Talk:Collatz conjecture/Archive 3

Archive 1

Archive 2

Archive 3

Longest runs

                  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) 

Generalized Collatz function

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)

Maximum residue

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)

History of the problem?

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)

PROOF OF THE COLLATZ CONJECTURE!

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)

‎This "proof" is wrong. Your most important error is that "even" and "power of two" is not the same. Yes, the theorem is obvious for powers of two, but not for even numbers which aren't a power of two. Nyh (talk)

Is this a solution?

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:

• Halving a number is really just logically shifting it towards the right by 1.
• Multiplying a number by 3 is really just logically shifting it to the left by 1! And then adding the original number onto itself.

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)

Not even wrong. There is a snowball's chance in hell that hand waving can solve such an intractable problem. Also, the conjecture concerns itself with multiplying by three and adding one, not just multiplying by three.--Jasper Deng (talk) 09:21, 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)

There is no such thing as a "proof by example". Per WP:NOTAFORUM, unless you have a serious peer-reviewed research paper that supports a positive solution to the problem, we are not going to discuss this on this talk page; you are free to discuss it more at WP:RDMA, but right now you clearly lack the rigor and depth of understanding required to even consider solutions to this problem (for one, addition of binary numbers is not identical to bitwise XOR).--Jasper Deng (talk) 09:36, 18 August 2018 (UTC)

The discussion above 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.

"Orbit" is undefined

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)

The term is now hyperlinked. 176.230.167.21 (talk) 10:00, 21 March 2019 (UTC)

Certain values for string lengths seem wrong

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)

I don't know Python or its treatment of large integers but the article is correct that the number of steps for 7,887,663,552,367 is 1563. Here is a PARI/GP verification which can be tested online at https://pari.math.u-bordeaux.fr/gp.html:

c(n) = x=n; s=0; while(x>1, if(x%2==0, x=x/2, x=3*x+1); s++); s
c(7887663552367)

It produces 1563. PrimeHunter (talk) 07:41, 15 June 2019 (UTC)

Error found, code fixed. My apologies.

Proof of the Collatz conjecture

https://www.researchgate.net/publication/296682209_Proof_of_the_Collatz_conjecture Chrisdecorte (talk) 01:07, 21 July 2019 (UTC)

Researchgate is not a reliable source. Wake me up when it's published in a legitimate mathematics journal. A hint: You will need a more rigorous proof technique than "no matter how big the numbers get, there will always be a chance". —David Eppstein (talk) 01:08, 21 July 2019 (UTC)

New proof of the convergence of the Collatz conjecture

https://doi.org/10.1155/2019/6814378 — Preceding unsigned comment added by Kamalbarghout (talk • contribs)

That's not a proof of anything. It says right at the start "approach towards a probabilistic proof". You might as well say "not really a not-really proof". And Hindawi is not a publisher of reliable sources. Their journals might not quite fit the bill of predatory publishing, but their quality control is too low and their spam too high for me to be comfortable having anything to do with them. —David Eppstein (talk) 16:39, 2 August 2019 (UTC)

Computational verification

According to this MSE answer, the record $10^{20}\approx 2^{66.4}$ is still valid as of August 2019. --DaBler (talk) 12:22, 19 August 2019 (UTC)

Pls look at it...

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

New Information to Add Question

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)

No. See WP:OR. We have to wait for material to be properly peer-reviewed and published elsewhere before it can be included here. —David Eppstein (talk) 23:25, 2 October 2019 (UTC)

Thank you for responding. I did have a mathematician higher up in the field check it, but again, it's not published anywhere. Jeanlovecomputers (talk) 02:48, 3 October 2019 (UTC)

Collatz conjecture don't only for natural numbers.

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

It appears to me that you are using floating-point arithmetic with about 52 bits of precision (53 bits is the standard double precision on PCs), and when the number gets large enough, it rounds off to an integer, which makes the results incorrect. Bubba73 ^{You talkin' to me?} 20:26, 3 October 2019 (UTC)

Also, a number like -0.49 is neither even nor odd, so how do you know which step of the process to use? Bubba73 ^{You talkin' to me?} 20:37, 3 October 2019 (UTC)

I use a long double for the format with about 80 bits of precision, this allows you to read large numbers correctly. Odd numbers are numbers that are not completely divisible by 2. (-0.49) is not completely divisible by 2.--Nikita fly (talk) 20:48, 3 October 2019 (UTC)

• • • That is not sufficient. Look at these two terms:
      • 36 1500946352969991.700000
      • 37 4502839058909976.000000
    • Anything with .7 after the decimal point, when multiplied by 3 will give something with .1 after the decimal point.
    • Assuming term 36 is correct and exact, term 37 should be 4502839058909976.1. The arithmetic you are using cannot represent it correctly. Your precision is limited to 52 or 53 bits (for the mantissa).

