Talk:Collatz conjecture

Semi-protected edit request on 16 March 2024
BEFORE: Eliahou (1993) proved that the period p of any non-trivial cycle is of the form AFTER: Eliahou (1993) proved that the period p of the next candidate for a non-trivial cycle is of the form

Idk, im not a mathematician, but i read the paper cited and the p that is used here seems to be only the current (1993) best candidate for a loop above m=2**39, but below 2**48. As it is earlier mentioned, this region has already been investigated, so not only the original statements "any" false, but the big letter equation is also irrelevant. 89.223.151.22 (talk) 07:55, 16 March 2024 (UTC)
 * Red information icon with gradient background.svg Not done: From what I understand, the paper states that all non-trivial cycles (if any exist) have a cardinality of the form $$301994a + 17087915b + 85137581c$$. Your proposed change would imply that a non-trivial cycle that is not the first one to be found might not have a cardinality of that form, which contradicts the theorem proved by the paper. Saucy[talk – contribs] 10:43, 25 April 2024 (UTC)

Semi-protected edit request on 8 April 2024
Explaining Convergence

While a mathematical formula or explanation-based proof may yet be difficult, it is possible to explain the underlying mechanism responsible for convergence by transforming the problem statement as followsCollatz Conjecture - Explaining the Convergence: a) For an odd number N, (3N+1) is always an even number, therefore the next step will always be a division by 2. Both these steps can be considered as a single operation, i.e. (3N+1)/2.

b) Sequential multiplication steps (3N+1)/2 may be considered as a single operation until an even number is obtained.

c) Sequential division (N/2) may be considered as a single operation until an odd number is obtained.

With the help of these transformations, it can be observed that for a starting odd number series, (2i-1)*2n-1,

a) Multiplication steps result in the even series (2i-1)*3n-1

b) The resulting superset of even values is represented by either {(6i-4), i ∈ ℕ} or by {(6i-1)*3n-1 and (6i-5)*3n-1, i & n ∈ ℕ}

c) The next set of odd numbers obtained through division steps is represented by {(6i-5)*22n-1]/3 and (6i-1)*22n+1-1]/3, i & n ∈ ℕ}

While mathematical traceability between the starting odd number and the next odd number obtained in step (c) above is difficult to maintain, this approach still helps understand the underlying mechanism leading to convergence.

To explain this, one may begin from the other end of the problem and perform (2N-1)/3 operations on an even series (instead of (3N+1)/2 on a starting odd series). The starting even series E1=1*22n-1, results in a set of odd numbers which can be multiplied by 2n or 2n-1 if the odd number belongs to series {6i-5} or {6i-1} respectively. Sequential performance of these inverse operations results in a hierarchy of even series as shown in this exhibit Hierarchy of Even Series. Rakesh Vajpai (talk) 06:22, 8 April 2024 (UTC)
 * We cannot use this material without a published reliable source. Personal blogs are not reliable sources for this purpose. —David Eppstein (talk) 07:15, 8 April 2024 (UTC)
 * Thanks David. I am not a mathematician and have no idea what it takes for such articles to be published.  I may not even be interested in doing so as my focus is on making these aspects known to people who may be attempting to solve this problem.  I have no interest in claiming any credit for the insights.  Therefore, if it cannot be published here, I will understand and will leave it at that.  Thanks once again.  Cheers Rakesh Vajpai (talk) 12:49, 10 April 2024 (UTC)

Collatz function for some known megaprimes
The project math101.guru/en/category/collatz/ contains the logs of the Collatz function for some of the top known megaprimes and their vicinities (can not add link due to spam filter). The data currently cover 15 out of Top 17 known megaprimes (except for #7 and #8 discovered in 2023). There are also some interesting graphs with numerical data available that may be added to other graphs in the article. As far as I could say, such big numbers (>45 in total) have never been tested for the validity of the Collatz conjecture. Re2000 (talk) 07:06, 14 April 2024 (UTC)


