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To: JMack

It will be interesting to see if someone can validate your proof. I’ve always thought the collatz conjecture was really cool. I wrote a few shell scripts that allowed you to play with it in different ways.

The first takes a number and gives you all of the steps, like so:

$ collatz 65
65 196 98 49 148 74 37 112 56 28 14 7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
Starting with the initial value of 65, it takes 27 steps to reach unity.

The second takes start/stop numbers.
$ collatz2 10 20
10 5 16 8 4 2 1
10 takes 6 steps, and 16 is the highest number reached.

11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
11 takes 14 steps, and 52 is the highest number reached.

12 6 3 10 5 16 8 4 2 1
12 takes 9 steps, and 16 is the highest number reached.

13 40 20 10 5 16 8 4 2 1
13 takes 9 steps, and 40 is the highest number reached.

14 7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
14 takes 17 steps, and 52 is the highest number reached.

15 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
15 takes 17 steps, and 160 is the highest number reached.

16 8 4 2 1
16 takes 4 steps, and 16 is the highest number reached.

17 52 26 13 40 20 10 5 16 8 4 2 1
17 takes 12 steps, and 52 is the highest number reached.

18 9 28 14 7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
18 takes 20 steps, and 52 is the highest number reached.

19 58 29 88 44 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
19 takes 20 steps, and 88 is the highest number reached.

20 10 5 16 8 4 2 1
20 takes 7 steps, and 20 is the highest number reached.

One of the interesting things I found out when doing analysis of the numbers is that there is clustering around certain numbers. For instance the number 9232 is the highest number reached in an amazing number of cases.


64 posted on 10/06/2021 6:45:04 PM PDT by zeugma (Stop deluding yourself that America is still a free country.)
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To: zeugma

If you have an odd number, multiplying it by 3 and adding one would produce an even number.

7 22 11

If you have an odd number, add to it half of the preceding number. Add one.

7 10 11

Don’t write the intermediate step.

7 11

You have an odd number. How many consecutive odd numbers follow it?

Wtite a number as a sum of powers of 2.


65 posted on 10/06/2021 7:02:44 PM PDT by TTFX ( )
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To: zeugma

It is messed up how it catches your attention, isn’t it?

If I am right, and you crack the hood on it, the rules are viewing the odd numbers as the sum of one and increasing powers of two. What is determining if the odd numbers go up or down is how many powers of two there are, and how close they are. So a number like 7, with two powers of two that are sequential, (1+2^1+2^2) will end up rising (to 11), while a number like five with one power of two, or a number like 13, with two powers of two that are separated (1+2^1+2^3) will decrease. The equations showing this are at the end of the proof.

9232 being the highest number for so many has to do with how the odd number before it is represented exponentially (3077 = 1+ 2^10 + 2^11) it is a product of only two exponents, and perfectly maxed out with only one added. If you jumped up to a base odd number of 6145, or (1 + 2^11 +2^12), and multiply it by three and add one (18,436), you will have the next plateau number I would bet.

I just want to see the problem put to bed now. It feels like all this work will have been for naught if there are still people out there wasting brain power on it when they could be discovering more interesting things.


66 posted on 10/06/2021 7:35:56 PM PDT by JMack
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