• • • See the links in my comment above and odd number. Bubba73 ^{You talkin' to me?} 21:09, 3 October 2019 (UTC)

Step 23 is the first error. It claims 313810595.590000 × 3 + 1 = 941431787.770001. Your sequence should always have two decimals and never reach an integer. It's possible you have about 80 bits of precision but I guess you fail to round to two decimals after each step, so the error accumulates. For a simpler example, start at 0.5. Every time you multiply by 3 and add 1, you should get something point 5, never reaching an integer. PrimeHunter (talk) 22:20, 3 October 2019 (UTC)

There is an 80-bit extended type (long double) available on some platforms, but only 64 bits are used for the mantissa. (The rest for exponent and sign.) That gives about 19 decimal digits of precision, which is more than his calculations show. Bubba73 ^{You talkin' to me?} 00:24, 4 October 2019 (UTC)

I speculated that the error accumulates by becoming larger on average after each step because he computes on an already inaccurate value and then rounds the result again. The error might e.g. start on the 19th digit, then move to the 18th digit after some steps, and so on. He only prints 6 decimals so maybe "313810595.590000" is actually something like 313810595.5900002222 in the computer after accumulating an error 22 times. Times 3 plus 1 is 941431787.7700006666 which prints as "941431787.770001" with 6 decimals. But maybe the error wouldn't accumulate fast enough for this. PrimeHunter (talk) 08:07, 4 October 2019 (UTC)

If I use the 80-bit floating point number (64-bit mantissa), my results start to differ at #23. If I use double, they agree until #31, then they differ in the 5th place after the decimal point. (And that indicates about 14 digits of precision, which indicates about 50 bits of precision.) Perhaps he is doing some decimal rounding at each step. Bubba73 ^{You talkin' to me?} 01:45, 5 October 2019 (UTC)

Relevance to mathematical theory

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)

You did see the relevant and sourced quotes from Erdős and Lagarias on this topic, near the top of the article, right? —David Eppstein (talk) 23:41, 10 November 2019 (UTC)

That's completely irrelevant here. Those quotes are about the difficulty of this problem, not about what specific significance it has to mathematical development. 37KZ (talk) 01:30, 11 November 2019 (UTC)

Incorrect. The significance is hinted at by the Erdős quote: it appears that we need to develop new methods in mathematics to make progress. —David Eppstein (talk) 01:35, 11 November 2019 (UTC)

I get that. My point is that the quote by Erdos says nothing about what specific mathematical theories would need to be advanced in order to solve this problem. Prime factorization theory is one such area (what happens to the prime factorization when 3n becomes 3n + 1?). --37KZ (talk) 12:21, 15 December 2019 (UTC)

Are you suggesting a change to the article, and if so, what is it? —JBL (talk) 13:55, 15 December 2019 (UTC)

It has been suggested and agreed by various accounts that it’s somehow tied to how addition and prime factorization interact. --37KZ (talk) 14:15, 15 December 2019 (UTC)

So what change are you suggesting to the article? --JBL (talk) 15:30, 15 December 2019 (UTC)

This is probably not a very relevant answer to your question, but the Collatz function can be used to multiply large integers. --DaBler (talk) 12:34, 16 May 2020 (UTC)

Experimental evidence

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)

Number 27 and 4:3

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 https://wiki.riteme.site/wiki/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)

Sorry, but this is considered WP:OR (a web page you wrote and a paper you wrote). Wikipedia does not publish such original work.--Bill Cherowitzo (talk) 19:48, 27 March 2020 (UTC)

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)

The article already contains the trajectory of the number 27. --JBL (talk) 20:47, 27 March 2020 (UTC)

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 origin The 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)

I understand your liking of this. Indeed, I do. But, alas, it is overwhelmingly WP:OR, so just isn’t going to happen. Indeed, if it were to happen by mistake, any of several editors would then undo it. The prohibition against WP:OR is that strong. JDAWiseman (talk) 14:15, 3 September 2020 (UTC)

Our Response to 4.2 A probabilistic heuristic:

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, $e_{max}$, in the Collatz sequence. $k=\left\lfloor {\frac {log(e_{max})}{log(2)}}\right\rfloor$. Therefore, $({\frac {3}{2}})^{\frac {t}{2}}*({\frac {3}{4}})^{\frac {t}{4}}*({\frac {3}{8}})^{\frac {t}{8}}*...*({\frac {3}{2^{k}}})^{\frac {t}{2^{k}}}\rightarrow 0$ as $t\rightarrow \infty$ ... 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)

Divisions by Two

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 $3x+1$ as well as in the generalized variant $kx+1$, where $k\in \mathbb {N} _{>0}$.

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:

${\hat {\alpha }}=\lfloor n*log_{2}k+log_{2}{x_{1}}\rfloor +1$

The parameter $n$ represents the count of odd numbers in the sequence and $x_{1}$ stands for the first odd number. Whenever the maximum ${\hat {\alpha }}$ 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:

${\bar {\alpha }}=\lfloor n*log_{2}k\rfloor +1$

Are these findings of interest for Wikipedia? We are looking forward to the discussion --C4ristian (talk) 16:51, 18 November 2020 (UTC)

Your article is published in a journal listed on Beall's list of predatory journals, and therefore is not a reliable source by Wikipedia standards. It cannot be used as a reference. —David Eppstein (talk) 07:12, 19 November 2020 (UTC)

Thank you very much for your reply. This argument seems to be a bit subjective to me. We could alternatively publish our findings on arXive or at publication servers at our universities. We would have imagined a discussion about the content not the formalities, however. Do you consider the findings interesting for the article? As far as I know these theorems have not been published yet. We are happy about any substantive feedback --C4ristian (talk) — Preceding undated comment added 07:30, 19 November 2020 (UTC)

Thank you also from my side for your comment. The great advantage of this community is the discussion on a content-related basis and the resulting chance to gain knowledge and progress. It may be that there exist unfortunate experiences with one or another journal, but does this mean content published there is no longer worth discussing? Let me give a counterexample: The paper published at https://www.scipress.com/IJPMS.19.10 has even be cited by an European Research Council (ERC) research under the EU’s Horizon 2020 research and innovation programme: https://arxiv.org/pdf/2007.06979.pdf