 * You may be interested in this thread. --DaBler (talk) 14:22, 15 April 2024 (UTC)
 * The numbers they discuss there are infinitesimal compared with the tested numbers in the project above. E.g., they discuss 26,000,000 -1, whereas the largest known megaprime tested in the project was 282,589,933 − 1. There is nothing of value in the discussion you mention, though it is definitely asking a valid question - "what is the largest number tested for the validity of the Collatz conjecture?" Re2000 (talk) 14:33, 18 April 2024 (UTC)


 * They calculated 210,000,000 + 1 in 48 minutes. --DaBler (talk) 18:57, 18 April 2024 (UTC)
 * This is still 272,589,033 smaller than the largest tested number. 87.236.191.246 (talk) 06:06, 19 April 2024 (UTC)

Collatz 2nd loop proven impossible in 5 steps of easy logic.
SIMPLIFIED PROOF A LOOP IS IMPOSSIBLE IN THE 3x+1 PROBLEM USING LOGICAL DEDUCTION WITH EASY TO FOLLOW IMAGES AND VOICEOVER IN A 4 MINUTE VIDEO ON YOUTUBE https://youtu dot be/_uugKBK1-l0?si=jq0PO8q_kJLLdNdd

Sean A Gilligan April 2023 refined and revised most recently on 22 May 2024

Abstract Prove the function x×3+1/2^n when repeated will always go to 1. The origin of the function is attributed to being proposed by German mathematician Lothar Collatz in 1937 [1] Take any odd number multiply it by 3 and add one then divide by 2 until one arrives at the next odd number, repeat as many times as possible, so far every number goes to one and loops between 1421. The task is to prove it always will, 2 possible exceptions have been hypothesised, one where a sequence returns to the same value of x and loops forever or where it rises eternally higher towards infinity. In this paper I prove such a hypothesised loop is impossible.

Introduction In order to have a loop in the 3x+1 problem the value of all rises VR must equal the value of all falls VF between the 1st and final  X(capital X) so that VR-VF=0 We can deduce whether that is possible by working backwards and using only full values of x's in 5 simple steps of logical deduction.

Where any 3x+1=y y is always even.

1. the 1st rise we cancel 2X between X and y-1 (which is 3X leaving X from 0 to X and 0 from X to 3X) with 2X in the final fall between the final y(fy) and X. This leaves fy-3X in the final fall between fy and X.  y is always even x is always odd so fy-3x is always odd

2. So we know we must get a net rise (between all y to x to y-1's) between the 1st y and the final y-1(fy-1) to cancel to 0

3)a) Between X and X, where any x is greater than the previous x then y=2x so we cancel 1x in the descent from y to x with 1x in the rise from x to 3x this leaves a net rise (NR) of 1x from x to 3x (or y-1) (and one value of x between 0 and x). 1x is always odd.

3. b) Where x is less than the previous x y=x×2^n(n greater than 1) in the fall we can cancel 2x in the rise from x to 2x with 2x in the fall from y to x leave a net fall (NF)=y-3x always odd (plus 1x between 0 and x). Add up all NR to a total net rise TNR of all values between x to y's and y to x's, add up all NF(not including fy-3X) to a total net fall TNF. Subtract one from the other to leave an overall net ascent ONA (from all y to x to y-1's). If ONA=fy-3X we could cancel one from the other to leave a final value of 0 between all x's and y's.

4. This would need an odd number of x's between the starting X and yf because we need an odd TNR minus an even TNF or an even TNR minus an odd TNF to leave an odd number equal to fy-3X. So when one is cancelled from the other we are left with 0 for all values between x and y's between Xand X.

