![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 |
2 1
3 7
6 8
7 16
9 19
18 20
25 23
27 111
54 112
73 115
97 118
129 121
171 124
231 127
313 130
327 143
649 144
703 170
871 178
1161 181
2223 182
2463 208
2919 216
3711 237
6171 261
10971 267
13255 275
17647 278
23529 281
26623 307
34239 310
35655 323
52527 339
77031 350
106239 353
142587 374
156159 382
216367 385
230631 442
410011 448
511935 469
626331 508
837799 524
1117065 527
1501353 530
1723519 556
2298025 559
3064033 562
3542887 583
3732423 596
5649499 612
6649279 664
8400511 685
11200681 688
14934241 691
15733191 704
31466382 705
36791535 744
63728127 949
127456254 950
169941673 953
226588897 956
268549803 964
537099606 965
670617279 986
1341234558 987
1412987847 1000
1674652263 1008
2610744987 1050
4578853915 1087
4890328815 1131
9780657630 1132
12212032815 1153
12235060455 1184
13371194527 1210
17828259369 1213
31694683323 1219
63389366646 1220
75128138247 1228
133561134663 1234 — Preceding
unsigned comment added by
Frank Klemm (
talk •
contribs)
23:04, 13 August 2016 (UTC)
The cycle 0→0 is listed as 'trivial'. This is misleading since for any number other than zero, the Collatz-function cannot reach zero. There is no positive or negative natural number other than zero such that C(n)=0. It's a special case (for n=0), and should be listed or omitted as such. Kleuske ( talk) 11:34, 15 September 2016 (UTC)
I found on Eric Roosendaal's page "On the 3x+1 Problem" the notion of "residue" for the ratio (2^h(n)/3^t(n))/n where h, t count the halving/tripling steps, which appears to be maximal (1.253...) for 993. (It is natural to expect that n ~ 2^h/3^t.) Is this an original result from this mathematician, and/or have others considered this quantity? — MFH: Talk 21:34, 8 May 2018 (UTC)
There is almost no history of the problem in the article, when was it first brought into serious mathematics? Naraht ( talk) 13:54, 12 February 2017 (UTC)
I support this request, it is referred to Collatz, Ulam and Kakutani (and Thwaites) but on their pages (if they exist) there is no reference to a publication and/or date which could hint on when the given person has considered this problem or made a conjecture. — MFH: Talk 22:48, 25 October 2017 (UTC)
It is not clear what serious maths is. I agree that it seems trivial. — Preceding unsigned comment added by 86.1.37.70 ( talk) 14:39, 1 June 2018 (UTC)
The COLLATZ CONJECTURE IS SO EASY TO EXPLAIN!. IT'S LIKE 1+1=?! The the number is a power of two, it will do divide by 2 and so continued, and will reach two, and then 1. If it is even number but not a power of two, it will do divide by 2 until it reaches an odd number. Then for the odd numbers, we've got to multiply it (Collatz is a strange guy, multiply by 3 OR ANY NUMBER), you still get an odd. So and one and you got an even. And restart the paragraph. :) — Preceding unsigned comment added by 223.72.58.54 ( talk) 13:47, 9 July 2018 (UTC)
The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.
It's due to a very simple property of binary arithmetic:
So you see, you've been tricked into thinking you're doing 3 steps, but really it's just 2 steps obfuscated, then 1 real step. So in this system, only even numbers (numbers with no LSB set) are shifted to the right (so no data loss occurs), then they are immidietely shifted to the left where they came from. So both steps DO NOTHING. NOTHING. Then you add the number! This logically will ALWAYS increase the number.. thus logically making you play this game over-and-over until the lower significant bits alternate as such 10101010.. or 01010101.. Once this pattern occurs, you have a number with a multiple of 10 (ends in decimal 0), which means the system is in it's 2nd last phase before death reducing by multiples of 10 until it hits 10. 10 is 1010 and 5 is 0101, so once 10 is shifted to the right it becomes 5, 5 is "temp stored" in the system, 5 gets shifted to the left again and becomes 10, which 10+5 = 15, and 15 is one less from 16, and once the number hits 16, this system will enter into it's final phase, it will right-shift infinitely until the LSB is set. And this means the number will always be 1.
You're really just trying a simple "test" against the number, if the test fails, like Penn & Teller, you are tricked into thinking something happens (logical right-shift, logical-left-shift), then something really happens (one simple addition), and you've got a slightly higher number with a slightly better chance at being.... duh duh duh duh a 1010 (10 decimal) pattern binary number... which will always reduce in that system to a 0101 (5 decimal) binary pattern number due to the fact that addition is just a logical XOR. You'll notice both 10 and 5 have alternating "Chaser Lights" pattern. Every other bit is set. When they are XOR'd together all bits will be set. When all bits are set, and you add 1, you'll get a Power of Two number. Because a Power of Two number is always just 1 bit that is 0-to-many right-shifts away from being a number 1, this simple system will always produce a number 1 no matter the input. It's super basic binary math.
Right? — Preceding unsigned comment added by 24.78.157.241 ( talk) 09:19, 18 August 2018 (UTC)
Whoops! Hey Jasper, my original idea still works. bare with me just a minute bud! I took a good look at what is happening in this system and all it does is ping-pong between two states. I'll walk through n = 12 (Wikipedia example). 12, 6, 3.. so in binary math, halving a number is logical shift right. With an even number, the system enters a state where it will shift the bits to the right until there is a bit in the LSB. Once there is a bit in the LSB, we LOGICALLY have a odd number. Multiply that ODD number by 3 and we must get an ODD number result. The +1 in the system makes it always an EVEN number. And since it's ALWAYS an even number... and an EVEN number is what started off this system.. it recursively does the same thing again.. and again.. and again. But the system itself can only do 1 of 2 things.. oscillate away from 1 for eternity (it wont), stick on the number 1, in which case the system halts. — Preceding unsigned comment added by 24.78.157.241 ( talk) 09:33, 18 August 2018 (UTC)
First reference is under a graphic caption. — Preceding unsigned comment added by 2601:18C:CD7F:6064:F4C9:9B76:6690:9E8F ( talk) 09:29, 2 November 2018 (UTC)
I do not claim to be a programming or maths genius. However, since I particularly enjoy this problem, I wrote a short script to calculate the size of the progression for a given number. I tested it to make sure it was correct by using the values on the article under the longest progression for a starting number part (in Examples), and some of them came back with very different step counts from what the article says they should be. I ran each value multiple times to be sure it wasn't just me creating the discrepancy. Most of the numbers I came up with do match the ones given on the article.
For less than 10 trillion the article says the number of steps for 7,887,663,552,367 is 1563. My script returned only 511. For less than 100 trillion, the article says the number of steps for 80,867,137,596,217 is 1662. My script returned only 385. For less than 1 quadrillion, the article says the number of steps for 942,488,749,153,153 is 1862. My script returned only 262. For less than 10 quadrillion, the article says the number of steps for 7,579,309,213,675,935 is 1958. My script returned only 296. For less than 100 quadrillion, the article says the number of steps for 93,571,393,692,802,302 is 2091. My script returned only 235.
In order to be sure, I've run other numbers through to see if it's a calculation limit. My computer was able to calculate 2^222 and correctly inform me it took 222 steps. I was able to run 67399866667876599486667537717549076684092861056351431202759025623046739986666787659948666753771754907668409286105635143120275902562304 and it took 711 steps. It appears that the magnitude is not, in fact, a limitation or source of error.
I looked at the citation given for the final number on that list, which I presume is the source for other numbers there. The website seems to be suspect as far as academic credibility is concerned. Again, I do not claim to have any academic background here, but my results do differ dramatically, and seem to be equally valid.
If I have missed something major, my sincerest apologies.
For those who wish to test my script, you may find it here. It is in Python 3.7.3, the latest version as of the time of writing. — Preceding unsigned comment added by 172.77.247.24 ( talk) 04:38, 15 June 2019 (UTC)
c(n) = x=n; s=0; while(x>1, if(x%2==0, x=x/2, x=3*x+1); s++); s c(7887663552367)
https://www.researchgate.net/publication/296682209_Proof_of_the_Collatz_conjecture Chrisdecorte ( talk) 01:07, 21 July 2019 (UTC)
https://doi.org/10.1155/2019/6814378 — Preceding unsigned comment added by Kamalbarghout ( talk • contribs)
According to this MSE answer, the record is still valid as of August 2019. -- DaBler ( talk) 12:22, 19 August 2019 (UTC)
This forum is for discussion of the Wikipedia article.
