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Old 2019-03-19, 12:36   #1
Cybertronic
 
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Default 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 2019-03-19 at 12:51
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Old 2019-03-19, 13:12   #2
Cybertronic
 
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Here in english:


A "prove" that the Collatz-sequenz 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 Collatz-sequenz we have a ratio of 4:2.
That follows we have over all in a six-member-block one 1/16 and one times 9.
Or ! round 9/16~56% is the last member of a predecessor-6-block.

Fact, lim 0.56^x -> 0 (specific ...4,2,1)
(x number of blocks,we required to continue the Collatz-sequenz)


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 2019-03-19 at 13:42
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Old 2019-03-19, 16:12   #3
CRGreathouse
 
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It's a good heuristic, but not a proof.
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Old 2019-03-19, 17:31   #4
Cybertronic
 
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Quote:
Originally Posted by CRGreathouse View Post
It's a good heuristic, but not a proof.

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 10-11-10-11 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 )
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Old 2019-03-20, 08:40   #5
Cybertronic
 
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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 2019-03-20 at 09:06
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