5. However here arises an inescapable inequality. For a loop to happen every value of x must be linked with no breaks in the chain so calculating from the lowest odd value of x in the loop as the starting X (Xn) we know the next odd value of x must be higher so the difference between X (Xn) and the next 3x or ((Xn+1)×3) must be an even number (Eg:between 31 and 47×3 we have 110) so if we cancel this value ((Xn+1)×3) from the final y instead of 3X we are still left with an odd number in the descent from the final y. However now we have an even number of odd x's between the 2nd y ((Xn+1)×3)+1 and the final y-1, this would be either an odd number of descents subtracted from an odd number of ascents or an even number of descents subtracted from an even number of ascents, which either would leave a net rise of an even number which can't cancel with the final y(fy) and X(n+1)×3 which is an odd number, so we can't get a net rise of 0.

So a 2nd loop is impossible in the 3x+1 problem, regardless of any value for X in any sequence within infinity.. This part of the conjecture is now ruled out as being impossible.

Having no address, no contacts in academic institutions and no bank account the peer review system is inaccessible. So it is in the public domain since May 2024. 83.137.6.162 (talk) 08:12, 8 June 2024 (UTC)


 * Sorry, Wikipedia is not the right place for Original Research.
 * Suggestion: team up with someone who does have access to the peer review system. Success! Uwappa (talk) 09:47, 8 June 2024 (UTC)
 * Thanks, that sounds easy and in theory should be, the comment no doubt from someone inside the academic system. That has been tried many times, as stated having no bank account means payment is not accessible, an endorsement requires someone willing to take time to read and think through of which there has been none found willing to do so without payment despite multiple requests. Even getting a meeting with busy people has been difficult, getting them to focus and read, listen without bias for 5 minutes impossible, requests for an endorsement for arxiv the same, people are clearly very busy and getting their time without money, bias or a "who knows who" is not as easy as it sounds, quite the opposite. Correction to your comment, it is not research it is absolute proof which you would see if you or anyone else would focus and read without bias for 5 minutes. It also begs the question... Why should a simple mathematical proof be held back from the public because the author(or his work) does not/cannot pay money to or become controlled by a hierarchical system? We saw this before with scientists like Galileo, Aristarchus and likely Tesla too. The lesson should be learned by now. If you don't understand the proof I suggest you focus and watch the video and follow the images for 5 minutes. The best we can hope is that the editors of the wiki page have scientific integrity enough to go with science and not bow down to a hierarchical monetary system which controls what or when truth reaches the public and discriminates in favour of richer people/nations, and against those with little or no money. Whatever such control is, it is not science.
 * If this is held back by such a hierarchical system that amounts to a colonial, anti science mindset, like it or not that's what it would be. Let's hope science is upheld over control. 185.114.163.227 (talk) 11:29, 8 June 2024 (UTC)
 * Sorry. Wikipedians will remove even the most brilliant absolute proof if not backed by a reliable secondary source. Wikipedia is not on the cutting edge, is not leading science. It trails science.
 * Your BREAKING PHYSICS youtube channel looks like a platform that suits you better. How about making more of the last 'A' (action) in AIDA_(marketing)? You do ask to share the video with anyone interested in mathematics and science. Why not take it one step further? Ask interested scientists to contact you so you can team up. Good luck! Uwappa (talk) 14:04, 8 June 2024 (UTC)
 * Thanks again, it's the easiest thing in the world to say something can't or won't be done and as for asking scientists to cooperate, you really have no idea how hard it is to get people to listen when you're not in the club. At the end of the day the essence of the proof is that 2 odd numbers added together cannot equate to an odd number. When something that simple needs someone with money to endorse it I guess it says a lot about the system and the people running it. It's sad for science, sad for humanity, I live in hope that common sense prevails somehow regardless. Thanks anyway for your interaction, even that is better than the wall of silence so often prevalent when the academic boat gets rocked. 83.137.6.166 (talk) 23:18, 8 June 2024 (UTC)

Please feel welcome to join 'this club'. No money or other people required to create a Wikipedia account. Still, face reality: Don't let these words deter you. Feel welcome. Join this club. Enjoy your own user page, your own talk page. Alternative: talk to local math students. They will understand you, adjudicate your proof and, once convinced, rub their hands with glee to take it up the hierarchy and rock a boat. Uwappa (talk) 12:31, 9 June 2024 (UTC)
 * 1) The top 25 of most popular Wikipedia pages contains very little, if any science. People mainly visit Wikipedia for sport results, news, movies, TV shows, artist fancruft, ...
 * 2) Original Research remains a big no no at Wikipedia. This is the wrong place to publish new proof, no matter how simple, true, brilliant, revolutionary...
 * 3) The academic use of Wikipedia is low. Wikipedia is not the right tool to rock an academic boat.