The appropriate place to discuss this is the Usenet group sci.math and I've transferred it there. You can reach it via Google Groups with subject Collatz Conjecture proof? —Preceding undated comment added 23:29, 15 June 2008
If someone has new information to add on here related to the Collatz conjecture but it hasn't been published elsewhere, can it still be added? It's not false information, but there's no source to confirm it. A proof for the new information can be provided. I just wanted to check first. I don't want to add it only to have it taken down shortly after. Jeanlovecomputers ( talk) 23:18, 2 October 2019 (UTC)
According to the Collatz conjecture, we can use numbers not only natural and only positive. In 252 steps on this conjecture, we can get 1 from the number (-0.49). It can also be said that this conjectureis valid not only for natural numbers, but also for rational numbers. In 436 steps on this conjecture, we can get 1 from the number 2.48. We can get 1 by this conjecture of our number(n), but only if n>(-0.5). It is proof for (-0.49): Number n: -0.49 1 -0.470000 2 -0.410000 3 -0.230000 4 0.310000 5 1.930000 6 6.790000 7 21.370000 8 65.110000 9 196.330000 10 589.990000 11 1770.970000 12 5313.910000 13 15942.730000 14 47829.190000 15 143488.570000 16 430466.710000 17 1291401.130000 18 3874204.390000 19 11622614.170000 20 34867843.510000 21 104603531.530000 22 313810595.590000 23 941431787.770001 24 2824295364.310002 25 8472886093.930007 26 25418658282.790020 27 76255974849.370056 28 228767924549.110170 29 686303773648.330570 30 2058911320945.991700 31 6176733962838.974600 32 18530201888517.922000 33 55590605665554.766000 34 166771816996665.310000 35 500315450989996.940000 36 1500946352969991.700000 37 4502839058909976.000000 38 2251419529454988.000000 39 1125709764727494.000000 40 562854882363747.000000 41 1688564647091242.000000 42 844282323545621.000000 43 2532846970636864.000000 44 1266423485318432.000000 45 633211742659216.000000 46 316605871329608.000000 47 158302935664804.000000 48 79151467832402.000000 49 39575733916201.000000 50 118727201748604.000000 51 59363600874302.000000 52 29681800437151.000000 53 89045401311454.000000 54 44522700655727.000000 55 133568101967182.000000 56 66784050983591.000000 57 200352152950774.000000 58 100176076475387.000000 59 300528229426162.000000 60 150264114713081.000000 61 450792344139244.000000 62 225396172069622.000000 63 112698086034811.000000 64 338094258104434.000000 65 169047129052217.000000 66 507141387156652.000000 67 253570693578326.000000 68 126785346789163.000000 69 380356040367490.000000 70 190178020183745.000000 71 570534060551236.000000 72 285267030275618.000000 73 142633515137809.000000 74 427900545413428.000000 75 213950272706714.000000 76 106975136353357.000000 77 320925409060072.000000 78 160462704530036.000000 79 80231352265018.000000 80 40115676132509.000000 81 120347028397528.000000 82 60173514198764.000000 83 30086757099382.000000 84 15043378549691.000000 85 45130135649074.000000 86 22565067824537.000000 87 67695203473612.000000 88 33847601736806.000000 89 16923800868403.000000 90 50771402605210.000000 91 25385701302605.000000 92 76157103907816.000000 93 38078551953908.000000 94 19039275976954.000000 95 9519637988477.000000 96 28558913965432.000000 97 14279456982716.000000 98 7139728491358.000000 99 3569864245679.000000 100 10709592737038.000000 101 5354796368519.000000 102 16064389105558.000000 103 8032194552779.000000 104 24096583658338.000000 105 12048291829169.000000 106 36144875487508.000000 107 18072437743754.000000 108 9036218871877.000000 109 27108656615632.000000 110 13554328307816.000000 111 6777164153908.000000 112 3388582076954.000000 113 1694291038477.000000 114 5082873115432.000000 115 2541436557716.000000 116 1270718278858.000000 117 635359139429.000000 118 1906077418288.000000 119 953038709144.000000 120 476519354572.000000 121 238259677286.000000 122 119129838643.000000 123 357389515930.000000 124 178694757965.000000 125 536084273896.000000 126 268042136948.000000 127 134021068474.000000 128 67010534237.000000 129 201031602712.000000 130 100515801356.000000 131 50257900678.000000 132 25128950339.000000 133 75386851018.000000 134 37693425509.000000 135 113080276528.000000 136 56540138264.000000 137 28270069132.000000 138 14135034566.000000 139 7067517283.000000 140 21202551850.000000 141 10601275925.000000 142 31803827776.000000 143 15901913888.000000 144 7950956944.000000 145 3975478472.000000 146 1987739236.000000 147 993869618.000000 148 496934809.000000 149 1490804428.000000 150 745402214.000000 151 372701107.000000 152 1118103322.000000 153 559051661.000000 154 1677154984.000000 155 838577492.000000 156 419288746.000000 157 209644373.000000 158 628933120.000000 159 314466560.000000 160 157233280.000000 161 78616640.000000 162 39308320.000000 163 19654160.000000 164 9827080.000000 165 4913540.000000 166 2456770.000000 167 1228385.000000 168 3685156.000000 169 1842578.000000 170 921289.000000 171 2763868.000000 172 1381934.000000 173 690967.000000 174 2072902.000000 175 1036451.000000 176 3109354.000000 177 1554677.000000 178 4664032.000000 179 2332016.000000 180 1166008.000000 181 583004.000000 182 291502.000000 183 145751.000000 184 437254.000000 185 218627.000000 186 655882.000000 187 327941.000000 188 983824.000000 189 491912.000000 190 245956.000000 191 122978.000000 192 61489.000000 193 184468.000000 194 92234.000000 195 46117.000000 196 138352.000000 197 69176.000000 198 34588.000000 199 17294.000000 200 8647.000000 201 25942.000000 202 12971.000000 203 38914.000000 204 19457.000000 205 58372.000000 206 29186.000000 207 14593.000000 208 43780.000000 209 21890.000000 210 10945.000000 211 32836.000000 212 16418.000000 213 8209.000000 214 24628.000000 215 12314.000000 216 6157.000000 217 18472.000000 218 9236.000000 219 4618.000000 220 2309.000000 221 6928.000000 222 3464.000000 223 1732.000000 224 866.000000 225 433.000000 226 1300.000000 227 650.000000 228 325.000000 229 976.000000 230 488.000000 231 244.000000 232 122.000000 233 61.000000 234 184.000000 235 92.000000 236 46.000000 237 23.000000 238 70.000000 239 35.000000 240 106.000000 241 53.000000 242 160.000000 243 80.000000 244 40.000000 245 20.000000 246 10.000000 247 5.000000 248 16.000000 249 8.000000 250 4.000000 251 2.000000 252 1.000000 Nikita fly ( talk) 20:18, 3 October 2019 (UTC) Chizhov Nikita 03.10.2019
Does someone have reliable sources as to this problem's relevance to important theoretical mathematical areas that deal with general properties of systems, such as number theory? Various users over at Reddit and StackExchange have discussed the relation between the prime factorizations of n and n+1 (for odd n the next number is 3n+1, so how does n's factorization affect 3n+1's factorization?), and mentioned the Riemann hypothesis in connection to this. Basically problems that help us advance our general understanding of mathematics or formulate new important branches of mathematics. 37KZ ( talk) 23:34, 10 November 2019 (UTC)
The article here reports that they have verified the validity of this conjecture for all numbers below 268. -- DaBler ( talk) 08:26, 2 July 2020 (UTC)
In order to improve the article
I ask you to consider the possibility of inserting into the article
--Collatz conjecture--
--in the Examples section
--number 27
my explanation of the behavior of the number 27
that is given on the page
/info/en/?search=User_talk:Eduard_Dyachenko
and the article
https://zenodo.org/record/4013334#.X1DhOcgzbIU
Dear editors, please do not delete as this is a discussion of a specific section of the article.
Eduard Dyachenko (
talk)
18:59, 27 March 2020 (UTC)E.Dyachenko(dyachenko.eduard@gmail.com)
You can check the table with number 27 manually, it is not at all difficult
Eduard Dyachenko (
talk)
20:18, 27 March 2020 (UTC)E.Dyachenko(dyachenko.eduard@gmail.com)
Trajectory in the form of a graph yes, but without explanation and calculations of its origin
E.Dyachenko(dyachenko.eduard@gmail.com)
without explanation and calculations of its originThe calculational rule for the trajectory is simple arithmetic: divide even numbers by 2, triple odd numbers and add 1. If you believe there is some deeper analysis that goes along with this particular trajectory, then we're back to the problem of WP:OR. -- JBL ( talk) 21:20, 27 March 2020 (UTC)
the table shows that the change in the peaks in the graph coincides with the change in the oddness from the form 4k+3 to 4k+1
which corresponds to a decrease in the "length of the number" in numeral system 4:3
E.Dyachenko(dyachenko.eduard@gmail.com) — Preceding unsigned comment added by Eduard Dyachenko ( talk • contribs) 21:43, 27 March 2020 (UTC)
which corresponds ...and that's the part that violates WP:OR. -- JBL ( talk) 00:00, 28 March 2020 (UTC)
I believe the table, posted along with the graph, is a worthy visualization how graph behaves.
This is a supporting element providing explanation to the picks of the graph.
E.Dyachenko (dyachenko.eduard@gmail.com)
— Preceding
unsigned comment added by
Eduard Dyachenko (
talk •
contribs)
10:39, 28 March 2020 (UTC)
WP:NOTFORUM |
---|
The following discussion has been closed. Please do not modify it. |
We let t be the number of trials it takes the Collatz sequence (orbit) of odd integers to converge. We can determine k by the largest positive even integer, , in the Collatz sequence. . Therefore, as ... That value, 0, represents the probability that the Collatz sequence of odd integers does not converge to one. Relevant Reference Link: A Brief Analysis of the Collatz Conjecture Therefore, we conclude the Collatz conjecture is true! — Preceding unsigned comment added by 96.76.246.142 ( talk) 16:10, 15 September 2020 (UTC) |
In our recently pusblished article we analyze the divisions by two that are performed within Collatz sequences: https://doi.org/10.18052/www.scipress.com/IJPMS.21.1
Aside from classical mathematical methods, we use techniques of data science. Based on the analysis of 10,000 sequences we show that the number of divisions by two lies within clear boundaries. The paper covers the Collatz problem both in it's original form as well as in the generalized variant , where .
Building on the results, we develop and prove the following equation, which calculates the maximum possible number of divisions by two for any given Collatz sequence of a certain length:
The parameter represents the count of odd numbers in the sequence and stands for the first odd number. Whenever the maximum is reached, a sequence leads to the result one, as conjectured by Lothar Collatz. Furthermore, we show how many divisions by two are required for a cycle of a specific length:
Are these findings of interest for Wikipedia? We are looking forward to the discussion -- C4ristian ( talk) 16:51, 18 November 2020 (UTC)
Thank you for your replies so far. We still hope to initiate a scientific debate, as we experienced in other open communities. There is no doubt that the Wikipedia policies are important. However, in many cases they leave room for interpretation. Let me give you two examples:
Not to be misunderstood: My main point is not the incorporation of our idea in the article. If there are mathematically or didactically sound reasons not to mention it, that is perfectly fine for us. Our goal is to contribute to the solution of the Collatz problem. For this reason, we have not edited the article directly, but have instead started this discussion. That is why we still hope to be treated fairly and get at least some scientific feedback -- C4ristian ( talk) 10:20, 21 November 2020 (UTC)
Self-published expert sources may be considered reliable when produced by an established expert on the subject matter, whose work in the relevant field has previously been published by reliable, independent publications.
