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