Different formulations as rewriting systems
Under the section "Other formulations of the conjecture", include some formulations of the Collatz conjecture as termination problems for rewriting systems. These formulations enable automated approaches to the Collatz conjecture by using methods that are developed for automatically proving termination of rewriting.

For this change, please add the following after the section "As a tag system" under "Other formulations of the conjecture".

As string rewriting systems
For this section, assume that the Collatz function is defined in the slightly modified form:

$$f(n) = \begin{cases} \frac{n}{2} &\text{if } n \equiv 0 \\[4px] \frac{3n+1}{2} & \text{if } n \equiv 1 \end{cases} \pmod{2}$$

The termination of the following string rewriting system is equivalent to the Collatz conjecture:

$$\begin{array}[t]{rcl} \mathtt{h} \mathtt{1} \mathtt{1} & \to & \mathtt{1} \mathtt{h} \end{array} \qquad \begin{array}{rcl} \mathtt{1} \mathtt{1} \mathtt{h} \mathtt{b} & \to & \mathtt{1} \mathtt{1} \mathtt{s} \mathtt{b} \\ \mathtt{1} \mathtt{s} & \to & \mathtt{s} \mathtt{1} \\ \mathtt{b} \mathtt{s} & \to & \mathtt{b} \mathtt{h} \end{array} \qquad \begin{array}{rcl} \mathtt{h} \mathtt{1} \mathtt{b} & \to & \mathtt{t} \mathtt{1} \mathtt{1} \mathtt{b} \\ \mathtt{1} \mathtt{t} & \to & \mathtt{t} \mathtt{1} \mathtt{1} \mathtt{1} \\ \mathtt{b} \mathtt{t} & \to & \mathtt{b} \mathtt{h} \end{array}$$

The above rewriting system represents each number in unary, and applications of its rules in any order to such a representation simulates the iterated applications of the Collatz function. Roughly speaking, this system simulates the execution of a Turing machine with cells that can be contracted or expanded, where $$\mathtt{h}$$, $$\mathtt{s}$$, $$\mathtt{t}$$ indicate the position of the head along with the state, $$\mathtt{b}$$ stands for the blank characters on the tape, and the symbol $$\mathtt{1}$$ is the unary digit.

An alternative string rewriting system that uses mixed binary–ternary representations of numbers is as follows:

$$\begin{array}{rcl} \mathtt{f} \triangleright & \to & \triangleright \\ \mathtt{t} \triangleright & \to & \mathtt{2} \triangleright \end{array} \qquad \begin{array}{rcl} \mathtt{f} \mathtt{0} & \to & \mathtt{0} \mathtt{f} \\ \mathtt{f} \mathtt{1} & \to & \mathtt{0} \mathtt{t} \\ \mathtt{f} \mathtt{2} & \to & \mathtt{1} \mathtt{f} \end{array} \qquad \begin{array}{rcl} \mathtt{t} \mathtt{0} & \to & \mathtt{1} \mathtt{t} \\ \mathtt{t} \mathtt{1} & \to & \mathtt{2} \mathtt{f} \\ \mathtt{t} \mathtt{2} & \to & \mathtt{2} \mathtt{t} \end{array} \qquad \begin{array}{rcl} \triangleleft \mathtt{0} & \to & \triangleleft \mathtt{t} \\ \triangleleft \mathtt{1} & \to & \triangleleft \mathtt{f} \mathtt{f} \\ \triangleleft \mathtt{2} & \to & \triangleleft \mathtt{f} \mathtt{t} \end{array}$$

In the above system, the symbols $$\mathtt{f}$$ and $$\mathtt{t}$$ represent the binary digits (0 and 1), while the symbols $$\mathtt{0}$$, $$\mathtt{1}$$, $$\mathtt{2}$$ represent the ternary digits. The symbols $$\triangleleft$$ and $$\triangleright$$ act as delimiters. The rules in the first column apply the Collatz function to a number represented in mixed base as long as the rightmost digit is binary. The remaining rules rewrite the representation (while preserving the value it represents) such that the rightmost digit is binary.