I noticed that there are several (sub)sections in this article that do not cite any sources, such as the recently-added " Connection to a sequence of different origin" section by Verihärö. Although I appreciate these insights into the conjecture, I am wondering whether this falls under original research? ― JochemvanHees ( talk) 10:17, 25 December 2020 (UTC)
The formulation of Collatz conjecture as a binary relation "→" is frequently used, e.g. in the section "Extensions to larger domains," subsection "Iterating on all integers" of this article. Also, it is relevant fact that slight modification of the formulation is provable, the proof is simple and can be used for educational purposes as argues an article recently published in peer-reviewed online journal Academia Letters (the article is accompanied by two public peer reviews, one of them from a professor of mathematics Shaun V. Ault).
So, I propose to add the subsection "As a binary relation" in the section "Other formulations of the conjecture" with this short text:
For any natural n and any odd number x, Collatz conjecture is equivalent to the statement about the transitive reflexive binary relation which satisfies conditions and .
Similar formulation represents weaker version of Collatz conjecture: it is proved [1] that for the transitive reflexive symmetric binary relation which satisfies conditions and .
93.72.108.33 ( talk) 08:57, 28 July 2021 (UTC)
References
Similar formulation represents weaker version of Collatz conjectureNo, it represents a completely different question that is connected to the Collatz conjecture only superficially. It has no business being in this article. Academia Letters is not a real journal, it is a self-publishing platform with a slight patina of peer review; I hope that you didn't pay them too much money. -- JBL ( talk) 11:14, 28 July 2021 (UTC)
I attempted to add a sentence about a 120 million yen prize being offered for a proof of the conjecture, but it was reverted because I chose prnewswire.com for the citation. I could have chosen mathprize.net, but that is the primary source and I thought secondary sources were preferred. In any case, such a large prize seems worthy of mention. 2601:C6:4100:F980:CD93:B16:285F:4B15 ( talk) 04:01, 31 July 2021 (UTC)
"The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined as follows: start with any positive integer n. Then each term is obtained from the previous term as follows: if the previous term is even, the next term is one half of the previous term. If the previous term is odd, the next term is 3 times the previous term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1."
Surely you mean "current term" and not "previous term". Basing the formula for the next term on the previous term rather than the current term would be totally weird. Like Crunchy Bananas weird. 2601:8C3:8201:51D0:9C7C:1257:CBA8:1B9B ( talk) 06:54, 3 January 2021 (UTC)
start with any positive integer n. Then each new term is obtained from the previous term as follows: if the previous term is even, the
nextnew term is one half of the previous term. If the previous term is odd, thenextnew term is 3 times the previous term plus 1.
18 August 2021
It seems to me that the consensus here is that the statement of the conjecture can be improved. I agreed, and on 12 August 2021 I edited the statement in the introduction, since at that time no one had taken any action. I have been editing the wikipedia anonymously (always) from public ip addresses for 20 years now. I follow the old advise "be bold". That is I just do it. There seem to be some editors that take pride of ownership and strive to control every little detail. It may be that David Eppstein is one such. I hope not, but if he is, I'll give up after my edit today. I have no time for edit wars.
Apparently, judging from his reversion summary:
("The sequence terminates when ... The sequence diverges to infinity.": really??)
he took issue with my clarification to include the truth that it is, in fact a CONJECTURE. No one knows that there is no number whose sequence will never arrive at unity or a previous value in the sequence. That is a possible outcome. In fact there are some results that show that if there is an independent loop (other than the trivial 4-2-1 loop) it must be at least 180,000,000,000 elements long.
Or perhaps he took issue with the idea that an infinite divergence is not strictly a termination. Fair enough, so today I reworded things slightly to appease the Gods. Might he not have simply done that himself? That is the point of a wiki. Why waste energy requiring this elaborate justification?
My edit is in fact a substantial clarification of the sloppy language that has prevailed, to say nothing of the awkward style ("following: .... following:": really??).
See Talk:Collatz conjecture/Archive 2#Attribution for credit to Collatz and 1937 date and HSM SE. Ain92 ( talk) 20:15, 27 August 2021 (UTC)
I think the series of lines
less than 10 is 9, which has 19 steps, less than 100 is 97, which has 118 steps, less than 1000 is 871, which has 178 steps,"
would be better in a table. Bubba73 You talkin' to me? 03:36, 29 August 2021 (UTC)
The description of this book on amazon seems to imply that a solution has been found. Does anyone know anything about this? Is this guy lying? https://www.amazon.com/Collatz-Conjecture-Solutions-Leong-Ying/dp/198075991X/ref=sr_1_3?crid=36AOENBGWHIL&dchild=1&keywords=collatz+conjecture&qid=1634074729&sr=8-3 — Preceding unsigned comment added by Skysong263 ( talk • contribs) 21:42, 12 October 2021 (UTC)
![]() | This help request has been answered. If you need more help, you can , contact the responding user(s) directly on their user talk page, or consider visiting the Teahouse. |
Is it not misleading to have the diagram at the top of the mobile version of the page skip the even numbers?02:38, 24 October 2021 (UTC) 147.134.101.106 ( talk)
Lawrencekhoo is advancing some rewrites of the lede, which have not been immediately accepted as useful by other editors. Lawrence, perhaps you want to make a case for your edits here on the talk page? Russ Woodroofe ( talk) 12:20, 1 December 2021 (UTC)
The unreliable citation is not a reason to remove this topic altogether; I am currently occupied but could someone please find some CREDIBLE sources and add back the good faith edits by Fmwithstuff Cassie Schebel, almost a savant. <3 ( talk) 23:51, 16 March 2022 (UTC)
Be careful, in table: -1 gives -1 and never -2. 2A04:CEC0:1192:22A0:EC1D:F521:5A80:D88B ( talk) 13:28, 28 March 2022 (UTC)
(1) No. (2). Not here. |
---|
The following discussion has been closed. Please do not modify it. |
When we look at the Collatz Conjecture, if a randomly chosen positive integer is even, it is divided by 2, and if the result is odd, the number is multiplied by 3 and added by 1; If the result is an even number, it is seen that dividing by 2 is continued. In this case, the assumption is made that the number will eventually reach 1, resulting in an infinite loop of 1-4-2-1. As a result of the investigation:
As a result of the experimental study:
- It has been determined that the highest total number of iterations is 350 and the number value is 77,031. - When we look at the iteration results of a randomly selected positive integer, it is seen that a maximum of 18 consecutive even numbers come in a row and this number value is 87,381. Slmyildirim79 ( talk) 07:57, 24 April 2022 (UTC) As a result: - It is seen that with the increase in the number values of the digits, the total number of iterations will also increase, and naturally, the probability of successive even numbers will increase. - Looking at the iteration results, it is seen that the probability of the number converging to 1 is higher than the probability of diverging to 1 because of the increase in consecutive even numbers. For the randomly selected positive integer values to increase, a "odd-odd-odd-odd..." loop must occur one after the other, which is impossible. Naturally, all positive integers will reach 1 after iterations. Therefore, the Collatz conjecture is valid for all positive integers. |
Is ac=0 an error here? If not, just what does that mean? John D. Goulden ( talk) 21:51, 1 December 2021 (UTC)
The section
A similar reasoning that accounts for the recent verification of the conjecture up to 2^68 leads to the improved lower bound 114208327604 (or 186265759595 without the "shortcut"). This lower bound is consistent with the above result, since 114208327604 = 17087915 × 361 + 85137581 × 1269.
is very unclear. What is the "similar reasoning?" What is the "shortcut"? Which citation supports this claim? 65.128.180.152 ( talk) 06:05, 15 June 2022 (UTC)
prompted
Illegibility is an issue. The boxes could be doubled in size.
Arrows might get enhanced by rounded corners or intermittent arrowheads.
Acyclicity is felt in your fingers following this graph but contested, after all it is still a mere conjecture.
How should the graph be improved and where ought it be prominently placed? IM Serious ( talk) 01:50, 24 January 2023 (UTC)
the snailwound walk from 25, or the overshooters 15, 23, 27. I also display the periodicities in rising or falling) is also extremely cryptic. -- JBL ( talk) 18:33, 24 January 2023 (UTC)
Can someone add this into the article if it is considered as relevant:
The behaviour of up and down of a Collatz sequence can be explained by using the binary system and n+(n+1)/2 instead of (3n+1)/2
Example 255 (with mod 4 = 3):
n = 11111111 (255) mod 4 = 3 +(n+1)/2 = 1000000 (128) number to add —————————— Result = 101111111 (383) mod 4 = 3
Example 27 (mod 4 = 3):
n = 11011 (27) mod 4 = 3 +(n+1)/2 = 11110 (14) number to add —————————— Result = 101001 (41) mod 4 = 1
Next iteration (with 41 mod 4 =1):
n = 101001 (41) mod 4 = 1 +(n+1)/2 = 10101 (21) number to add —————————— Result = 111110 (62) mod 4 = 1 -> halving: 11111 (31) mod 4 = 3
We can say:
Additionally there are 4 rules for reversing the Collatz sequences:
Source:
Thanks for your help! Chears from Germany ( iovialis)...
-- Iovialis ( talk) 20:39, 3 April 2022 (UTC)
1. I had a similar idea last month: rather than calling numbers "iceballs" or something, binary tricks could solve this thing.
2. It appears to me, 101, 11011, 1110111 and so on are the hardest, the longest to crunch.
3. It is possible to represent the operation with a handful of operations a Turing machine can do:
At this rate, it's not arithmetics-like calculating, it's more like sorting while shifting the array of the number (in binary) to the left. 2A00:1FA0:231:2C7A:0:6C:BD22:BF01 ( talk) 14:40, 18 July 2022 (UTC)
Bakuage Co., Ltd, a Japanese company, offers 120 million yen ( c. USD 919,280 as of March 2023) for the resolution of Collatz cojeture. [1] When I tried to add this, it was taken down and I was advised to at least discuss it here first.