That's why I'm so convinced that we all make progress and gain benefit by dealing with the findings in terms of content. There is also a university version of the paper available (but this is again just a formality): https://opus4.kobv.de/opus4-ohm/frontdoor/index/index/docId/620 Please let us see together if the limits (described above by logarithm) are basically relevant and provide an impulse for further research. If really the formality is the key, then of course we are able to upload the paper into arXiv. —Sultanow (talk) 09:12, 19 November 2020 (UTC)

Articles “published” on preprint servers and your individual university websites would likewise not be reliable sources — you should read the link David Eppstein shared to understand that this formality is of high importance on Wikipedia. More substantively, though, no, I doubt very much that this result will be mentioned in the Wikipedia article any time soon, for many reasons (the failure to meet WP:RS being the easiest to articulate briefly). —JBL (talk) 11:55, 19 November 2020 (UTC)

To be honest, I am a little surprised. The Wiki article currently refers to Terence Tao's WordPress website, which in turn refers to an arXiv paper. In the past I have published a number of articles on German Wikipedia. The discussions there seem more open and content-oriented to me. In Wiki DE, universities are at least as reliable a source as public websites. The name of the university does not play a major role there. It seems to me that in Wiki EN the name of the author and the institution are of primary importance. If there are reasons why you do not find the theorems interesting, we are open to any discussion. As I wrote in my previous post, we have hoped to get scientific feedback, even if it would require some words to articulate C4ristian (talk) 13:51, 19 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:

• Beall's list has been closed for years and has been highly controversial (see Wikipedia article). Is it up to date? Is it representative?
• Is arXiv a reliable source?

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)

You seem confused about the purpose of Wikipedia: it is not a venue for promoting, publishing, or getting feedback on your research. (If you wanted those things, you probably should have submitted your work to the kind of academic journal that subjects papers to a rigorous review process.) With respect to Terry Tao, per WP:RSSELF, 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.
Here is a basic rule of thumb: if a person comes to Wikipedia and their sole purpose is to mention/reference/promote their own work, they are going to get a frosty reception. If, on the other hand, and expert comes to Wikipedia and writes (from their position of expertise) about foundational material in the field, or about the most important aspects of various topics, they are likely to be welcomed warmly. Occasionally, the two descriptions I just gave will be in tension with each other, but that is clearly not the case in this instance. --JBL (talk) 20:14, 22 November 2020 (UTC)

If self-promotion had been our main goal, we wouldn't have started this discussion. Our question was "Are these findings of interest for Wikipedia?". That implies that the answer can be no. We hoped to get a discussion based on arguments and not on a list that has been closed for years. Beall's List was shut down in 2017 due to its controversies and is now being maintained by an "anonymous postdoctoral European researcher". Do you consider this a reliable source? --C4ristian (talk) 21:20, 22 November 2020 (UTC)

I think you have received a pretty clear answer to your question. --JBL (talk) 22:22, 22 November 2020 (UTC)

I agree with others that the answer is "no" to the question "are these findings of interest for Wikipedia", for reason of limited impact (at least so far). Russ Woodroofe (talk) 22:25, 22 November 2020 (UTC)

Thank you, I accept this argument, of course --C4ristian (talk) 23:22, 22 November 2020 (UTC)

@JBL: I am not criticising your decision, but the underlying rationale and the way it was expressed --C4ristian (talk) 23:22, 22 November 2020 (UTC)

I have left a message on your talk page on this matter --C4ristian (talk) 12:09, 23 November 2020 (UTC)

Potential original research?

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)

Indeed - I have reverted the recent addition as obviously in violation of WP:OR. —JBL (talk) 13:20, 25 December 2020 (UTC)

Subsection "As a binary relation" in the section "Other formulations of the conjecture"

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 $n\rightarrow 1$ about the transitive reflexive binary relation $\rightarrow$ which satisfies conditions $2n\rightarrow n$ and $x\rightarrow 3x+1$.

Similar formulation represents weaker version of Collatz conjecture: $\forall n\in \mathbb {N}$ it is proved that $n\leftrightarrow 1$ for the transitive reflexive symmetric binary relation $\leftrightarrow$ which satisfies conditions $2n\leftrightarrow n$ and $n\leftrightarrow 3n+1$.

93.72.108.33 (talk) 08:57, 28 July 2021 (UTC)

References

Similar formulation represents weaker version of Collatz conjecture No, 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)

You have a right to POV, but it seems a bit contradictory. "Completely different" things can't be "superficially connected" too. Collatz conjecture modification above in its formulation is obviously similar to the original. As for Academia Letters online journal, don't confuse it with whole Academia portal. It even has a page at Publons. Academia Letters publish articles only after positive peer reviews, and they publish the peer reviews along with articles. Anyone can check who recommended article for publication and for what reasons. Their open access publishing business model isn't predatory, because you can publish your articles free-of-charge in Academia (not peer-reviewed) and in Academia Letters (peer reviewed). Only additional services such as personal website, mass download of articles, or detailed reading analytics are paid. I like the Academia Letters experiment because it helps to engage wider circle of peer reviewers than limited pools of classic journals. Multidisciplinary nature of the edition allows to transcend boundaries of narrow academic discourses. Of course, among tens of articles I reviewed only few were publishable. My suggestions to reject bad-quality articles were supported by editors. But I think everyone deserves objective explanation of her or his mistakes, at least several sentences useful for further improvement. Generally, Academia organized effective process of publishing and reviewing; it is admirable contribution to development of academic communication. --93.72.108.33 (talk) 14:24, 28 July 2021 (UTC)

That the paper is published in Academia Letters is not necessarily a reason not to include it as a source. However, there is no evidence of impact (it would be surprising if there were for a 2021 paper), and this is such a reason. I don't see it as WP:DUE to include this, at least at this time. Russ Woodroofe (talk) 15:23, 28 July 2021 (UTC)

Is WP:DUE standard applicable here? I think it isn't a question of majority/minority viewpoints whether a modified conjecture is right or wrong. It is proved, and nobody claim the proof is wrong. The proof is so simple that almost anybody can understand it.

