Posted on 10/06/2021 12:32:40 PM PDT by JMack
Thank you! I like that tag line, too.
Ask Penny. She may be able to help you.
https://www.youtube.com/watch?v=6gA-ZzVgmRc
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.
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.
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.
# Wtite a number as a sum of powers of 2.
Uh, it’s a shell script, not a program. :-)
Here’s another version that bails out on any number lower than itself. You can put together the whole path by jumping from the last number in the string to where it was calculated.
1
2 1
3 10 5 16 8 4 2
4 2
5 16 8 4
6 3
7 22 11 34 17 52 26 13 40 20 10 5
8 4
9 28 14 7
10 5
11 34 17 52 26 13 40 20 10
One of the thing I find to be interesting is that there doesn’t seem to be much rhyme or reason to the number of steps it takes for any series of consecutive numbers. In the below you can see that 27 is a big outlier.
zprc:~ :) collatz2 10 30 | grep step
10 takes 6 steps, and 16 is the highest number reached.
11 takes 14 steps, and 52 is the highest number reached.
12 takes 9 steps, and 16 is the highest number reached.
13 takes 9 steps, and 40 is the highest number reached.
14 takes 17 steps, and 52 is the highest number reached.
15 takes 17 steps, and 160 is the highest number reached.
16 takes 4 steps, and 16 is the highest number reached.
17 takes 12 steps, and 52 is the highest number reached.
18 takes 20 steps, and 52 is the highest number reached.
19 takes 20 steps, and 88 is the highest number reached.
20 takes 7 steps, and 20 is the highest number reached.
21 takes 7 steps, and 64 is the highest number reached.
22 takes 15 steps, and 52 is the highest number reached.
23 takes 15 steps, and 160 is the highest number reached.
24 takes 10 steps, and 24 is the highest number reached.
25 takes 23 steps, and 88 is the highest number reached.
26 takes 10 steps, and 40 is the highest number reached.
27 takes 111 steps, and 9232 is the highest number reached.
28 takes 18 steps, and 52 is the highest number reached.
29 takes 18 steps, and 88 is the highest number reached.
30 takes 18 steps, and 160 is the highest number reached.
# 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.
Ya, I can see that. It’s a great thing to show to kids though. They really dig how it works. Anything that gets kids interested in math is a good thing IMO.
The steps are calculable, I am sure because you can calulate them from the powers of 2 in a number. I just did the math as Collatz sees it on 9932 from 3077 and the drop to 577.
The way Collatz sees it is 3077 is (1+2^2+2^10+2^11)
If you put that in 3x+1, and multiply it, and represent 3 as (1+2) it will look like this:
(1+2)(1+2^2+2^10+2^11) +1
(1 + 2 + 2^2 + 2^3 + 2^10 + 2^11 + 2^11 + 2^12)+1
4 + 2^2 + 2^3 + 2^10 + 2^11 + 2^11 + 2^12
2^2 + 2^2 + 2^3 + 2^10 + 2^11 + 2^11 + 2^12
At this point the same exponents combine, bumping up the exponent by one, so two 2^2s add together and produce a 2^3, which produces two 2^3’ that add and produce a 2^4, and so on. After that, it looks like
2^4 + 2^10 + 2^13
This shows both the high number, the next low odd number, and the number of divisions, if you draw out the 2^4 like this:
2^4 x (1+2^6+2^9)
Four divisions, down to (1+2^6+2^9), which is 577.
The key is every time you draw out the 2^x, those exponents are all reduced, even if the number pops up occasionally as nearby exponents combine. Because once the combine and rise, they will all go down inexorably.
True. This one was deadly though. It would hook people bad. I got caught.
I think the stock answer is https://arxiv.org/archive/math
My dad is a retired math professor. I sent him your proof. He may have a suggestion. I guess I’ll bookmark this thread to post if he has a suggestion.
It was enjoyable reading your proof.
Thank you for sharing it here.
I hope it gets recognized.
I see you found a way to divide by 0 on the second page!
(just kidding)
=)
Thank you!
Thank you.
Strangely I cannot get anyone to look at it and either tell me I am wrong somewhere or show me what it is they don’t understand.
Thank you for reading it, and the kind words.
I did look at it. I suspect many did. I think one has to get immersed in it to evaluate it.
Well, the answer is intuitively obvious. The tricky part is describing that answer mathematically.
I understand. I appreciate you taking a look.
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