Is this a trustable claim? If it is, then we should add it to the article. A long while ago, someone added this claim to the german Wikipedia page on Collatz conjecture. So if it is not trustable, then it's about time to get it removed. Felixsj ( talk) 19:42, 23 March 2023 (UTC)
References
The edits concerning the work of Hercher (2023) (minimum number of elements in a non trivial cycle, minimum number of k in a k-cycle, see https://cs.uwaterloo.ca/journals/JIS/VOL26/Hercher/hercher5.pdf) were reverted. Why? The work was published in a well known and accepted journal (Journal of Integer Sequences) with peer review. Why list the old and outdated numbers? 2A02:8108:1280:1C96:486F:207B:A7EC:A19B ( talk) 23:20, 24 March 2023 (UTC)
This edit may have been made accidentally. The edits of the last few days and weeks looked more like an attack. -- B wik ( talk) 07:03, 27 March 2023 (UTC)
I solved it I'm just a kid but I solved it 3x+1=googleplex I did it 2001:BB6:2BA0:F800:9DA6:D8A7:1346:1F3B ( talk) 16:00, 10 December 2022 (UTC)
Someone should update the Collatz Conjecture page, a proof has been proven impossible. An inequality makes a loop impossible, the final descent from 3x+1/2^n to 3X (where X Is the initial X) can only be a 12n-5. The net value of all ascents minus all descents between the 1st 3X+1 and final 3x must be the same value, a 12n-5, for a loop but it can't be, it can only be a 12n-1,12n+1,12n+3,12n-3 or 12n+5, it can't be a 12n-5. This 9 minute video explains it step by step using simple logical deduction. youtu dot be/O_oRnFuRfRM Sean g 2001:BB6:1A0E:9E58:2D84:F0F:4025:87CE ( talk) 08:52, 24 April 2023 (UTC)
I assume that What Goes for 3x + 1, Goes as well for Px + 1, where P stands for any prime number. algorithm is then as follows: Given a number N divide N by all prime Numbers < P (where possible) Multiply resulting number by P, and add 1
after Few iterations result will be (and stay) 1
Exactly like in the 3x + 1 problem Derijan ( talk) 11:15, 9 May 2023 (UTC)
Example for the above take P =17, and the dividers 2, 3, 5, 7, and 11 lets take a random number N, for example 917
start: 917 Divided by 7: 131 Multiplied by 17 +1 and divided by 2: 1114 Divided by 2: 557 Multiplied by 17 +1 and Divided by 2’ 4735 Divided by 5: 947 Multiplied by 17 +1 and Divided by 2: 8050 Divided by 2: 4025 Divided by 5: 805 Divided by 5: 161 Divided by 7: 23 Multiplied by 17 +1 and Divided by 2: 196 Divided by 7: 28 Divided by 7: 4 Divided by 2: 2 divided by 2: 1 Derijan ( talk) 12:35, 9 May 2023 (UTC)
I have an idea. Instead of thinking of natural numbers and arithmetics of "reaching 1", we could maybe try a binary pattern.
![]() | This
edit request to
Collatz conjecture has been answered. Set the |answered= or |ans= parameter to no to reactivate your request. |
I believe I've found a number with a longer chain than 3732423 which is 7464847. The chain length is 598. 2600:1700:D4CE:9400:A40B:B33D:EF86:A834 ( talk) 03:28, 2 August 2023 (UTC)
An illustration says: "The same plot on the left but on log scale, so all y values are shown. The first thick line towards the middle of the plot corresponds to the tip at 27, which reaches a maximum at 9232."
27 is literally 3*3*3, and the conjecture is based on 3n+1. Therefore, the article could use a "trinary" representation. 81.89.66.133 ( talk) 08:56, 8 August 2023 (UTC)
The second and last paragraph in the section Stopping times is as follows:
"In 2019, Terence Tao improved this result by showing, using logarithmic density, that almost all Collatz orbits are descending below any given function of the starting point, provided that this function diverges to infinity, no matter how slowly. Responding to this work, Quanta Magazine wrote that Tao "came away with one of the most significant results on the Collatz conjecture in decades"."
The phrase "are descending" does not strike me as using an appropriate tense to reflect a mathematical statement. Things either exist or not; statements are either true or not. 2601:200:C000:1A0:2897:C654:842B:EC14 ( talk) 00:02, 6 August 2022 (UTC)
The section Syracuse function lets I denote the set of all odd integers, but its conclusion is only about positive odd integers.
I hope someone will fix this.
The tag was added with only a vague comment that the article "is beyond the reach on non-experts". I'm not sure if the article can be significantly simplified without sacrificing its overall quality. Perhaps the introduction could give a numerical example, like . -- Hugo Spinelli ( talk) 15:24, 5 October 2023 (UTC)
At the end of the Modular restrictions section, there is the line:
"For example, the only surviving residues mod 32 are 7, 15, 27, and 31."
But I have reason to believe that this is incorrect and the correct surviving residues mod 32 are 7, 9, 15, 27, 30, 31.
I would like to add a Citation Needed to the line. Brentonb ( talk) 21:06, 20 October 2023 (UTC)
def f(m, b):
if b % 2 == 0:
return m//2, b//2
return 3*m//2, (3*b + 1)//2
def fk(m, b, k):
if k == 1:
return f(m, b)
return fk(*f(m, b), k - 1)
d = set()
for p in range(1, 5 + 1):
for b in range(2**p):
m2, b2 = fk(2**p, b, p)
if m2 < 2**p:
d.add((2**p, b))
# Verify surviving residues mod N (must be a power of 2)
N = 32
for k in range(N):
for a, b in d:
if k % a == b:
break
else:
print(k)
— Hugo Spinelli ( talk) 20:17, 22 October 2023 (UTC)
Terence Tao has worked on this problem quite a bit. He has held two notable public lectures about the problem: in 2021 and 2023. He mentions some things about the problem which are not mentioned here. Most importantly, if the conjecture were true, than we would some theorem about the difference between powers of two and those of three – which we do know, but demand quite sophisticated techniques to prove (see The Collatz conjecture, Littlewood-Offord theory, and powers of 2 and 3). Felixsj ( talk) 11:24, 1 November 2023 (UTC)
like the Collatze conjecture, there may be another one as below: input an integer number as 'n', then:
m = mod(n,3);
if m == 0 n = n/3; elseif m == 1 n = 4*n - 1; else n = 4*n + 1; end
this procedure will end up to 1 for any number (!)
Hadijavadi ( talk) 11:11, 2 November 2023 (UTC)
Step #3 (clearing trailing 0s) should be done as step #1. Why? We obviously want to clear the trailing 0s from even numbers, but for odd numbers, clearing the trailing 0s from an odd number does nothing (there aren't any), hence it's safe to do. This also removes the restriction on the input of the algorithm (currently only allows odd numbers). The algorithm converges onto 4 (0b100) instead of 1 since it clears 0s at the beginning of the loop, not the end.
--
The edits I'm asking:
Move the #3 up to #1 in the algorithm.
The machine will perform the following three steps on any odd number until only one 1 remains:
->
The machine will perform the following three steps on any binary number until 4 remains:
--
As a side note, this code does something similar to the abstract machine, in that it clears trailing zeroes in one step. "x // (x & -x)" clears trailing zeros. It works because "(x & -x)" extracts the lowest set bit which is then divided out.
x = 27 while x != 4: x = 3 * (x // (x & -x)) + 1 print(bin(x), x)
"clear trailing zeroes. add 2x+1 to x. repeat"
x = 27 while x != 4: x //= (x & -x) x += (x << 1 | 1) print(bin(x), x)
2605:A601:A629:3300:7738:24BD:85A0:9FC7 ( talk) 05:47, 13 November 2023 (UTC)
A006577: Number of halving and tripling steps to reach 1 in '3x+1' problem, or -1 if 1 is never reached.
If n is 0 or negative, then A006577(n) = -1.
0: 0, 0, 0, 0, 0, 0, 0, ...
-1: -2, -1, -2, -1, -2, -1, ...
-3: -8, -4, -2, -1, -2, -1, -2, -1, ...
-5: -14, -7, -20, -10, -5, -14, -7, -20, -10, -5, -14, -7, -20, -10, -5, ...
-6: -3, -8, -4, -2, -1, -2, -1, -2, -1, ...