And what do you think about including to the Wikipedia article the short description of Collatz conjecture formulation as a binary relation, without any mentions about the Academia Letters publication or the modified conjecture it describes? Such formulation is well-known and it is already used in the article, there are arrows in the table in subsection "Iterating on all integers" without explanations what the arrows mean. The binary relation formulation I proposed to add just above the table will give neccessary explanation. --93.72.108.33 (talk) 16:02, 28 July 2021 (UTC)

You'd need sourcing. Well-known by who? Russ Woodroofe (talk) 16:06, 28 July 2021 (UTC)

There is definitely no chance of your contribution being added to the article, please stop. --JBL (talk) 16:38, 28 July 2021 (UTC)

To put it another way: even if we believed in the reliability of its publication, and in the relevance to this problem, Google Scholar lists over 1000 publications containing the phrase "Collatz conjecture". We currently cite 37 of them. What evidence do you have that the results in this new publication are among the most significant 5% of results on this topic? —David Eppstein (talk) 22:07, 28 July 2021 (UTC)

New prize being offered

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)

If reliable secondary sources report on it, then it will become worthy of mention (per WP:DUE). prnewswire just reprints press releases, it is not better than a promotional primary source. --JBL (talk) 10:18, 31 July 2021 (UTC)

Incorrect statement of conjecture

"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)

Agreed. It should be "term" or "current term" instead of "previous term". Lemondevon (talk) 16:50, 13 January 2021 (UTC)

The phrase is a standard English construction: each term of the sequence 1, 2, 4, 8, 16, ... is twice the previous term, each year of the Trump administration was worse than the previous year, etc. "Preceding" could also be used, with the same meaning, but it's a bit more formal and wouldn't hold up to repetition as well. --JBL (talk) 17:53, 13 January 2021 (UTC)

Also, it is conventional on WP that new talk sections go at the bottom, rather than the top; I have relocated this one. --JBL (talk) 17:54, 13 January 2021 (UTC)

No, your examples have only 'term' and 'previous term' in them; while in the article text you have 'previous term', 'term' and 'next term' which is not precise. — Preceding unsigned comment added by Lemondevon (talk • contribs) 00:07, 9 April 2021 (UTC)

Maybe it would be less ambiguous if we used "new term" instead of "next term"? For example like this:

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 next new term is one half of the previous term. If the previous term is odd, the next new term is 3 times the previous term plus 1.

―JochemvanHees (talk) 12:42, 9 April 2021 (UTC)

Yes, that sounds like a good compromise. What is your preference? Alternatively just leave out 'new' as well: "Then each term is obtained from the previous term as follows: if the previous term is even, then the term is one half of the previous term. If the previous term is odd, then the term is 3 times the previous term plus 1." Lemondevon (talk) 19:51, 9 April 2021 (UTC)

I am also happy with "new" in place of "next". --JBL (talk) 19:53, 9 April 2021 (UTC)

Your version would work as well. The only nitpick is that in "if the previous term is even, then the term (...)", it's possible to interpret "the term" as referring to the aforementioned "previous term". But I don't think that would be a real issue, either of them work. ―JochemvanHees (talk) 21:54, 9 April 2021 (UTC)

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??).

The full statement of the problem is already given immediately below, in the "Statement of the problem". The first paragraph of the article is not an appropriate place to repeat a repeated, redundant, and overwhelming copy of the same full statement of the problem, already given immediately below. It is a place to summarize briefly, sometimes even in a sloppy way (if the sloppiness can be corrected later), in order to give readers a quick idea of what the article is about. —David Eppstein (talk) 22:43, 18 August 2021 (UTC)

The 1937 date seems to be wrong

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)

Table?

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)

I second this. It reads a bit awkward. Fynsta (talk) 11:50, 29 August 2021 (UTC)

---

Leong Ying

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)

I don't think we should be paying a lot of attention to a self-published book on this topic by a fringe physicist, unless it gets a lot of independent and reliably-published attention. —David Eppstein (talk) 21:54, 12 October 2021 (UTC)

Discussion question

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)

Not an edit request and not really a request for help editing Wikipedia.

The behavior for even numbers is regular and, since it always results in a lower number, relatively uninteresting for exploring the behavior of the sequence. The caption mentions that even numbers are omitted, so I would not call it misleading. — jmcgnh^{(talk) (contribs)} 03:12, 24 October 2021 (UTC)

I thought maybe the caption might be hidden on mobile but I just checked and, at least on my Android phone, it's still visible. On the other hand, you have to scroll quite a way through the first paragraph, unsolved box, and illustration before you get to it, and it's in tiny print, so I see how it could easily have been overlooked. —David Eppstein (talk) 05:52, 24 October 2021 (UTC)

Lede

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)

I was trying to make the lede more grammatical and more readable to the lay reader. Surely the people here can see the problems with the language? How about changing the the first sentence to:

"The Collatz conjecture is a conjecture in mathematics that concerns sequences, which are defined as follows:"

If that's not ok, I'd welcome it if someone else can give it a stab. LK (talk) 03:47, 2 December 2021 (UTC)

"The Collatz conjecture is a conjecture in mathematics that concerns sequences defined as follows:" Fynsta (talk) 08:15, 2 December 2021 (UTC)

