Patterns of Collatz
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- Опубликовано: 23 июн 2023
- The rules of the Collatz conjecture are simple, yet it remains unsolved since its creation in 1937. This video will not attempt to prove or disprove the conjecture but will analyse the patterns and visualisations that arise from its properties.
#Collatz #SoME3
Some of the math animations used in this video was created using Manim - www.manim.community/
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References:
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• The Simplest Math Prob...
• UNCRACKABLE? The Colla...
• Collatz Conjecture in ...
en.wikipedia.org/wiki/Collatz...
mathworld.wolfram.com/Collatz...
www.dcode.fr/collatz-conjecture
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Наука
Amazing job dissecting the problem.
Interesting fact about your findings:
7+32x-1 = -25
27+32x-1 = -5
31+16x-2 = -1
These numbers are all part of the loops in the negative number range.
These videos deserve SOO many more views. They are so good and are even better than Numberphile and other math channels. Keep making these videos please!
Your videos are criminally underrated! Please keep uploading ❤
Thanks for your kind words!
Your channel is so underrated! Continue the good work!
Thanks!
fascinating video!
Glad you enjoyed it :)
Thank you for showing us how beautiful math can be.🥰🥰
Another reason we can ignore the case where a sequence goes below the starting number when there is a lower number in the repeating loop is because the lower number in the loop will already give a counterexample and we only need the counter with the smallest starting number
That was a fascinating explanation! top notch quality content, great work
Thank you for your comment!
A new pattern I found with triangles-> Something I found to think about: there are triangles of twins( numbers one apart in the Collatz)a represents the top number of the triangle and below it is a+1 and the restriction is a+2 is prime.Twins are vertical in these triangles and reach a connection in the same number of moves as the rest of the twin ex: 20,21 both reach 16 after 3 iterations. I’m not sure why 144, 145, and 146 have this behavior, reaching 94 after several moves. The triangles “start” with a(2^n -1)+(a-2), where the start are the values on the left. PS. a=-1 is an exception but is a triangle . What happens in the middle and on are that when you apply the Collatz sequence to it the twins and connections form. I think I am the first to find these “Collatz Triangles” and that they are helpful to the proof. There are patterns with a+1= 0 and 2 mod4 but I’ll leave those for someone to figure out. -Isaac D.
This is the reason why you should use (3x+1)/2 as the odd rule, counting as 1 step. Then you will see that all combinations of s steps repeats with a period of 2^s.
Then you can generalize the patterns like this: If a number n goes below itself in s steps, then every number (2^s)+n will also go below itself in s steps.
This is beautiful!
5:03 1) their indexes are divisible by two (2x)
So you end up at indexes of type 3x. (2steps for 2+8x'th odd, ending up at 3+12x'th odd,4+8x'th odd, would therefore end up at 9+18x and so on but after 4steps)
2)indexes of type 3x(not divisible by two) always point to ones of type 4x+1(where x is now a different offset) therefore you can apply 4x+1->x(3 steps but even, so check other branch a step back) (or 4(x)+3->4(3x-1/2)+6(x-1/2)+3(non- odd math makes it hard) but then you are at an even odd x and not at an odd odd x and you need to check step 1)
3)leaving you at number smaller than the initial by ~3/4 (you can derive the exact value)
What's your point, the conjecture is already true because you can't construct a number with infinite power as infinity is not a number.
So you stop increasing after n steps and go into a spiral down. The same 1241 loop but with the end being smaller than the initial and longer in length.(to be peer reviewed)
It's hard to keep two numbering systems parallelly in your head while describing complex topics. So that's the excuse for the potential mistakes.
What i mean to say is you have only 3 outcomes, ((((5+6x)*2)-1)/3)=3+4x
(((1+6x)*4)-1)/3=1+8x
And everything else, where 4(y)=(y) where y=3x+1 going down to step 1 or 2
The first increases at (3/2)^n
The second decreases at (3/4)^n
The third decreases as 1/(4^n)
Where n stops at indivisibility and you have to swap n and x with smaller n or x.
The problem seems to be intimately related to binary and also to powers of 3 which tend to give long sequences. Eg 9 to 28 to 14 to 7 to 22 to 11 to 34 to 17 to 52 to 26 to 13 to 40 to 20 to 10 to 5 to 16... like turning a power of 3 into a power of 2 at some point.
Hey, thanks for your comment. Yeah.. I agree there is definitely something about powers of 3 and longer sequences. I think one of the difficulties with this approach and also the approach in the video is finding some generalisations that apply to all numbers...
👍👍👍
N= positive odd number. N changes to (3N+1)/2
(3N+1)/2 could be:
1- (3N+1)/2 = positive odd integer
2- (3N+1)/2 = positive even integer
1- assume (3N+1)/2 = positive odd integer, Since N = positive odd integer, it could be a value for (3N+1)/2
So, (3N+1)/2 = N
3N + 1 = 2N
3N - 2N = -1
N = -1, which contradicts with N = positive odd integer.
So, the assumption (3N+1)/2 = positive odd integer is false.
2- assume (3N+1)/2 = positive even integer, Since N + 1 = positive even integer, it could be a value for (3N+1)/2
So, (3N+1)/2 = N + 1
3N + 1 = 2N + 2
3N - 2N = 2 - 1
N = 1, which does not contradict with N = positive odd integer.
So, the assumption (3N+1)/2 = positive even integer is true, and (3N+1)/2 will change to smaller value (3N+1)/4 < N, getting toward the destination 1.
If N = 1, (3N+1)/4 will equal 1, which is a term within the destination loop 1 → 2 → 1.
So, N = positive odd integer, just changes to (3N+1)/2 = positive even integer, which changes to (3N+1)/4 < N, getting toward the destination 1. So, Collatz conjecture is true.
Eng. Mahmoud Attalla.
WhatsApp: +20 1112669096.
Next video: How to prove the conjecture?
Problem 3n+1 solved! ruclips.net/video/ewZnfVO3wX8/видео.html
N= positive odd number.
N changes to (3N+1)/2
(3N+1)/2 could be:
1- (3N+1)/2 = positive odd integer
2- (3N+1)/2 = positive even integer
1- assume (3N+1)/2 = positive odd integer.
Since N = positive odd integer, it could be a value for (3N+1)/2
So, (3N+1)/2 = N
3N + 1 = 2N
3N - 2N = -1
N = -1, which contradicts with N = positive odd integer.
So, the assumption (3N+1)/2 = positive odd integer is false.
2- assume (3N+1)/2 = positive even integer.
Since N + 1 = positive even integer, it could be a value for (3N+1)/2
So, (3N+1)/2 = N + 1
3N + 1 = 2N + 2
3N - 2N = 2 - 1
N = 1, which does not contradict with N = positive odd integer.
So, the assumption (3N+1)/2 = positive even integer is true, and (3N+1)/2 will change to smaller value (3N+1)/4 < N, getting toward the destination 1.
If N = 1, (3N+1)/4 will equal 1, which is a term within the destination loop 1 → 2 → 1.
So, N = positive odd integer, just changes to (3N+1)/2 = positive even integer, which changes to (3N+1)/4 < N, getting toward the destination 1. So, Collatz conjecture is true.
Eng. Mahmoud Attalla.
WhatsApp: +20 1112669096.
That proof is false
I don’t see what ur logical reasoning is the first part
Bc ur first assumption is not logical all subsequent logic is flawed