-9: -26, -13, -38, -19, -56, -28, -14, -7, -20, -10, -5, -14, -7, -20, -10, -5, -14, -7, -20, -10, -5, ... 2A00:6020:A123:8B00:3913:1297:6B6B:CCEF ( talk) 13:19, 14 December 2023 (UTC)
This function is mentioned in the French page and avoid testing parity. [ [2]] Japarthur ( talk) 09:02, 8 March 2024 (UTC)
This article could have used an animation for trinary numbers, say, a .GIF of 4975 being broken down. 81.89.66.133 ( talk) 13:01, 2 February 2024 (UTC)
Collatz Conjecture Solution The Collatz conjecturela is one of the most famous unsolved problems in mathematics. The conjecture asks whether repeating two simple arithmetic operations will eventually transform every positive integer into 1. /info/en/?search=Collatz conjecture The solution in simple words, all number made out of 1. Like 1=1 2=1+1 3=1+1+1 4=1+1+1+1| 5=1+1+1+1+1 Etc. I am saying not only 1 is repetitive but 4,2,1 is repetitive. 3x+1 in if x=1 then, 3(1)+1= 4, then as per rules 4/2 =2 then 2/2=1 means 4,2,1 Now if x=2 then 3(2)+1=7 then as per rules 3(7)+1=22, then 22/2= 11, then 3(11)+1= 34 then 34/2= 16, 16/2=8, 8/2=4, 4/2=2, 2/2=1. Now if x=3 then 3(3)+1=10, 10/2 = 5, 3(5)+1=16, 16/2=8, 8/2=4, 4/2=2, 2/2=1| If we see in all solutions starting from one of the small integers 4,2,1 is repetitive. Because in x=3 there is ans 5. allAll ISSN:3006-4023 (Online),JournalofArtificiallntelligence GeneralScience (JAIGS)86 GaurangkumarPatel20894 ( talk) 07:01, 5 April 2024 (UTC)
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 |
2 1
3 7
6 8
7 16
9 19
18 20
25 23
27 111
54 112
73 115
97 118
129 121
171 124
231 127
313 130
327 143
649 144
703 170
871 178
1161 181
2223 182
2463 208
2919 216
3711 237
6171 261
10971 267
13255 275
17647 278
23529 281
26623 307
34239 310
35655 323
52527 339
77031 350
106239 353
142587 374
156159 382
216367 385
230631 442
410011 448
511935 469
626331 508
837799 524
1117065 527
1501353 530
1723519 556
2298025 559
3064033 562
3542887 583
3732423 596
5649499 612
6649279 664
8400511 685
11200681 688
14934241 691
15733191 704
31466382 705
36791535 744
63728127 949
127456254 950
169941673 953
226588897 956
268549803 964
537099606 965
670617279 986
1341234558 987
1412987847 1000
1674652263 1008
2610744987 1050
4578853915 1087
4890328815 1131
9780657630 1132
12212032815 1153
12235060455 1184
13371194527 1210
17828259369 1213
31694683323 1219
63389366646 1220
75128138247 1228
133561134663 1234 — Preceding
unsigned comment added by
Frank Klemm (
talk •
contribs)
23:04, 13 August 2016 (UTC)
The cycle 0→0 is listed as 'trivial'. This is misleading since for any number other than zero, the Collatz-function cannot reach zero. There is no positive or negative natural number other than zero such that C(n)=0. It's a special case (for n=0), and should be listed or omitted as such. Kleuske ( talk) 11:34, 15 September 2016 (UTC)
I found on Eric Roosendaal's page "On the 3x+1 Problem" the notion of "residue" for the ratio (2^h(n)/3^t(n))/n where h, t count the halving/tripling steps, which appears to be maximal (1.253...) for 993. (It is natural to expect that n ~ 2^h/3^t.) Is this an original result from this mathematician, and/or have others considered this quantity? — MFH: Talk 21:34, 8 May 2018 (UTC)
There is almost no history of the problem in the article, when was it first brought into serious mathematics? Naraht ( talk) 13:54, 12 February 2017 (UTC)
I support this request, it is referred to Collatz, Ulam and Kakutani (and Thwaites) but on their pages (if they exist) there is no reference to a publication and/or date which could hint on when the given person has considered this problem or made a conjecture. — MFH: Talk 22:48, 25 October 2017 (UTC)
It is not clear what serious maths is. I agree that it seems trivial. — Preceding unsigned comment added by 86.1.37.70 ( talk) 14:39, 1 June 2018 (UTC)
The COLLATZ CONJECTURE IS SO EASY TO EXPLAIN!. IT'S LIKE 1+1=?! The the number is a power of two, it will do divide by 2 and so continued, and will reach two, and then 1. If it is even number but not a power of two, it will do divide by 2 until it reaches an odd number. Then for the odd numbers, we've got to multiply it (Collatz is a strange guy, multiply by 3 OR ANY NUMBER), you still get an odd. So and one and you got an even. And restart the paragraph. :) — Preceding unsigned comment added by 223.72.58.54 ( talk) 13:47, 9 July 2018 (UTC)
The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.
It's due to a very simple property of binary arithmetic:
So you see, you've been tricked into thinking you're doing 3 steps, but really it's just 2 steps obfuscated, then 1 real step. So in this system, only even numbers (numbers with no LSB set) are shifted to the right (so no data loss occurs), then they are immidietely shifted to the left where they came from. So both steps DO NOTHING. NOTHING. Then you add the number! This logically will ALWAYS increase the number.. thus logically making you play this game over-and-over until the lower significant bits alternate as such 10101010.. or 01010101.. Once this pattern occurs, you have a number with a multiple of 10 (ends in decimal 0), which means the system is in it's 2nd last phase before death reducing by multiples of 10 until it hits 10. 10 is 1010 and 5 is 0101, so once 10 is shifted to the right it becomes 5, 5 is "temp stored" in the system, 5 gets shifted to the left again and becomes 10, which 10+5 = 15, and 15 is one less from 16, and once the number hits 16, this system will enter into it's final phase, it will right-shift infinitely until the LSB is set. And this means the number will always be 1.
You're really just trying a simple "test" against the number, if the test fails, like Penn & Teller, you are tricked into thinking something happens (logical right-shift, logical-left-shift), then something really happens (one simple addition), and you've got a slightly higher number with a slightly better chance at being.... duh duh duh duh a 1010 (10 decimal) pattern binary number... which will always reduce in that system to a 0101 (5 decimal) binary pattern number due to the fact that addition is just a logical XOR. You'll notice both 10 and 5 have alternating "Chaser Lights" pattern. Every other bit is set. When they are XOR'd together all bits will be set. When all bits are set, and you add 1, you'll get a Power of Two number. Because a Power of Two number is always just 1 bit that is 0-to-many right-shifts away from being a number 1, this simple system will always produce a number 1 no matter the input. It's super basic binary math.
Right? — Preceding unsigned comment added by 24.78.157.241 ( talk) 09:19, 18 August 2018 (UTC)
Whoops! Hey Jasper, my original idea still works. bare with me just a minute bud! I took a good look at what is happening in this system and all it does is ping-pong between two states. I'll walk through n = 12 (Wikipedia example). 12, 6, 3.. so in binary math, halving a number is logical shift right. With an even number, the system enters a state where it will shift the bits to the right until there is a bit in the LSB. Once there is a bit in the LSB, we LOGICALLY have a odd number. Multiply that ODD number by 3 and we must get an ODD number result. The +1 in the system makes it always an EVEN number. And since it's ALWAYS an even number... and an EVEN number is what started off this system.. it recursively does the same thing again.. and again.. and again. But the system itself can only do 1 of 2 things.. oscillate away from 1 for eternity (it wont), stick on the number 1, in which case the system halts. — Preceding unsigned comment added by 24.78.157.241 ( talk) 09:33, 18 August 2018 (UTC)
First reference is under a graphic caption. — Preceding unsigned comment added by 2601:18C:CD7F:6064:F4C9:9B76:6690:9E8F ( talk) 09:29, 2 November 2018 (UTC)
I do not claim to be a programming or maths genius. However, since I particularly enjoy this problem, I wrote a short script to calculate the size of the progression for a given number. I tested it to make sure it was correct by using the values on the article under the longest progression for a starting number part (in Examples), and some of them came back with very different step counts from what the article says they should be. I ran each value multiple times to be sure it wasn't just me creating the discrepancy. Most of the numbers I came up with do match the ones given on the article.
For less than 10 trillion the article says the number of steps for 7,887,663,552,367 is 1563. My script returned only 511. For less than 100 trillion, the article says the number of steps for 80,867,137,596,217 is 1662. My script returned only 385. For less than 1 quadrillion, the article says the number of steps for 942,488,749,153,153 is 1862. My script returned only 262. For less than 10 quadrillion, the article says the number of steps for 7,579,309,213,675,935 is 1958. My script returned only 296. For less than 100 quadrillion, the article says the number of steps for 93,571,393,692,802,302 is 2091. My script returned only 235.
In order to be sure, I've run other numbers through to see if it's a calculation limit. My computer was able to calculate 2^222 and correctly inform me it took 222 steps. I was able to run 67399866667876599486667537717549076684092861056351431202759025623046739986666787659948666753771754907668409286105635143120275902562304 and it took 711 steps. It appears that the magnitude is not, in fact, a limitation or source of error.
I looked at the citation given for the final number on that list, which I presume is the source for other numbers there. The website seems to be suspect as far as academic credibility is concerned. Again, I do not claim to have any academic background here, but my results do differ dramatically, and seem to be equally valid.
If I have missed something major, my sincerest apologies.
For those who wish to test my script, you may find it here. It is in Python 3.7.3, the latest version as of the time of writing. — Preceding unsigned comment added by 172.77.247.24 ( talk) 04:38, 15 June 2019 (UTC)
c(n) = x=n; s=0; while(x>1, if(x%2==0, x=x/2, x=3*x+1); s++); s c(7887663552367)
https://www.researchgate.net/publication/296682209_Proof_of_the_Collatz_conjecture Chrisdecorte ( talk) 01:07, 21 July 2019 (UTC)
https://doi.org/10.1155/2019/6814378 — Preceding unsigned comment added by Kamalbarghout ( talk • contribs)
According to this MSE answer, the record is still valid as of August 2019. -- DaBler ( talk) 12:22, 19 August 2019 (UTC)
This forum is for discussion of the Wikipedia article.