I think the first sentence should at least say something specific to the Collatz conjecture, rather than just stating something vague that it is about sequences. None of the versions under discussion do that, but Lawrencekhoo's versions made it even vaguer. So I prefer the earlier text, but I hope it might be possible to draft a first sentence that is more specific than the current one (which just says that it is about integer sequences) without getting overlong. We can save room by not repeating that it is a conjecture. Maybe something like "The Collatz conjecture in mathematics asks whether all positive integers reach one when certain simple arithmetic operations are performed repeatedly on them."? —David Eppstein (talk) 08:57, 2 December 2021 (UTC)

Alternatively "The Collatz conjecture in mathematics asks whether a certain sequence of simple integer operations will always reach one, regardless of at what integer the operations begin"? "All positive integers reach one" seems a little strange to me. Then the next sentence can begin "More precisely, we start with any positive integer n" and continue from there. Some care may be needed to distinguish the sequence of operations from the sequence of numbers. Russ Woodroofe (talk) 13:04, 2 December 2021 (UTC)

Some care is needed to state positivity. The Collatz iteration has other loops for negative numbers. —David Eppstein (talk) 17:46, 2 December 2021 (UTC)

Yes, would be better as "regardless of at what positive integer the operations begin". Russ Woodroofe (talk) 18:32, 2 December 2021 (UTC)

Hmmm. "The Collatz conjecture in mathematics asks whether a certain procedure involving simple arithmetic operations will always arrive at the number one, no matter which positive integer the procedure begins with"? The phrase "reach one" sounds awkward to me, for some reason. XOR'easter (talk) 22:08, 3 December 2021 (UTC)

Phrased that way, it sounds like you're asking whether a certain function (somehow defined procedurally) is the constant-1 function, which misses the point. "The Collatz conjecture in mathematics asks whether repeating certain simple arithmetic operations will eventually transform every positive integer into one"? —David Eppstein (talk) 00:01, 4 December 2021 (UTC)

That's better than mine. I think I like it the best of everything suggested so far, actually. XOR'easter (talk) 03:15, 4 December 2021 (UTC)

I don't like the word "transform". The number isn't really transformed, it is talking about the result of a sequence (of functions applied to the number). But that can get too wordy. Bubba73 ^{You talkin' to me?} 06:51, 4 December 2021 (UTC)

Award

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)

But please, if you do find such sources, absolutely do not do what Fmwithstuff did: the lead section should summarize the body of the article, it's not a collection area for every PR stunt related to this topic. --JBL (talk) 23:56, 16 March 2022 (UTC)

Wikipedia is a place for all good information; information about that award belongs on this page. Cassie Schebel, almost a savant. <3 (talk) 00:33, 17 March 2022 (UTC)

Incorrect. See WP:INDISCRIMINATE: "merely being true, or even verifiable, does not automatically make something suitable for inclusion in the encyclopedia". —David Eppstein (talk) 00:39, 17 March 2022 (UTC)

thats not quite what that section is talking about... its relevant, and similarly relevant things appear in the article; the movie incendies, for example. Also; that isn't how good is defined. beyond being true, "good" information is appropriate and desirable. Cassie Schebel, almost a savant. <3 (talk) 02:40, 17 March 2022 (UTC)

Yes, the movie is mentioned in the body but not the lead section -- exactly like I requested. And then for some reason you objected to the straightforward point that "just dump it in the lead" is not an appropriate way to include new minor information. It is unlikely that this content is "good" (it has all the markings of a pointless PR stunt), and the sourcing and placement of the addition were definitely not "good"; but also you should probably not be lecturing vastly more experienced editors about the basic principles of Wikipedia. --JBL (talk) 13:28, 17 March 2022 (UTC)

The unreliable citation is absolutely a reason to remove the topic altogether. Even if better sources existed (I found none), the topic would not merit inclusion in the introduction. XOR'easter (talk) 04:06, 17 March 2022 (UTC)

I oppose any mention. https://bakuage.com/en/about/ at the official company site says Employees 0 and Capital 3,000,000 JPY. I'm very sceptical they would actually pay out 120 million JPY. The alleged prize sounds like a PR stunt. PrimeHunter (talk) 16:13, 17 March 2022 (UTC)

Extention to larger domain

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)

The new additional conjecture of the Collatz conjecture.

(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: If a randomly selected positive integer is an even number, the result of dividing the number by 2; The number will be either an odd number or an even number. In addition, if a randomly chosen number is odd, if the number is multiplied by 3 and added by 1, the result will be an even number. Therefore, when the iteration results are examined, it is seen that no positive integer, except for the number 4, continues in a successive "odd-even-odd-even..." cycle in the whole iteration sequence. In the iteration sequence of any positive integer, at least 2 consecutives even numbers will appear. As a result of the experimental study: As a result of applying the Collatz conjecture to a randomly selected positive integer, there is no "odd-even-odd-even..." loop in the entire iteration sequence of the randomly selected positive integer (except the number 4). When the numbers from 1 to 100,000 are examined: - 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.

