Another fascinating video. I first came across this conjecture in Douglas Hofstadter's seminal work."Godel, Escher, Bach: An Eternal Golden Braid". It's amazing that what appears to be a simple conjecture, which can be easily understood by a six rear old, is so intractable.

excellent! very clear and informative, thank you.

***is nothing more then the use of a property of power of 2 , a method to bring a number to a form((2)^n) but I think you can do it only using 1 or 3 to cover it for all integers , for n>3 maybe can cover only any Integers , you can try!***

Havent heard about Polya and Mertens conjectures - Mertens wasnt defined here but i found that it is based on Mertens function which uses Móbius function :) and it was stated by some guy called Stjeltes (Dutch) who wrote a letter to Hermitte :) - everybody knows Hermitian transformation or polynomials :) or a Hermitian array in quantum physics.

Terence Tao has recently made some progress on the conjecture.

@rylaczero3740

4 жыл бұрын

Yes, he showed that Collatz conjecture is true for 'almost' all numbers.

@brendawilliams8062

2 жыл бұрын

You have a 375 and a 133 that are competitors.

@nosuchthing8

Жыл бұрын

Doubtful

@maynardtrendle820

Жыл бұрын

Me too...and I don't even know who this Tao guy is! Real bush league combinatorics. I'm an Erdősian, m'self. ...aaand, tip the fedora- I'm out. 🐢

Sir I have one doubt tht 1 should come r not or should we get another number

Collatz Cycle(s) explained: kzread.info/dash/bejne/hHyfqtRqlZmdh6g.html

Peculiar that 73 also reaches 9232 at most, and takes 115 steps to reach 1.

Thanks for the vids so much. Your videos help me look much of a mathematician in front of my friends.

@discovermaths

4 жыл бұрын

Glad we could help!

@kulbinderkaur4780

2 жыл бұрын

@@discovermaths kzread.info/dron/HNja6OgnOzqbO5DoJkVs0g.html Here is the solution for COLLATZ CONJECTURE•

What is the part that is not solved: whether or not there is a number that will continue forever and not descend down to the loop; or the equation for finding the number of iterations of each number used?

@brendawilliams8062

2 жыл бұрын

A loop needs to be a slam dunk or triangulation in the hidden layers of an algorithmic triangulation will have a problem. This is what I feel it’s about. I am no mathmatician but just see it that way. An hour glass can stretch it’s neck and you’d like to figure where you are at and if you can call home.

@wprandall2452

2 жыл бұрын

@@brendawilliams8062 I don't understand the first sentence. What is "algorithmic triangulation""

@brendawilliams8062

2 жыл бұрын

@@wprandall2452 when you want to make use of 100079917 x 1000564416 x .. you need triangles. You can’t just fly in that

@wprandall2452

2 жыл бұрын

@@brendawilliams8062 Oh! I see. Actually I don't. I read up on triangulation method. I think I've used some of that. Collatz Conjecture is multi-faceted, and there are other answers to be found besides whether all progressions go down to the loop or continue on into outer space. The answer is that all progressions go down to the loop. There are a number of reasons why, and I'm not sure I want to tell them at the moment. Now I used a spreadsheet to find some of the answers, and I don't see how anyone can do anything with Collatz Conjecture without it. I'm not sure what program you're using or by what method you're approaching the problem. Have you gained any ground on it?

@brendawilliams8062

2 жыл бұрын

@@wprandall2452 it’s wonderful fun to study. No one ever knows it all. Thankyou for your reply. We are all a work in progress. I have looked at a few yt channels. Actually odd and even and the Collatz conjecture has never interested me. I resist triangles I choose rather other motion.

if even devide by 2. if odd +1. will also ends with 1 right

@Caldermologist

2 жыл бұрын

Since this would grow far slower at any given odd number, it would certainly reach one.

Can you make video about geometry (visualisation) of complex numbers arithmetic (+/-, ×/÷, ^/sqrt...)? And of course,👍for nice work.

@discovermaths

4 жыл бұрын

Thanks, Sanel. We'll be putting up two introductory videos on complex numbers in December. However, you may have to wait till 2020 for one that is specifically about geometric representation.

@kulbinderkaur4780

2 жыл бұрын

kzread.info/dron/HNja6OgnOzqbO5DoJkVs0g.html Here is the solution for COLLATZ CONJECTURE•

If we change the odd number rule, the 1 instead of 3 it can be proven easily, guess such problems they depend on how well you believe

@kulbinderkaur4780

2 жыл бұрын

@@awebmate kzread.info/dron/HNja6OgnOzqbO5DoJkVs0g.html Here is the solution for COLLATZ CONJECTURE•

something I stumbled across with the collar conjecture, is I think its proof for finding Prime numbers in an infinite system ,because of the relation of 2 and 3. Here is my conjecture, which I have yet destroyed... but it seems to work.. 100%.. but I think, perhaps I lean to wards .. 70 is percent confident in it. So it goes like this. 2 to the power of any prime number subtract one from the result. That number is either a prime or a factors into primes. You could basically then find infinitely larger pies ,simply by multiplying in this form.. IE 2 to the power of (largest known prime) subtract one from the result. its either prime or factors into primes. Which I think would make me a winner of finding the math languages largest pie, since this process can be repeated for each result that is found... from primes. Other numbers work also, but often... you have to subtract one and divide the answer by 3. which gives a prime, also... but so far... the result is pretty alarming... and pretty accurate... for hitting primes or a number that factors into primes, which makes sense to the language relation of 2 and 3. Thanks. :D

@learningbalance1226

Жыл бұрын

Your process works since gcd(2^n - 1, 2^m - 1) = 2^gcd(n,m) - 1.