The appropriate place to discuss this is the Usenet group sci.math and I've transferred it there. You can reach it via Google Groups with subject Collatz Conjecture proof? —Preceding undated comment added 23:29, 15 June 2008
If someone has new information to add on here related to the Collatz conjecture but it hasn't been published elsewhere, can it still be added? It's not false information, but there's no source to confirm it. A proof for the new information can be provided. I just wanted to check first. I don't want to add it only to have it taken down shortly after. Jeanlovecomputers ( talk) 23:18, 2 October 2019 (UTC)
According to the Collatz conjecture, we can use numbers not only natural and only positive. In 252 steps on this conjecture, we can get 1 from the number (-0.49). It can also be said that this conjectureis valid not only for natural numbers, but also for rational numbers. In 436 steps on this conjecture, we can get 1 from the number 2.48. We can get 1 by this conjecture of our number(n), but only if n>(-0.5). It is proof for (-0.49): Number n: -0.49 1 -0.470000 2 -0.410000 3 -0.230000 4 0.310000 5 1.930000 6 6.790000 7 21.370000 8 65.110000 9 196.330000 10 589.990000 11 1770.970000 12 5313.910000 13 15942.730000 14 47829.190000 15 143488.570000 16 430466.710000 17 1291401.130000 18 3874204.390000 19 11622614.170000 20 34867843.510000 21 104603531.530000 22 313810595.590000 23 941431787.770001 24 2824295364.310002 25 8472886093.930007 26 25418658282.790020 27 76255974849.370056 28 228767924549.110170 29 686303773648.330570 30 2058911320945.991700 31 6176733962838.974600 32 18530201888517.922000 33 55590605665554.766000 34 166771816996665.310000 35 500315450989996.940000 36 1500946352969991.700000 37 4502839058909976.000000 38 2251419529454988.000000 39 1125709764727494.000000 40 562854882363747.000000 41 1688564647091242.000000 42 844282323545621.000000 43 2532846970636864.000000 44 1266423485318432.000000 45 633211742659216.000000 46 316605871329608.000000 47 158302935664804.000000 48 79151467832402.000000 49 39575733916201.000000 50 118727201748604.000000 51 59363600874302.000000 52 29681800437151.000000 53 89045401311454.000000 54 44522700655727.000000 55 133568101967182.000000 56 66784050983591.000000 57 200352152950774.000000 58 100176076475387.000000 59 300528229426162.000000 60 150264114713081.000000 61 450792344139244.000000 62 225396172069622.000000 63 112698086034811.000000 64 338094258104434.000000 65 169047129052217.000000 66 507141387156652.000000 67 253570693578326.000000 68 126785346789163.000000 69 380356040367490.000000 70 190178020183745.000000 71 570534060551236.000000 72 285267030275618.000000 73 142633515137809.000000 74 427900545413428.000000 75 213950272706714.000000 76 106975136353357.000000 77 320925409060072.000000 78 160462704530036.000000 79 80231352265018.000000 80 40115676132509.000000 81 120347028397528.000000 82 60173514198764.000000 83 30086757099382.000000 84 15043378549691.000000 85 45130135649074.000000 86 22565067824537.000000 87 67695203473612.000000 88 33847601736806.000000 89 16923800868403.000000 90 50771402605210.000000 91 25385701302605.000000 92 76157103907816.000000 93 38078551953908.000000 94 19039275976954.000000 95 9519637988477.000000 96 28558913965432.000000 97 14279456982716.000000 98 7139728491358.000000 99 3569864245679.000000 100 10709592737038.000000 101 5354796368519.000000 102 16064389105558.000000 103 8032194552779.000000 104 24096583658338.000000 105 12048291829169.000000 106 36144875487508.000000 107 18072437743754.000000 108 9036218871877.000000 109 27108656615632.000000 110 13554328307816.000000 111 6777164153908.000000 112 3388582076954.000000 113 1694291038477.000000 114 5082873115432.000000 115 2541436557716.000000 116 1270718278858.000000 117 635359139429.000000 118 1906077418288.000000 119 953038709144.000000 120 476519354572.000000 121 238259677286.000000 122 119129838643.000000 123 357389515930.000000 124 178694757965.000000 125 536084273896.000000 126 268042136948.000000 127 134021068474.000000 128 67010534237.000000 129 201031602712.000000 130 100515801356.000000 131 50257900678.000000 132 25128950339.000000 133 75386851018.000000 134 37693425509.000000 135 113080276528.000000 136 56540138264.000000 137 28270069132.000000 138 14135034566.000000 139 7067517283.000000 140 21202551850.000000 141 10601275925.000000 142 31803827776.000000 143 15901913888.000000 144 7950956944.000000 145 3975478472.000000 146 1987739236.000000 147 993869618.000000 148 496934809.000000 149 1490804428.000000 150 745402214.000000 151 372701107.000000 152 1118103322.000000 153 559051661.000000 154 1677154984.000000 155 838577492.000000 156 419288746.000000 157 209644373.000000 158 628933120.000000 159 314466560.000000 160 157233280.000000 161 78616640.000000 162 39308320.000000 163 19654160.000000 164 9827080.000000 165 4913540.000000 166 2456770.000000 167 1228385.000000 168 3685156.000000 169 1842578.000000 170 921289.000000 171 2763868.000000 172 1381934.000000 173 690967.000000 174 2072902.000000 175 1036451.000000 176 3109354.000000 177 1554677.000000 178 4664032.000000 179 2332016.000000 180 1166008.000000 181 583004.000000 182 291502.000000 183 145751.000000 184 437254.000000 185 218627.000000 186 655882.000000 187 327941.000000 188 983824.000000 189 491912.000000 190 245956.000000 191 122978.000000 192 61489.000000 193 184468.000000 194 92234.000000 195 46117.000000 196 138352.000000 197 69176.000000 198 34588.000000 199 17294.000000 200 8647.000000 201 25942.000000 202 12971.000000 203 38914.000000 204 19457.000000 205 58372.000000 206 29186.000000 207 14593.000000 208 43780.000000 209 21890.000000 210 10945.000000 211 32836.000000 212 16418.000000 213 8209.000000 214 24628.000000 215 12314.000000 216 6157.000000 217 18472.000000 218 9236.000000 219 4618.000000 220 2309.000000 221 6928.000000 222 3464.000000 223 1732.000000 224 866.000000 225 433.000000 226 1300.000000 227 650.000000 228 325.000000 229 976.000000 230 488.000000 231 244.000000 232 122.000000 233 61.000000 234 184.000000 235 92.000000 236 46.000000 237 23.000000 238 70.000000 239 35.000000 240 106.000000 241 53.000000 242 160.000000 243 80.000000 244 40.000000 245 20.000000 246 10.000000 247 5.000000 248 16.000000 249 8.000000 250 4.000000 251 2.000000 252 1.000000 Nikita fly ( talk) 20:18, 3 October 2019 (UTC) Chizhov Nikita 03.10.2019
Does someone have reliable sources as to this problem's relevance to important theoretical mathematical areas that deal with general properties of systems, such as number theory? Various users over at Reddit and StackExchange have discussed the relation between the prime factorizations of n and n+1 (for odd n the next number is 3n+1, so how does n's factorization affect 3n+1's factorization?), and mentioned the Riemann hypothesis in connection to this. Basically problems that help us advance our general understanding of mathematics or formulate new important branches of mathematics. 37KZ ( talk) 23:34, 10 November 2019 (UTC)
The article here reports that they have verified the validity of this conjecture for all numbers below 268. -- DaBler ( talk) 08:26, 2 July 2020 (UTC)
In order to improve the article
I ask you to consider the possibility of inserting into the article
--Collatz conjecture--
--in the Examples section
--number 27
my explanation of the behavior of the number 27
that is given on the page
/info/en/?search=User_talk:Eduard_Dyachenko
and the article
https://zenodo.org/record/4013334#.X1DhOcgzbIU
Dear editors, please do not delete as this is a discussion of a specific section of the article.
Eduard Dyachenko (
talk)
18:59, 27 March 2020 (UTC)E.Dyachenko(dyachenko.eduard@gmail.com)
You can check the table with number 27 manually, it is not at all difficult
Eduard Dyachenko (
talk)
20:18, 27 March 2020 (UTC)E.Dyachenko(dyachenko.eduard@gmail.com)
Trajectory in the form of a graph yes, but without explanation and calculations of its origin
E.Dyachenko(dyachenko.eduard@gmail.com)
without explanation and calculations of its originThe calculational rule for the trajectory is simple arithmetic: divide even numbers by 2, triple odd numbers and add 1. If you believe there is some deeper analysis that goes along with this particular trajectory, then we're back to the problem of WP:OR. -- JBL ( talk) 21:20, 27 March 2020 (UTC)
the table shows that the change in the peaks in the graph coincides with the change in the oddness from the form 4k+3 to 4k+1
which corresponds to a decrease in the "length of the number" in numeral system 4:3
E.Dyachenko(dyachenko.eduard@gmail.com) — Preceding unsigned comment added by Eduard Dyachenko ( talk • contribs) 21:43, 27 March 2020 (UTC)
which corresponds ...and that's the part that violates WP:OR. -- JBL ( talk) 00:00, 28 March 2020 (UTC)
I believe the table, posted along with the graph, is a worthy visualization how graph behaves.
This is a supporting element providing explanation to the picks of the graph.
E.Dyachenko (dyachenko.eduard@gmail.com)
— Preceding
unsigned comment added by
Eduard Dyachenko (
talk •
contribs)
10:39, 28 March 2020 (UTC)
WP:NOTFORUM |
---|
The following discussion has been closed. Please do not modify it. |
We let t be the number of trials it takes the Collatz sequence (orbit) of odd integers to converge. We can determine k by the largest positive even integer, , in the Collatz sequence. . Therefore, as ... That value, 0, represents the probability that the Collatz sequence of odd integers does not converge to one. Relevant Reference Link: A Brief Analysis of the Collatz Conjecture Therefore, we conclude the Collatz conjecture is true! — Preceding unsigned comment added by 96.76.246.142 ( talk) 16:10, 15 September 2020 (UTC) |
In our recently pusblished article we analyze the divisions by two that are performed within Collatz sequences: https://doi.org/10.18052/www.scipress.com/IJPMS.21.1
Aside from classical mathematical methods, we use techniques of data science. Based on the analysis of 10,000 sequences we show that the number of divisions by two lies within clear boundaries. The paper covers the Collatz problem both in it's original form as well as in the generalized variant , where .
Building on the results, we develop and prove the following equation, which calculates the maximum possible number of divisions by two for any given Collatz sequence of a certain length:
The parameter represents the count of odd numbers in the sequence and stands for the first odd number. Whenever the maximum is reached, a sequence leads to the result one, as conjectured by Lothar Collatz. Furthermore, we show how many divisions by two are required for a cycle of a specific length:
Are these findings of interest for Wikipedia? We are looking forward to the discussion -- C4ristian ( talk) 16:51, 18 November 2020 (UTC)
Thank you for your replies so far. We still hope to initiate a scientific debate, as we experienced in other open communities. There is no doubt that the Wikipedia policies are important. However, in many cases they leave room for interpretation. Let me give you two examples:
Not to be misunderstood: My main point is not the incorporation of our idea in the article. If there are mathematically or didactically sound reasons not to mention it, that is perfectly fine for us. Our goal is to contribute to the solution of the Collatz problem. For this reason, we have not edited the article directly, but have instead started this discussion. That is why we still hope to be treated fairly and get at least some scientific feedback -- C4ristian ( talk) 10:20, 21 November 2020 (UTC)
Self-published expert sources may be considered reliable when produced by an established expert on the subject matter, whose work in the relevant field has previously been published by reliable, independent publications.