Cycle Length

Is ac=0 an error here? If not, just what does that mean? John D. Goulden (talk) 21:51, 1 December 2021 (UTC)

It means that at least one of a and c is zero (see Zero-product property), or equivalently that the formula for the period can be simplified to one of two forms with either the a term or the c term removed. —David Eppstein (talk) 22:01, 1 December 2021 (UTC)

• I think I know how. Once X>1 = 2^N, it never becomes odd** by dividing by 2; it never does the 3X+1 thing. 81.89.66.133 (talk) 08:14, 17 March 2022 (UTC)
  • Like I stated above with "X>1", it "never becomes odd" until it is just one digit 1. 81.89.66.133 (talk) 08:14, 17 March 2022 (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)

CollatzOdds.svg

01:27, 24 January 2023‎ IM Serious

‎bold graph replacement better illustrating the structure, feel free to add margins to the graph on commons undo Tag: Reverted

prompted

01:31, 24 January 2023‎ User:David_Eppstein

‎Undid good faith revision 1135335502 by IM Serious (talk). New image is illegible at displayed size, hard to read because of arrows going every which way, and fails to make the acyclicity of the graph visually apparent. undothank Tag: Undo

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)

Your phrasing presupposes that this illustration should be improved and prominently placed. You are asking the wrong question. The right question is: what information if any is inadequately illustrated in the current article, and what is the best way to improve the existing illustrations or add more in order to better present that information? —David Eppstein (talk) 02:02, 24 January 2023 (UTC)

Yes, the current picture kinda hides the crazy hailstoning resisting an inductive proof which i display in the snailwound walk from 25, or the overshooters 15, 23, 27. I also display the periodicities in rising or falling, and having no or infinite preimages, from above only or from above and below, and so on. IM Serious (talk) 09:11, 24 January 2023 (UTC)

It is extremely difficult to extract any information at all from this image. Your description of what information you think is displayed (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)

Explanation for up and down and 4 rules for reversing Collatz sequences

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:

• n mod 4 = 3 let the net result (the number divided by 2 until it is odd) grow, until the iteration reach a number with mod 4 = 1
• n mod 4 = 1 let the net result (the number divided by 2 until it is odd) shrink, until the iteration reach a number with mod 4 = 3 (or the number is 1)

Additionally there are 4 rules for reversing the Collatz sequences:

1. Any number n, which leads with the Collatz-rules to the end loop 4-2-1, multiplied by 2x (x from 1 to ∞), will also lead to 4-2-1; the result will be even and you can either repeat this instruction or, if the result’s mod 6 = 4, you can subtract 1 and divide its result by 3 for a new number (which will be odd, thus, this instruction or instruction 2 can be applied, or depending on the result‘s mod 3, instruction 3 or 4)
2. Any odd number n, which leads with the Collatz-rules to the end loop 4-2-1, multiplied by 4 and then added 1, will also lead to 4-2-1; and because the result is again an odd number, this instruction or instruction 1 can be applied, or depending on the result‘s mod 3, instruction 3 or 4
3. Any odd number n with n mod 3 = 2, which leads with the Collatz-rules to the end loop 4-2-1, subtracted (n+1)/3, will also lead to 4-2-1; and because the result is again an odd number, instruction 1 or 2 can be applied or, depending on the result‘s mod 3, this instruction or instruction 4
4. Any odd number n with n mod 3 = 1, which leads with the Collatz-rules to the end loop 4-2-1, added (n-1)/3, will also lead to 4-2-1; and because the result is again an odd number, instruction 1 or 2 can be applied or, depending on the result‘s mod 3, this instruction or instruction 3

Source:

• How to proof the Collatz conjecture - an approach by reversing the sequences

Thanks for your help! Chears from Germany (iovialis)...

--Iovialis (talk) 20:39, 3 April 2022 (UTC)

You can't use 3n-1.de as a source. It gives a unbelievably flawed elementery proof of Collatz conjecture. And its author is utterly clueless when it comes to advenced mathematics. Felixsj (talk) 17:16, 17 February 2023 (UTC)

All these formulas are already contained in section "Optimizations" --B wik (talk) 18:26, 18 February 2023 (UTC)

Russian anonymous jumps in.

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:

• divide even numbers by 2 means "leave zeroes behind"
• multiply by 3 means "sum two adjacent digits for whole number"
• X+1... well, summing.

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's money offer

Bakuage Co., Ltd, a Japanese company, offers 120 million yen (c. USD 919,280 as of March 2023) for the resolution of Collatz cojeture. 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)

Copying my post from Talk:Collatz conjecture/Archive 3#Award: "I oppose any mention. https://bakuage.com/en/about/ at the official company site says Employees 0 and Capital 3,000,000 JPY. I'm very sceptical they would actually pay out 120 million JPY. The alleged prize sounds like a PR stunt."

A year later the about page still says Employees 0 and Capital 3,000,000 JPY. PrimeHunter (talk) 19:58, 23 March 2023 (UTC)

Oppose This is obviously a publicity stunt by a company that expects it will never have to pay the prize, and probably does not have the resources to pay if it were claimed.

Regarding the sources, the first four sources are all just copies of the same press release and are not independent coverage. The fifth source consists of two vague sentences that don't even mention Bakuage. The sixth source is a Youtube video in which Tao briefly mentions the prize, again without naming the company. This does not add up to coverage in reliable sources. CodeTalker (talk) 20:17, 23 March 2023 (UTC)

Thank you for your replies. I see that I was probably fooled by a PR stunt.

I'll try to get the claim removed from the german page. Felixsj (talk) 20:28, 23 March 2023 (UTC)

References

Why were the edits with new statements from a *peer reviewed paper* undone?