I have some time to consider it a headache.

kzread.info/dash/bejne/lGyKrLivebrLfdI.html este canal dice tener la solución a la conjetura. Está por publicarlo.

That's not Paul Erdös!

@discovermaths

4 жыл бұрын

You're quite right, Patrick - it's Lothar Collatz. My apologies, I should have made that clear.

@patrickt.2136

4 жыл бұрын

@@discovermaths Of course I should have recognized our Collatz! 😃

Its because numbers are infinite but formulas break the path. Think of numbers as flowing in a straight line and think of the Collatz conjecture as the formula that can be put in the infinitely forward moving path to break it off into a new path. When it gets back to 1, it means its back to flowing forward. I believe this can be done with multiple different formulas, there is nothing special about this one.

@usg4357

Жыл бұрын

Also to define the length of the formula and how long it would take to get back to infinity, you can change the value to go negative rather than positive, continuously, from your chosen number to get back to the number you originally started from. And apply this enough times, you can get to -1 and so on. We’re just bending infinity and think its something to be proud of. If numbers always move forward 0 to infinity, then that means we’ve just made up these formulas to suit our needs. There is nothing special we are doing. People spend their time on this and they don’t realise time is fleeting. If the formula works for an even number and an odd number, that means it will work with any number you choose. Why is this so hard to understand for people?

@usg4357

Жыл бұрын

Infinity is too big for us to comprehend. Thats why I say, if it works with an even number and an odd number, then it will work with any number your mind can comprehend. Depending on the number you choose, it could take years but because it must always get back to 1, it will always work. Value of your number + Collatz conjecture = will always get back to 1 Think of even and odd as the pillars that confirm this. If all factors that can affect the numbers have been taken into account (factors being even and odd), then there is no need to test it further. You can do it with any number but im sure even the worlds best computers will take some time to get back to 1. It could take infinite amount of years for the computer if the number is great enough. Because the greatest number we can think of doesn’t really matter, infinity trumps everything. We are creating an infinite loop by applying the Collatz conjecture to a number. The only thing that would throw a wrench into “infinity” would be something like the Collatz conjecture, so there is no point to it. I’m not sure where it can be used for real world applications, but in my mind its pretty straight forward. Of course I could be wrong but I challenge anyone to prove me wrong.

Where can I get the 500 dollar reward? 0.... if you read my paper is the short analysis of collatz...only the tail has a cycle...

@ismaelfortunado

3 ай бұрын

I am writing the 120 pages proof and 2000 articles@rocknrolladube

@ismaelfortunado

3 ай бұрын

@rocknrolladube I am writing a 120 page proof and 2000 articles...

Why they stop at 1

@Caldermologist

2 жыл бұрын

It is convenient to say it ends there. If you keep going you just loop through 1 - 4 - 2 and back to 1 for all eternity. Few have time to wait that long.

I can get the nice prove

thing is there is no upper bound for the length of a collatz sequence. The proof of this is trivial. a collatz sequence of any arbitrary length n can be generated by the number 2 to the power n. So doesn't this prove Collatz?

@alfredclark994

3 ай бұрын

No. You need to show that the Collatz Conjecture applies to all the integers not just the small infinite set you identified. Alf.

Take any binary representation of a number. If the number is even all will be shifted right. (Example 1000010 becomes 0100001) now look at what happens to a section of a binary number being multiplied by 3 or 11 in binary. In trachtenburg terms put down the first number (1) and then for all other numbers add the right neighbour. We see that for any mid section of a binary number will eventually fill with 1’s (Note the section remains the same just shifted right as leading zeros are removed) When all the binary numbers reach 1 as they must eventually do ( 01 * 11 = 11 ) adding 1 to this will result in all the 1’s to become 0 while the final leftmost 0 becomes 1. All zeros will be removed according to the divide by 2 rule leaving only 1 as the answer. Thus proving the Collatz Conjecture. Your welcome!

@nosuchthing8

Жыл бұрын

Remember the right shift, or divide by two, moves the number down, while the miltiply by 3 and add one moves the number up. How do we know we will always hit an even number that is a multiple of 2, allowing it to fall back to 1?

I've solved the Collatz conjecture. Its simple. The secret is you are not adding integers. You are adding halfs. The Collatz conjecture is just a geometric series of 1/2+1/4+1/8...=1.

@nosuchthing8

Жыл бұрын

It's not that type of infinite sequence.

The most boring conjecture ever, useless.

Solved - is not difficult at all