The Simplest Math Problem No One Can Solve - Collatz Conjecture
Video Overview & Insights
The Collatz Conjecture is the simplest math problem no one can solve — it is easy enough for almost anyone to understand but notoriously difficult to solve. This video is sponsored by Brilliant. The first 200 people to sign up via https://brilliant.org/veritasium get 20% off a yearly subscription.
Why not prove that every number can be the written in the form 3n+1 not from infinity but from 1.
And that set equal to R ad infinity
Special thanks to Prof. Alex Kontorovich for introducing us to this topic, filming the interview, and consulting on the script and earlier drafts of this video.
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So these examples were whole numbers. What happens if you start using numbers with increasingly long decimal strings?
References:
Lagarias, J. C. (2006). The 3x+ 1 problem: An annotated bibliography, II (2000-2009). arXiv preprint math/0608208. — https://ve42.co/Lagarias2006
What is there to solve?
Lagarias, J. C. (2003). The 3x+ 1 problem: An annotated bibliography (1963–1999). The ultimate challenge: the 3x, 1, 267-341. — https://ve42.co/Lagarias2003
Tao, T (2020). The Notorious Collatz Conjecture — https://ve42.co/Tao2020
The fact that anyone would waste time on this when the explanation is so simple is ridiculous. I mean u created a set of rules that result in that outcome then question why it results in that outcome... umm. It's the rules u applied. If we add another rule... say that when 4 or 1 are reached you add 96. You could cycle it back down again.
So yeah i can see why if any mathematician were to say they were working on it theyd be seen as insane... but i think theres a more politically incirrect term for it... "retarded".
Again.. the rules ensure the inevitable result and the loop. Thats the outcome the rules create. Duh. Again... duh.
A. Kontorovich and Y. Sinai, Structure Theorem for (d,g,h)-Maps, Bulletin of the Brazilian Mathematical Society, New Series 33(2), 2002, pp. 213-224.
A. Kontorovich and S. Miller Benford's Law, values of L-functions and the 3x+1 Problem, Acta Arithmetica 120 (2005), 269-297.
Now I can understand the song 4,2 ka 1. 1,2 ka 4 my name is chhagan😂😂
A. Kontorovich and J. Lagarias Stochastic Models for the 3x + 1 and 5x + 1 Problems, in "The Ultimate Challenge: The 3x+1 Problem," AMS 2010.
Tao, T. (2019). Almost all orbits of the Collatz map attain almost bounded values. arXiv preprint arXiv:1909.03562. — https://ve42.co/Tao2019
its solved.
because mathmaticians find the mean of both results if its a infinitely repeating loop . this means that
implement notation DESMOS
1+1-1+1-1+1... = mean(1 , 2) = 1.5.
1-1+1-1+1-1+1... = 0.5
well. lets apply it here.
implement notation DESMOS
ticker(off , z -> collatz(z) , 1)
collatz(c) = {c/2 = round(c/2) : c/2 , 3c+1}
z=7
now. this will simulate the same exact collatz conjecture. in 1 millisecond a lap
since the ticker is off, it wont even do that. if we set it to on, then z would fry your cpu clocks and wont stop till it is solved.
we get a repeating 1 , 2 , and 4.
lets find the mean of that
mean(1 , 2 , 4) = (1+2+4)/3 = 7/3 = 2.33333333...
so there you go mathmaticians you happy now? a gen alpha kid solved such a question and barely anyone could do that
Conway, J. H. (1987). Fractran: A simple universal programming language for arithmetic. In Open problems in Communication and Computation (pp. 4-26). Springer, New York, NY. — https://ve42.co/Conway1987
The Manim Community Developers. (2021). Manim – Mathematical Animation Framework (Version v0.13.1) [Computer software]. https://www.manim.community/
Rayo's number's parity is permanently and provably unknowable, therefore the Collatz rule cannot be meaningfully applied to it, therefore it exists outside the conjecture's domain, and a conjecture with cases it cannot address is incomplete, not universally true
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What are we actually solving for 3x + 1? Why do we want it to be true or false. What does true mean and what does false mean for a collatz coverture?
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Written by Derek Muller, Alex Kontorovich and Petr Lebedev
3x + 1 = -⅓
Animation by Ivy Tello, Jonny Hyman, Jesús Enrique Rascón and Mike Radjabov
Filmed by Derek Muller and Emily Zhang
Ik the guys grandson
Edited by Derek Muller
SFX by Shaun Clifford
16:42 i realized that im watching this as some disgussion of nerds over marvel heroes or etc 😅😂 this is not science . Its “science based on our maded system” . Its like i will say Hi and you People will try to play with characters Hi with All possibilities 😅 its same as geeks do - they argue over fiction systems . Who would win Hill or thor ? Their strenght and etc is system based on fiction . And I think I jave an answer why People last 100-150 years didnt actually didnt learn anything new . Its career and system of “science” . Basically you sit there and researching and trying find a way how to fit into “known” circles of carrierist and their theories . You find a sweetspot and you win Nobel prize 😅 but this shouldnt be science . Its researching or data researcher. True science should be based on finding out what is universe how its made and etc not this . This is like We have light . We dont ser atoms so When We flashlight it We see reflections of light and works of EM. Okay cause of our math needs 26 dimensions We will say its not a particle 😅😂
Additional video supplied by Getty Images
Produced by Derek Muller, Petr Lebedev and Emily Zhang
10:44 man i think too much time given to “magic of Numbers” simple reason why everything comes from some Numbers its structure . Its not Numbers but structure of Numbers and math 😅 so if you want to solve it - you have to invent new structure to a math 😅
3d Coral by Vasilis Triantafyllou and Niklas Rosenstein — https://ve42.co/3DCoral
Coral visualisation by Algoritmarte — https://ve42.co/Coral
thats why i made a solution like graham did for his problem! have fun reading all of this
Sinteplexionsis — My Ultimate Number for the 4-2-1 Problem
I created Sinteplexionsis to solve an extremely hard recursive problem called the 4-2-1 problem (a massive extension of ideas like the Collatz conjecture). The core power of this number is that it is so enormous and self-referential that it can recover itself. Even if you divide it, multiply it, or damage it in any way, its recursive structure allows it to rebuild and restore its full value. Lower levels like Sintex already have this recovery ability, and it grows stronger at every stage.