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)

Pinging @Frietjes: who performed this revert. —David Eppstein (talk) 00:40, 25 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 reverted myself. I found this edit while cleaning up WP:Database reports/Transclusions of non-existent templates, since the edit introduced Template:≤ which is not a valid template. Frietjes (talk) 14:14, 27 March 2023 (UTC)

3x+1

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)

Hundreds of people with a very limited understanding of math has also claimed so. You are most likely not an exception. Felixsj (talk) 17:02, 17 February 2023 (UTC)

Straght outta Wikipedia:Humorton 81.89.66.133 (talk) 08:07, 3 April 2023 (UTC)

Collatz page needs update, loop proven impossible

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)

No, certainly not. The reasons for this are obvious but they begin with WP:RS. --JBL (talk) 17:50, 24 April 2023 (UTC)

You seem to be talking about Xn+Y "general" formula, not the OG 3n+1 array. 81.89.66.133 (talk) 13:25, 28 April 2023 (UTC)

Problem is, "normal" 4-->2-->1-->4-->2-->1-->4 loop also counts. 81.89.66.133 (talk) 13:29, 28 April 2023 (UTC)

Please try publishing your ideas as a paper. Also, please try creating an account to use its talk page for a chat on the topic - maybe I'll make some graphs for you81.89.66.133 (talk) 10:26, 26 June 2023 (UTC)

Generalization of Collatz conjecture

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)

Great, but are you saying there should be a section in the article about extending the concept to all primes, not just 3? And if so, can you find papers written about his generalization? Dhrm77 (talk) 13:07, 9 May 2023 (UTC)

Thans for Your comment.

in fact, I have been looking for Such papers for a while, but could not find one. Therefor I wondered whether I was the first finding out that… what I can hardly believe but I would of course be very proud.

anyway, I am convinced that my generalization is correct, but it definitely needs more investigation.

Adding it to the wiki page could trigger off such investigations Derijan (talk) 13:26, 9 May 2023 (UTC)

5x + 1 and 7x + 1 were tested to 10¹⁶ in https://sweet.ua.pt/tos/px+1.html. PrimeHunter (talk) 15:07, 9 May 2023 (UTC)

That is very interesting. Thank you.

however, my algoritme for Px + 1 Seems to be a bit different than Yours. In each iteration I Divide by Every prime number smaller than P (where possible) before multiplying by “P” and adding 1.

And if my assumption is correct, than the 3x + 1 would be juist a particular case of the Px + 1 conjecture Derijan (talk) 15:33, 9 May 2023 (UTC)

That is also effectively what my link does. Maybe you mean that if the trajectory is written then you would include fewer numbers. It doesn't change whether you reach 1. A small script quickly found that 11x + 1 has the cycle 17, 188, 94, 47, 518, 259, 37, 408, 204, 102, 51. It's in OEIS:A057614. 13x + 1 has the cycle 19, 248, 124, 62, 31, 404, 202, 101, 1314, 657, 219, 73, 950, 475, 95. It's in OEIS:A057522. 15:52, 9 May 2023 (UTC)PrimeHunter (talk)

Maybe a revised conjecture could be that you always reach a cycle but there may be multiple cycles, meaning some of them don't include 1. PrimeHunter (talk) 15:55, 9 May 2023 (UTC)

You are completely right. I wonder why I haven’t Seen that myself.

not sure if there should be a reference to the fact that a cycle would be reached in any case. In fact, that Goes without saying. The beauty of the 3x + 1 is only that such cycle is around the number 1, meaning that you never reach the same number > 1, although not proven yet. Derijan (talk) 09:28, 10 May 2023 (UTC)

It's not known whether 3x + 1 always reaches a cycle. See Collatz conjecture#Stopping times. PrimeHunter (talk) 12:31, 10 May 2023 (UTC)

Generalization in binary form

I have an idea. Instead of thinking of natural numbers and arithmetics of "reaching 1", we could maybe try a binary pattern.

• 3*N means N + 2*N ---- or "N + 10N" in binary, in base-two.
• Imagine it now as 2 identical braids laying each next to other. But with a little, 1-pearl-place shift. (see radix)
• Imagine both braids to have a pearl for binary "1" and nothing, an empty space for binary "0"
• Imagine a new braid being sewn out of combining those two. Rules: Sewing 2 pearls together destroys/removes one pearl and shifts one pearl up by one space (binary "1+1=10") where is also may collide with another pearl and so on.
• Imagine the next 3N+1 operation as "sewing 2 braids together" + "add a pearl"
• As for "divide an even number by 2", imagine it as fraying away a single-space piece of the braid
• In a sense, 2^N numbers (such as 16, 32, 128, 256, 2048, 65536 etc.) are on their way to reach 1-2-4-1-2-4-1... loop AND those can be represented with only one pearl + some empty space.
• Partial generalization Try imagining 3N+1 or 9N+1 or ((1+2^nat)N+1) as the same 2 braids, but with a different shift. 81.89.66.133 (talk) 10:23, 26 June 2023 (UTC)
• General question: is it possible to fool such a mechanism to give you more than one pearl in your braid? (for example, once you make a number that fools the sewing computer into looping, you claim a many-pearl braid as a reward because "the client is always right") 81.89.66.133 (talk) 10:23, 26 June 2023 (UTC)

  This talk-page is not a forum for discussion of the article subject, nor for making conjectures or attempted proofs related to it. --JBL (talk) 17:24, 26 June 2023 (UTC)

  OK, saved a spreadsheet with this idea of future publishing somewhere else. You see, I tried to adapt the Collatz's task to someone who is good at patterns, but bad at calculus e.g. cannot handle that "advanced math full of sigmas and curly braces". 81.89.66.133 (talk) 11:19, 28 June 2023 (UTC)

Semi-protected edit request on 2 August 2023

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)

 Not done: it's not clear what changes you want to be made. Please mention the specific changes in a "change X to Y" format and provide a reliable source if appropriate. Cannolis (talk) 07:30, 2 August 2023 (UTC)

Needs "trinary" representation

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)

Strange tense to use for a mathematical theorem

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)

It is true that statements are either true or not, but this is a partial result. He doesn't show that it is true, but he gives evedence for it. The Collatz conjecture claims that the stopping time is always finite, whereas Tao kinda showed that the stopping time isn't "too big" (which is weaker but still remarkable). Felixsj (talk) 17:11, 17 February 2023 (UTC)

I think that nice "almost all" phrase is also okay, since math-people are trying to find a possible exception.