I noticed that there are several (sub)sections in this article that do not cite any sources, such as the recently-added " Connection to a sequence of different origin" section by Verihärö. Although I appreciate these insights into the conjecture, I am wondering whether this falls under original research? ― JochemvanHees ( talk) 10:17, 25 December 2020 (UTC)
The formulation of Collatz conjecture as a binary relation "→" is frequently used, e.g. in the section "Extensions to larger domains," subsection "Iterating on all integers" of this article. Also, it is relevant fact that slight modification of the formulation is provable, the proof is simple and can be used for educational purposes as argues an article recently published in peer-reviewed online journal Academia Letters (the article is accompanied by two public peer reviews, one of them from a professor of mathematics Shaun V. Ault).
So, I propose to add the subsection "As a binary relation" in the section "Other formulations of the conjecture" with this short text:
For any natural n and any odd number x, Collatz conjecture is equivalent to the statement about the transitive reflexive binary relation which satisfies conditions and .
Similar formulation represents weaker version of Collatz conjecture: it is proved [1] that for the transitive reflexive symmetric binary relation which satisfies conditions and .
93.72.108.33 ( talk) 08:57, 28 July 2021 (UTC)
References
Similar formulation represents weaker version of Collatz conjectureNo, it represents a completely different question that is connected to the Collatz conjecture only superficially. It has no business being in this article. Academia Letters is not a real journal, it is a self-publishing platform with a slight patina of peer review; I hope that you didn't pay them too much money. -- JBL ( talk) 11:14, 28 July 2021 (UTC)
I attempted to add a sentence about a 120 million yen prize being offered for a proof of the conjecture, but it was reverted because I chose prnewswire.com for the citation. I could have chosen mathprize.net, but that is the primary source and I thought secondary sources were preferred. In any case, such a large prize seems worthy of mention. 2601:C6:4100:F980:CD93:B16:285F:4B15 ( talk) 04:01, 31 July 2021 (UTC)
"The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined as follows: start with any positive integer n. Then each term is obtained from the previous term as follows: if the previous term is even, the next term is one half of the previous term. If the previous term is odd, the next term is 3 times the previous term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1."
Surely you mean "current term" and not "previous term". Basing the formula for the next term on the previous term rather than the current term would be totally weird. Like Crunchy Bananas weird. 2601:8C3:8201:51D0:9C7C:1257:CBA8:1B9B ( talk) 06:54, 3 January 2021 (UTC)
start with any positive integer n. Then each new term is obtained from the previous term as follows: if the previous term is even, the
nextnew term is one half of the previous term. If the previous term is odd, thenextnew term is 3 times the previous term plus 1.
18 August 2021
It seems to me that the consensus here is that the statement of the conjecture can be improved. I agreed, and on 12 August 2021 I edited the statement in the introduction, since at that time no one had taken any action. I have been editing the wikipedia anonymously (always) from public ip addresses for 20 years now. I follow the old advise "be bold". That is I just do it. There seem to be some editors that take pride of ownership and strive to control every little detail. It may be that David Eppstein is one such. I hope not, but if he is, I'll give up after my edit today. I have no time for edit wars.
Apparently, judging from his reversion summary:
("The sequence terminates when ... The sequence diverges to infinity.": really??)
he took issue with my clarification to include the truth that it is, in fact a CONJECTURE. No one knows that there is no number whose sequence will never arrive at unity or a previous value in the sequence. That is a possible outcome. In fact there are some results that show that if there is an independent loop (other than the trivial 4-2-1 loop) it must be at least 180,000,000,000 elements long.
Or perhaps he took issue with the idea that an infinite divergence is not strictly a termination. Fair enough, so today I reworded things slightly to appease the Gods. Might he not have simply done that himself? That is the point of a wiki. Why waste energy requiring this elaborate justification?
My edit is in fact a substantial clarification of the sloppy language that has prevailed, to say nothing of the awkward style ("following: .... following:": really??).
See Talk:Collatz conjecture/Archive 2#Attribution for credit to Collatz and 1937 date and HSM SE. Ain92 ( talk) 20:15, 27 August 2021 (UTC)
I think the series of lines
less than 10 is 9, which has 19 steps, less than 100 is 97, which has 118 steps, less than 1000 is 871, which has 178 steps,"
would be better in a table. Bubba73 You talkin' to me? 03:36, 29 August 2021 (UTC)
The description of this book on amazon seems to imply that a solution has been found. Does anyone know anything about this? Is this guy lying? https://www.amazon.com/Collatz-Conjecture-Solutions-Leong-Ying/dp/198075991X/ref=sr_1_3?crid=36AOENBGWHIL&dchild=1&keywords=collatz+conjecture&qid=1634074729&sr=8-3 — Preceding unsigned comment added by Skysong263 ( talk • contribs) 21:42, 12 October 2021 (UTC)
![]() | This help request has been answered. If you need more help, you can , contact the responding user(s) directly on their user talk page, or consider visiting the Teahouse. |
Is it not misleading to have the diagram at the top of the mobile version of the page skip the even numbers?02:38, 24 October 2021 (UTC) 147.134.101.106 ( talk)
Lawrencekhoo is advancing some rewrites of the lede, which have not been immediately accepted as useful by other editors. Lawrence, perhaps you want to make a case for your edits here on the talk page? Russ Woodroofe ( talk) 12:20, 1 December 2021 (UTC)
The unreliable citation is not a reason to remove this topic altogether; I am currently occupied but could someone please find some CREDIBLE sources and add back the good faith edits by Fmwithstuff Cassie Schebel, almost a savant. <3 ( talk) 23:51, 16 March 2022 (UTC)
Be careful, in table: -1 gives -1 and never -2. 2A04:CEC0:1192:22A0:EC1D:F521:5A80:D88B ( talk) 13:28, 28 March 2022 (UTC)
(1) No. (2). Not here. |
---|
The following discussion has been closed. Please do not modify it. |
When we look at the Collatz Conjecture, if a randomly chosen positive integer is even, it is divided by 2, and if the result is odd, the number is multiplied by 3 and added by 1; If the result is an even number, it is seen that dividing by 2 is continued. In this case, the assumption is made that the number will eventually reach 1, resulting in an infinite loop of 1-4-2-1. As a result of the investigation:
As a result of the experimental study:
- It has been determined that the highest total number of iterations is 350 and the number value is 77,031. - When we look at the iteration results of a randomly selected positive integer, it is seen that a maximum of 18 consecutive even numbers come in a row and this number value is 87,381. Slmyildirim79 ( talk) 07:57, 24 April 2022 (UTC) As a result: - It is seen that with the increase in the number values of the digits, the total number of iterations will also increase, and naturally, the probability of successive even numbers will increase. - Looking at the iteration results, it is seen that the probability of the number converging to 1 is higher than the probability of diverging to 1 because of the increase in consecutive even numbers. For the randomly selected positive integer values to increase, a "odd-odd-odd-odd..." loop must occur one after the other, which is impossible. Naturally, all positive integers will reach 1 after iterations. Therefore, the Collatz conjecture is valid for all positive integers. |
Is ac=0 an error here? If not, just what does that mean? John D. Goulden ( talk) 21:51, 1 December 2021 (UTC)
The section
A similar reasoning that accounts for the recent verification of the conjecture up to 2^68 leads to the improved lower bound 114208327604 (or 186265759595 without the "shortcut"). This lower bound is consistent with the above result, since 114208327604 = 17087915 × 361 + 85137581 × 1269.
is very unclear. What is the "similar reasoning?" What is the "shortcut"? Which citation supports this claim? 65.128.180.152 ( talk) 06:05, 15 June 2022 (UTC)
prompted
Illegibility is an issue. The boxes could be doubled in size.
Arrows might get enhanced by rounded corners or intermittent arrowheads.
Acyclicity is felt in your fingers following this graph but contested, after all it is still a mere conjecture.
How should the graph be improved and where ought it be prominently placed? IM Serious ( talk) 01:50, 24 January 2023 (UTC)
the snailwound walk from 25, or the overshooters 15, 23, 27. I also display the periodicities in rising or falling) is also extremely cryptic. -- JBL ( talk) 18:33, 24 January 2023 (UTC)
Can someone add this into the article if it is considered as relevant:
The behaviour of up and down of a Collatz sequence can be explained by using the binary system and n+(n+1)/2 instead of (3n+1)/2
Example 255 (with mod 4 = 3):
n = 11111111 (255) mod 4 = 3 +(n+1)/2 = 1000000 (128) number to add —————————— Result = 101111111 (383) mod 4 = 3
Example 27 (mod 4 = 3):
n = 11011 (27) mod 4 = 3 +(n+1)/2 = 11110 (14) number to add —————————— Result = 101001 (41) mod 4 = 1
Next iteration (with 41 mod 4 =1):
n = 101001 (41) mod 4 = 1 +(n+1)/2 = 10101 (21) number to add —————————— Result = 111110 (62) mod 4 = 1 -> halving: 11111 (31) mod 4 = 3
We can say:
Additionally there are 4 rules for reversing the Collatz sequences:
Source:
Thanks for your help! Chears from Germany ( iovialis)...
-- Iovialis ( talk) 20:39, 3 April 2022 (UTC)
1. I had a similar idea last month: rather than calling numbers "iceballs" or something, binary tricks could solve this thing.
2. It appears to me, 101, 11011, 1110111 and so on are the hardest, the longest to crunch.
3. It is possible to represent the operation with a handful of operations a Turing machine can do:
At this rate, it's not arithmetics-like calculating, it's more like sorting while shifting the array of the number (in binary) to the left. 2A00:1FA0:231:2C7A:0:6C:BD22:BF01 ( talk) 14:40, 18 July 2022 (UTC)
Bakuage Co., Ltd, a Japanese company, offers 120 million yen ( c. USD 919,280 as of March 2023) for the resolution of Collatz cojeture. [1] When I tried to add this, it was taken down and I was advised to at least discuss it here first.
