20190319, 12:36  #1 
Jan 2007
Germany
2^{2}×7×13 Posts 
Collatz 3x+1 problem
Hello members !
Yesterday I got an idea that solve maybe this problem. This idea is wrote in german, but maybe it is understandable. Here the Link to post no. 40 (pzktupel , thats me) https://matheplanet.de/matheplanet/n...rt=40#p1752861 Thanks for feedback. regards Norman (pzktupel) Okay, I try it to write in english....later more Last fiddled with by Cybertronic on 20190319 at 12:51 
20190319, 13:12  #2 
Jan 2007
Germany
364_{10} Posts 
Here in english:
A "prove" that the Collatzsequenz ends ever with 4,2,1. My idea is. Take a natural Number N and transform into the dual expression. Consider only the last 2 digits. We get 4 combinations. There are 00,01,10 and 11 We have 2 Collatz regulations : 3x+1 for odd numbers and 0.5x for even numbers. Now: combination > after Collatz regulations ..00 > ..00 or ..10 ..01 > ..00 ..10 > ..01 or ..11 ..11 > ..10 You see easy, there are 4 ways to halve the successor and only 2 ways to increase threefold the predecessor (+1). Over all members in a Collatzsequenz we have a ratio of 4:2. That follows we have over all in a sixmemberblock one 1/16 and one times 9. Or ! round 9/16~56% is the last member of a predecessor6block. Fact, lim 0.56^x > 0 (specific ...4,2,1) (x number of blocks,we required to continue the Collatzsequenz) Example: N=91 We get : 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 Total:92 numbers 59 are even 33 are odd Ratio: ~2:1 ... it is also round 1~ 91 * 3^33 / 2^59 More examples: N=131071 , odd: 80 , even: 144: 131071*3^80/2^144 ~1 (0.8..) 144*9/16~81 interations (80 is it) N=89898989 , odd: 98, even: 182 N*3^98/2^182~1 (0.8...) 182*9/16~102 interations ( 98 is it ) Its works :) Conclution: Start with an number N and the sequenz goes to 4,2,1, because 3^(2x) is smaller than 2^(4x). That was my way. Thanks. Last fiddled with by Cybertronic on 20190319 at 13:42 
20190319, 16:12  #3 
Aug 2006
3×1,993 Posts 
It's a good heuristic, but not a proof.

20190319, 17:31  #4 
Jan 2007
Germany
2^{2}×7×13 Posts 
Here, I will show, that never run a sequenz endless ... Addition: There are 2 possible critical endless loops. First critical loop: This run go to infinite. (0.5 * 3 * 0.5 * 3 *...) 10>11>10>11> .. Every number N=4k+2 ends with ..10 Every number N=4k+3 ends with ..11 start with ..10: N=4k+2 I:halve it, to get a number ..11 : N=2k+1 II:3N+1 it , to get a number ..10 : N=6k+4 III:halve it, to get a number ..11 : N=3k+2 IV:3N+1 it , to get a number ..10 : N=9k+7 V:halve it, to get a number ..11 : N=4.5k+3.5 There are 4 strings for k: k=0,4,8,... "I" fails, because 2k+1 is not member of 4k+3 for ending ..11 k=1,5,9,... "V" fails, because 4.5k+3.5 is not member of 4k+3 for ending ..11 k=2,6,10,.. "I" fails, because 2k+1 is not member of 4k+3 for ending ..11 k=3,7,11,.. "V" fails, because 4.5k+3.5 is not member of 4k+3 for ending ..11 So, it is show, that for every k never run this loop 10111011 to infinite. Second critical loop: 10>01>00>10>01>00>... This loop can not run to infinite. ( 0.5 * 3 * 0.5 * 0.5 * 3 * ....~ 3^n/8^n) > members get a smaller value ) 
20190320, 08:40  #5 
Jan 2007
Germany
2^{2}·7·13 Posts 
One loop I forgot
3rd critical loop: This can go to infinite. (0.5 * 3 * 0.5 * 0.5 * 3 ...) 10>01>00>10>11>10.. start with ..10: N=4k+2 I:halve it, to get a number ..01 : N=2k+1 II:3N+1 it , to get a number ..00 : N=6k+4 III:halve it, to get a number ..10 : N=3k+2 IV:halve it, to get a number ..11 : N=1.5k+1 V:3N+1 it, to get a number ..10 : N=4.5k+5 k=0,4,8,... "V" fails, because 4.5k+5 is not member of 4k+2 for ending ..10 k=1,5,9,... "I" fails, because 2k+1 is not member of 4k+1 for ending ..01 k=2,6,10,.. "III" fails, because 3k+2 is not member of 4k+2 for ending ..10 k=3,7,11,.. "II" fails, because 6k+4 is not member of 4k for ending ..00 For all k's , this loop can not run to infinite. Summary: possible dual combination > after Collatz regulations ..00 > ..00 or ..10 ..01 > ..00 ..10 > ..01 or ..11 ..11 > ..10 There are 2 critical loops where more increased threefold than halved. (condition for undless run) These are: loop 1: (0.5 * 3 * 0.5 * 0.5 * 3 ...) 10>01>00>10>11>10.. (9/8>1) loop 2: (0.5 * 3 * 0.5 * 3 *...) 10>11>10>11> 10.. (3/2)>1 Both I showed that for all numbers this is not possible. All other loops have more halved than increased threefold. ( numbers get smaller and smaller und end to 4,2,1. ) Last fiddled with by Cybertronic on 20190320 at 09:06 
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