It's like a detective novel of a whodunit kind: in succh a story, its man character need to grasp some clues... A LOT of clues... and make the verdict. Terence Tao's verdict is, if such an exception exists, it's has to be defined with a really weird formula. 81.89.66.133 (talk) 08:06, 3 April 2023 (UTC)

Not so strange if "are descending" is copula+adjective rather than a progressive verb. —Tamfang (talk) 04:09, 17 April 2023 (UTC)

The Collatz orbit is the numbers that constitute the sequence using the Syracuse function. Descending means that at some point, the sequence reaches one or more numbers that dip below the origin number. Almost all (which is a defined mathematical term meaning basically : not infinite, few and far appart). Should someone find that *all* sequences for all natural numbers eventually go below the origin point, the Collatz conjecture would be verified... What Tao shows is that the conjecture has a high probability to be true. It is not a definite resolution, but an advance in the right direction. Ivan Scott Warren (talk) 17:36, 18 August 2023 (UTC)

Error re Syracuse function

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.

 Done: [1]. --Hugo Spinelli (talk) 11:17, 1 October 2023 (UTC)

WP:TECHNICAL tag

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 $5\rightarrow 16\rightarrow 8\rightarrow 4\rightarrow 2\rightarrow 1$. -- Hugo Spinelli (talk) 15:24, 5 October 2023 (UTC)

I've removed the tag; Sundayclose is welcome to come here and leave meaningful comments about the article. JBL (talk) 17:31, 5 October 2023 (UTC)

Surviving residues mod 32

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)

 Not done That line is correct. For example, 9 can't be a surviving residue because it is 1 mod 4. The article even mentions that it must be 3 mod 4. A citation would be welcome, but it's not required for simple calculations used only as examples (see WP:CALC). If you wish to check that the surviving residues are indeed 7, 15, 27 and 31, run the following Python script:

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)

I see. Thank you Brentonb (talk) 21:30, 23 October 2023 (UTC)

On what more we can add

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)

I would rather rely on secondary sources for that. The article is already a bit long. Could you give a summary of what else you think should be added? We could create a new section of related results or consequences, like in Riemann hypothesis § Consequences, or use Collatz conjecture § Other formulations of the conjecture. By the way, his quote in the second paragraph that the Collatz conjecture is "unlikely be proven by current technology" is noteworthy in my opinion. —Hugo Spinelli ^(talk) 16:22, 1 November 2023 (UTC)

another conjecture

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)

This is called Lu Pei's conjecture: http://www.numbertheory.org/php/lu_pei_3branch.html --DaBler (talk) 16:46, 3 November 2023 (UTC)

Improving As an abstract machine that computes in base two

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)

Same person coming back, a month later (coincidentally). Normally someone replies within this timeframe, no? Otherwise IMO the article shouldn't be locked. 49.205.149.156 (talk) 21:51, 13 December 2023 (UTC)

 Not done. The algorithm is correct as it is. Moving step #3 to #1 would require to change the halting condition to a weird one, so there's no clear advantage. The article gives some emphasis on the sequence restricted to odd numbers, so the algorithm feels natural this way. —Hugo Spinelli ^(talk) 07:31, 14 December 2023 (UTC)

0 and negative numbers = 1 is never reached

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)

Worth mentioning h(x)=(7x+2-(5x+2)cosπx))/4 ?

This function is mentioned in the French page and avoid testing parity. [[2]] Japarthur (talk) 09:02, 8 March 2024 (UTC)

See the section Collatz_conjecture#Iterating_on_real_or_complex_numbers. --JBL (talk) 17:50, 8 March 2024 (UTC)

The article could have used a visualization in base 3 (in trinary calculation)

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)

Please clarify and expand that. —Tamfang (talk) 19:56, 27 March 2024 (UTC)

It’s solved by Gaurangkumar Patel

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.https://wiki.riteme.site/wiki/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)

WP:NOR Felixsj (talk) 15:37, 5 April 2024 (UTC)

Yes. This talk page is only for discussing improvements to the article, which can only be based on reliably published sources. If you believe you have a solution, this is not the place for publicizing it, nor for getting it published, nor for encouraging people to explain why it does not constitute a valid proof. —David Eppstein (talk) 20:03, 5 April 2024 (UTC)

But it’s true data it’s about giving right information to public we don’t fool them. GaurangkumarPatel20894 (talk) 02:21, 7 April 2024 (UTC)

The way to get information to the public is to publish it in respectable academic journals, first. —David Eppstein (talk) 03:26, 7 April 2024 (UTC)

Felixs are David are right. Bubba73 ^{You talkin' to me?} 03:37, 7 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*2^2n-1, results in a set of odd numbers which can be multiplied by 2ⁿ or 2^n-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 2^6,000,000 -1, whereas the largest known megaprime tested in the project was 2^82,589,933 − 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 2^10,000,000 + 1 in 48 minutes. --DaBler (talk) 18:57, 18 April 2024 (UTC)

This is still 2^72,589,033 smaller than the largest tested number. 87.236.191.246 (talk) 06:06, 19 April 2024 (UTC)

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)

 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)

Collatz 2nd loop proven impossible in 5 steps of easy logic.

Wikipedia is not an appropriate venue in which to publish your original research. --JBL (talk) 17:31, 9 June 2024 (UTC)
The following discussion has been closed. Please do not modify it.
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: 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, ... 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... The academic use of Wikipedia is low. Wikipedia is not the right tool to rock an academic boat. 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)