opinion or a test. If we take the sequences modulo 6 it's easy to see that every number reaches 4, therefore if we have N=6n+4 --> N'=6n'+4 and we focus on the change in the coefficient n-->n' we get a new formula that is similar enough to the original one. To verify the original up to N it should be enough to check the new up to N/2. Not enough characters for details... – Carsaxy Sep 16 '22 at 21:36
  • @Carsaxy Can you please explain more in the chat room? – DaBler Sep 17 '22 at 08:12
  • @DaBler sure, I just did. – Carsaxy Sep 17 '22 at 09:54
  • (Just for the record, from another comment of mine) "Recently, the user @DaBler has announced $\chi_{2022}=645 \cdot 2^{60}$. 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)" (see comment at https://math.stackexchange.com/a/3127885/) – Gottfried Helms Jan 08 '23 at 14:26
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    @GottfriedHelms As of January 6, 2023, my project was able to verify the validity of the Collatz conjecture for all numbers less than $660 × 2^{60}$ (≈ $2^{69.37}$). See here for current status. – DaBler Jan 08 '23 at 15:23
  • I've to correct an error (source unclear yet) to my prev. comment. The given $N$ is wrong, my simple method doesn't disprove "a cycle of that $N$" ; a recalculation plus more heuristics give even an improvement to $N=1'546'344'187'744'718'534'528$ (using $\chi_{2022}=645 \cdot 2^{60}$ – Gottfried Helms Jan 09 '23 at 14:37
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    It seems, that for $\chi_{2301}=660 \cdot 2^{60}$ the longest disprovable general cycle (not small-m "m-cycle") is $N=1'582'282'699'088'924'797'319$ (odd steps). However this is only heuristic so far, I see this now as challenge, to be improved... – Gottfried Helms Jan 09 '23 at 17:08
  • Using $\chi_{2301} $ I've got one larger $N$: $N=1'582'305'694'802'731'032'874$ cannot be the length of a cycle. After that about $41 e9$ candidates remain, which would need 22 hrs to test... – Gottfried Helms Jan 09 '23 at 21:15
  • @GottfriedHelms Is your simple method published somewhere? Where can I read how to extend it for new $\chi$? – DaBler Jan 11 '23 at 14:17
  • Dabler - I just remember I had recently posted an Q&A on this here : https://math.stackexchange.com/q/3138566/1714 The explanation in the answer there is much more extended than my comments here - the required shortness in comments spoils the precision of text unfortunately and is prone to needless small errors in notation :-( – Gottfried Helms Jan 11 '23 at 15:12
  • I'm currently trying to find the largest disprovable cycle-length on base of your $\chi = 645 \times 2^{60}$ Manually I found (15'th optimaziation) $N_{15}=1,546,344,201,699,447,217,164$ but in view of an upper bound for such an $N$ by $N_\chi = 1,546,344,201,779,196,168,856$ it is hopeless to perform the tests on all sequential $N$ in between. (...) – Gottfried Helms Jul 10 '23 at 19:27
  • (...) However, I'm just trying some new idea for an automatable method based on convgts of cont.frac. $\log_2(3)$, perhaps I can say more tomorrow. If I get this to work I'll put the maximal $N$-values for your $\chi_{22}=645 \times 2^{60}$ and the new $\chi_{23}=2^{70}$ here. – Gottfried Helms Jul 10 '23 at 19:31
  • By heuristics (mainly based on generalized convgts of cont frac) I conjecture now for $\chi_{2307}=2^{70}$ the largest length $N$ for which a cycle can be disproved with the simple method is now $N_{2307(1)}=2,454,971,259,794,877,266,896$ – Gottfried Helms Aug 23 '23 at 12:20
  • (Sorry @DaBler - still don't have a fully functional automatic determination; still need manually tuning of finding optimal guessings. The final open space for $N$ is still $6 \times 10^10$ and likely cannot be done sequentially in reasonable computing time.) – Gottfried Helms Aug 23 '23 at 12:24
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    @GottfriedHelms Look, I've updated my answer. – DaBler Nov 03 '23 at 11:53