Is this a trustable claim? If it is, then we should add it to the article. A long while ago, someone added this claim to the german Wikipedia page on Collatz conjecture. So if it is not trustable, then it's about time to get it removed. Felixsj ( talk) 19:42, 23 March 2023 (UTC)
References
The edits concerning the work of Hercher (2023) (minimum number of elements in a non trivial cycle, minimum number of k in a k-cycle, see https://cs.uwaterloo.ca/journals/JIS/VOL26/Hercher/hercher5.pdf) were reverted. Why? The work was published in a well known and accepted journal (Journal of Integer Sequences) with peer review. Why list the old and outdated numbers? 2A02:8108:1280:1C96:486F:207B:A7EC:A19B ( talk) 23:20, 24 March 2023 (UTC)
This edit may have been made accidentally. The edits of the last few days and weeks looked more like an attack. -- B wik ( talk) 07:03, 27 March 2023 (UTC)
I solved it I'm just a kid but I solved it 3x+1=googleplex I did it 2001:BB6:2BA0:F800:9DA6:D8A7:1346:1F3B ( talk) 16:00, 10 December 2022 (UTC)
Someone should update the Collatz Conjecture page, a proof has been proven impossible. An inequality makes a loop impossible, the final descent from 3x+1/2^n to 3X (where X Is the initial X) can only be a 12n-5. The net value of all ascents minus all descents between the 1st 3X+1 and final 3x must be the same value, a 12n-5, for a loop but it can't be, it can only be a 12n-1,12n+1,12n+3,12n-3 or 12n+5, it can't be a 12n-5. This 9 minute video explains it step by step using simple logical deduction. youtu dot be/O_oRnFuRfRM Sean g 2001:BB6:1A0E:9E58:2D84:F0F:4025:87CE ( talk) 08:52, 24 April 2023 (UTC)
I assume that What Goes for 3x + 1, Goes as well for Px + 1, where P stands for any prime number. algorithm is then as follows: Given a number N divide N by all prime Numbers < P (where possible) Multiply resulting number by P, and add 1
after Few iterations result will be (and stay) 1
Exactly like in the 3x + 1 problem Derijan ( talk) 11:15, 9 May 2023 (UTC)
Example for the above take P =17, and the dividers 2, 3, 5, 7, and 11 lets take a random number N, for example 917
start: 917 Divided by 7: 131 Multiplied by 17 +1 and divided by 2: 1114 Divided by 2: 557 Multiplied by 17 +1 and Divided by 2’ 4735 Divided by 5: 947 Multiplied by 17 +1 and Divided by 2: 8050 Divided by 2: 4025 Divided by 5: 805 Divided by 5: 161 Divided by 7: 23 Multiplied by 17 +1 and Divided by 2: 196 Divided by 7: 28 Divided by 7: 4 Divided by 2: 2 divided by 2: 1 Derijan ( talk) 12:35, 9 May 2023 (UTC)
I have an idea. Instead of thinking of natural numbers and arithmetics of "reaching 1", we could maybe try a binary pattern.
![]() | This
edit request to
Collatz conjecture has been answered. Set the |answered= or |ans= parameter to no to reactivate your request. |
I believe I've found a number with a longer chain than 3732423 which is 7464847. The chain length is 598. 2600:1700:D4CE:9400:A40B:B33D:EF86:A834 ( talk) 03:28, 2 August 2023 (UTC)
An illustration says: "The same plot on the left but on log scale, so all y values are shown. The first thick line towards the middle of the plot corresponds to the tip at 27, which reaches a maximum at 9232."
27 is literally 3*3*3, and the conjecture is based on 3n+1. Therefore, the article could use a "trinary" representation. 81.89.66.133 ( talk) 08:56, 8 August 2023 (UTC)
The second and last paragraph in the section Stopping times is as follows:
"In 2019, Terence Tao improved this result by showing, using logarithmic density, that almost all Collatz orbits are descending below any given function of the starting point, provided that this function diverges to infinity, no matter how slowly. Responding to this work, Quanta Magazine wrote that Tao "came away with one of the most significant results on the Collatz conjecture in decades"."
The phrase "are descending" does not strike me as using an appropriate tense to reflect a mathematical statement. Things either exist or not; statements are either true or not. 2601:200:C000:1A0:2897:C654:842B:EC14 ( talk) 00:02, 6 August 2022 (UTC)
The section Syracuse function lets I denote the set of all odd integers, but its conclusion is only about positive odd integers.
I hope someone will fix this.
The tag was added with only a vague comment that the article "is beyond the reach on non-experts". I'm not sure if the article can be significantly simplified without sacrificing its overall quality. Perhaps the introduction could give a numerical example, like . -- Hugo Spinelli ( talk) 15:24, 5 October 2023 (UTC)
At the end of the Modular restrictions section, there is the line:
"For example, the only surviving residues mod 32 are 7, 15, 27, and 31."
But I have reason to believe that this is incorrect and the correct surviving residues mod 32 are 7, 9, 15, 27, 30, 31.
I would like to add a Citation Needed to the line. Brentonb ( talk) 21:06, 20 October 2023 (UTC)
def f(m, b):
if b % 2 == 0:
return m//2, b//2
return 3*m//2, (3*b + 1)//2
def fk(m, b, k):
if k == 1:
return f(m, b)
return fk(*f(m, b), k - 1)
d = set()
for p in range(1, 5 + 1):
for b in range(2**p):
m2, b2 = fk(2**p, b, p)
if m2 < 2**p:
d.add((2**p, b))
# Verify surviving residues mod N (must be a power of 2)
N = 32
for k in range(N):
for a, b in d:
if k % a == b:
break
else:
print(k)
— Hugo Spinelli ( talk) 20:17, 22 October 2023 (UTC)
Terence Tao has worked on this problem quite a bit. He has held two notable public lectures about the problem: in 2021 and 2023. He mentions some things about the problem which are not mentioned here. Most importantly, if the conjecture were true, than we would some theorem about the difference between powers of two and those of three – which we do know, but demand quite sophisticated techniques to prove (see The Collatz conjecture, Littlewood-Offord theory, and powers of 2 and 3). Felixsj ( talk) 11:24, 1 November 2023 (UTC)
like the Collatze conjecture, there may be another one as below: input an integer number as 'n', then:
m = mod(n,3);
if m == 0 n = n/3; elseif m == 1 n = 4*n - 1; else n = 4*n + 1; end
this procedure will end up to 1 for any number (!)
Hadijavadi ( talk) 11:11, 2 November 2023 (UTC)
Step #3 (clearing trailing 0s) should be done as step #1. Why? We obviously want to clear the trailing 0s from even numbers, but for odd numbers, clearing the trailing 0s from an odd number does nothing (there aren't any), hence it's safe to do. This also removes the restriction on the input of the algorithm (currently only allows odd numbers). The algorithm converges onto 4 (0b100) instead of 1 since it clears 0s at the beginning of the loop, not the end.
--
The edits I'm asking:
Move the #3 up to #1 in the algorithm.
The machine will perform the following three steps on any odd number until only one 1 remains:
->
The machine will perform the following three steps on any binary number until 4 remains:
--
As a side note, this code does something similar to the abstract machine, in that it clears trailing zeroes in one step. "x // (x & -x)" clears trailing zeros. It works because "(x & -x)" extracts the lowest set bit which is then divided out.
x = 27 while x != 4: x = 3 * (x // (x & -x)) + 1 print(bin(x), x)
"clear trailing zeroes. add 2x+1 to x. repeat"
x = 27 while x != 4: x //= (x & -x) x += (x << 1 | 1) print(bin(x), x)
2605:A601:A629:3300:7738:24BD:85A0:9FC7 ( talk) 05:47, 13 November 2023 (UTC)
A006577: Number of halving and tripling steps to reach 1 in '3x+1' problem, or -1 if 1 is never reached.
If n is 0 or negative, then A006577(n) = -1.
0: 0, 0, 0, 0, 0, 0, 0, ...
-1: -2, -1, -2, -1, -2, -1, ...
-3: -8, -4, -2, -1, -2, -1, -2, -1, ...
-5: -14, -7, -20, -10, -5, -14, -7, -20, -10, -5, -14, -7, -20, -10, -5, ...
-6: -3, -8, -4, -2, -1, -2, -1, -2, -1, ...
-9: -26, -13, -38, -19, -56, -28, -14, -7, -20, -10, -5, -14, -7, -20, -10, -5, -14, -7, -20, -10, -5, ... 2A00:6020:A123:8B00:3913:1297:6B6B:CCEF ( talk) 13:19, 14 December 2023 (UTC)
This function is mentioned in the French page and avoid testing parity. [ [2]] Japarthur ( talk) 09:02, 8 March 2024 (UTC)
This article could have used an animation for trinary numbers, say, a .GIF of 4975 being broken down. 81.89.66.133 ( talk) 13:01, 2 February 2024 (UTC)
Collatz Conjecture Solution The Collatz conjecturela is one of the most famous unsolved problems in mathematics. The conjecture asks whether repeating two simple arithmetic operations will eventually transform every positive integer into 1. /info/en/?search=Collatz conjecture The solution in simple words, all number made out of 1. Like 1=1 2=1+1 3=1+1+1 4=1+1+1+1| 5=1+1+1+1+1 Etc. I am saying not only 1 is repetitive but 4,2,1 is repetitive. 3x+1 in if x=1 then, 3(1)+1= 4, then as per rules 4/2 =2 then 2/2=1 means 4,2,1 Now if x=2 then 3(2)+1=7 then as per rules 3(7)+1=22, then 22/2= 11, then 3(11)+1= 34 then 34/2= 16, 16/2=8, 8/2=4, 4/2=2, 2/2=1. Now if x=3 then 3(3)+1=10, 10/2 = 5, 3(5)+1=16, 16/2=8, 8/2=4, 4/2=2, 2/2=1| If we see in all solutions starting from one of the small integers 4,2,1 is repetitive. Because in x=3 there is ans 5. allAll ISSN:3006-4023 (Online),JournalofArtificiallntelligence GeneralScience (JAIGS)86 GaurangkumarPatel20894 ( talk) 07:01, 5 April 2024 (UTC)