Patterns of Collatz

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
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References:
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• The Simplest Math Prob...
• UNCRACKABLE? The Colla...
• Collatz Conjecture in ...
en.wikipedia.o...
mathworld.wolf...
www.dcode.fr/c...
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[@Indie_Dennis]

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.

[@andrejlochman6672]

Your videos are criminally underrated! Please keep uploading ❤

[@DcubedJ]

Thanks for your kind words!

[@pythonforintermediates8566]

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!

[@catsdontboot8735]

Your channel is so underrated! Continue the good work!

[@DcubedJ]

Thanks!

[@st3althyone]

Thank you for showing us how beautiful math can be.🥰🥰

[@AlgoRhythmsWithAyush]

That was a fascinating explanation! top notch quality content, great work

[@DcubedJ]

Thank you for your comment!

[@TheLuckySpades]

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

[@johndickinson82]

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.

At least the specific cycle at 1:31 can easily be ruled out: 9c/4+5/2=c would have to match for it. A calculator quickly spits out -2 (or error, when restricted to positive integers).

[@Matrix_Architect]

fascinating video!

[@DcubedJ]

Glad you enjoyed it :)

[@ГеоргиГеоргиев-с3г]

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.

[@ГеоргиГеоргиев-с3г]

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.

[@parisi.]

This is beautiful!

[@mahmoudattalla2972]

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: [phone].

[@riadsouissi]

👍👍👍

[@jonorgames6596]

Next video: How to prove the conjecture?

[@kurmetsultan]

Problem 3n+1 solved! kzbin.info/www/bejne/m6i9n5mMhJiqjpo

[@mahmoudattalla2972]

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: [phone].

[@ruaidhridoylelynch5522]

That proof is false

[@ruaidhridoylelynch5522]

I don’t see what ur logical reasoning is the first part

[@ruaidhridoylelynch5522]

Bc ur first assumption is not logical all subsequent logic is flawed