Rephrasing the Collatz conjecture as termination of string rewriting enables, at least in principle, using automated approaches to try to prove the conjecture thanks to the methods developed for automatically proving termination of rewriting.

46.196.80.203 (talk) 13:49, 17 June 2024 (UTC)

Not done: This seems a bit WP:TOOSOON to me. It's based on a paper that's less than a year old with few citations. PianoDan (talk) 22:32, 18 June 2024 (UTC)


 * The paper by Zantema is from 2005, so it is ~19 years old. The other paper by Yolcu, Aaronson, and Heule is actually ~3 years old; the conference version of their paper is here: https://doi.org/10.1007/978-3-030-79876-5_27. The latter work was also featured in 2021 in popular science magazines, for instance: https://www.technologyreview.com/2021/07/02/1027475/computers-ready-solve-this-notorious-math-problem. 46.196.80.203 (talk) 09:52, 24 June 2024 (UTC)
 * Red information icon with gradient background.svg Not done for now: please establish a consensus for this alteration before using the template. I still don't think it's warranted based on the level of citations it has received.  Given that there is disagreement from at least one editor that this material should be included in the article, the "Edit request template" procedure no longer applies, since it's only for non-controversial edits.  Instead, you can attempt to build a consensus on this page that the material should be included. PianoDan (talk) 16:29, 24 June 2024 (UTC)
 * I fail to see why the citation count of the source material should be a significant basis for the decision of whether to include some content on the page, but let's assume it should be: Arguably the most similar content currently visible on the page is Liesbeth De Mol's 2-tag system from the 2008 paper. According to Google Scholar, that paper has 36 citations since 2008. Zantema's paper has 64 citations since 2005, and YAH's (conference) paper has 12 citations since 2021. To my understanding, the edit decision should be made based on the significance of the content and whether it is interesting enough. I think those rewriting systems are somewhat more interesting than the tag system since they enable an entirely new approach (at least in principle) towards a proof of the conjecture. 46.196.80.203 (talk) 11:09, 25 June 2024 (UTC)

Binary number of trailing ones equals number of following increases
The chapter Collatz_conjecture Example:
 * 1) already states that the number of trailing binary zeroes equals the number of next repeated divisions (/2). That is obvious.
 * 2) but it does not yet state that the number of trailing binary ones equals the next number of increments, (*3+1)/2, as the number of trailing ones decreases by one with each increment.
 * 19: 10011 (2 next increments)
 * 29 : 11101 (1 next increment)
 * 44: 101100 (no immediate increment) The trailing 1100 indicates: after two decrements (22, 11), two increments will follow (17, 26).

The source (BOM - Beauty Of Mathematics) claims at https://www.youtube.com/watch?v=UZH3P8Ey_C8 Any mathematician known to have picked this up? Any secondary source known?
 * not to be a mathematician just working on it as a hobby, so a primary source.
 * this does not proof the conjecture, but it might lead to one.

The graph already shows the total number of binary ones in brackets.

Is there any graph available that shows the number of trailing ones? Uwappa (talk) 07:46, 25 June 2024 (UTC)


 * There does not seem to be a graph yet that shows how a Collatz sequence impacts trailing bits of binary numbers.
 * Created a graph for Collatz_conjecture showing how the Collatz conjecture 'nibbles' on trailing bits, binary ones and zeroes:
 * Collatz conjecture sequence for 27 with decreasing binary trailing bits.png Uwappa (talk) 13:28, 14 July 2024 (UTC)