I didn’t just make an existing number bigger by adding digits. I created entirely new mathematical numbers built from the previous ones, using the full hierarchy as raw material.
The Full Hierarchy
Sintax = ((G^G)^TREE(3))^G * TREE(3)
then
Sintaplex = Sintax(Sintax(...(Sintax)...)) [Sintax times]
then
Sintaplexa = Sintaplex(Sintaplex(...(Sintaplex)...)) [Sintaplex times]
then
Sintox = Sintaplexa^Sintaplexa
then
Sintoplex = Sintox(Sintox(...(Sintox)...)) [Sintox times]
then
Sintoplexis = Sintoplex process repeated Sintoplex times
then
Sintex = Sintoplexis process repeated Sintoplexis times
then
Sinteplex = All previous steps (1 to 7) from the beginning, using Sintex as the base, Sintex times
then
Sinteplexis = do this thing from 1-8 sinteplex times
then
Sinteplexion = now do this thing from 1-9 sinteplexis times then do it again and again all of these works each one is 0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001% of what we want when we reach sinteplexis% we get sinteplexion
then
Sinteplexionsis = i maybe surpassed that guy who made the biggest number in a video even if i didnt lets take his number is at 0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001% when we reach sinteplexion% (multiplying his number if bigger than mine itself he will have to multiply his bigger number with that big number and so on every time he does this he gets0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001% when he reaches sinteplexion% we do that to the power of sinteplexion% we will get sinteplexionsis
Proof that Sinteplexionsis Surpasses Robinson’s Number (ꙮ – the Multiocular O)
In the YouTube video by Stepheniscool titled "I Created The World's LARGEST Number", the creator defines Robinson’s Number (the multiocular O ꙮ). He uses "Robinson’s multiplier" to remove decimal points, creates infinite 9s, then builds higher and higher base systems where each new base uses the previous largest number as its digits. This process is repeated infinitely many times.
Critical flaw: Once the construction requires completing truly infinite steps, it stops being a well-defined finite number. Infinity cannot be "reached" and assigned a specific numerical value. Robinson’s Number therefore collapses into an ill-defined concept rather than a concrete, comparable integer.
Sinteplexionsis, by contrast, is built through explicit, recursive, finite (though insanely large) steps. It remains fully definable. Even if Robinson’s Number were somehow valid and temporarily larger, Sinteplexionsis treats it as an astronomically small percentage (far smaller than 10^{-something ridiculous}), repeatedly absorbs and multiplies it, and finally raises the result to the power of Sinteplexion%.
This self-referential domination ensures Sinteplexionsis overtakes and renders Robinson’s Number negligible.
Self-Recovery Property
Sinteplexionsis is designed so that even if divided or multiplied by known giants like Graham’s number (G) or TREE(3), it can recover its full value due to its deep recursive structure. Finding or reconstructing Sinteplexionsis after any such operation becomes another hellish computational challenge — far beyond what current mathematics can handle.
Conclusion
Sinteplexionsis wins by construction.
It is vastly larger than Robinson’s Number (ꙮ), remains a well-defined finite number, possesses powerful self-recovery abilities, and was purpose-built to solve extreme recursive problems like the 4-2-1 problem.
This is my entry into the world of truly massive numbers.
More User Perspectives
Ok I don’t understand what’s there to solve?
@Latesttechs9 because it keeps going up
@Ribbonplayswhite46/15 easy or if your boring 3.06 barred
So it’s 10.20
i tried using a calculator to see what happens to exactly 300 quintillion, and it eventually fell down to 1.
@Benjamin-m8vHad this video open for almost 4 years in an inactive tab. Finally watched it.
I also feel like I have found the answer.
I tried it for a long time but last find it is difficult for me to make some progress😂
@JackMath-i7lThe binary number works on the logic that every number can be written in 2 powers
@ytusersk1:29 well, this is a choice 1 is not odd or even algebraically. 1 Is the identity.
@erikolsen13335 + 5² + 5³ + 5⁴ + ... <
6 + 6² + 6³ + 6⁴ + ... <
7 + 7² + 7³ + 7⁴ + ... <
8 + 8² + 8³ + 8⁴ + ... <
9 + 9² + 9³ + 9⁴ + ...
Therefore,
Legendre's conjecture holds for numbers from primes to numbers less than 1.
Given two prime numbers p and q,
when p < q,
there exists a prime number q such that p < p² < (p+1)² < q. In this case, Legendre's conjecture is rejected, and this is one counterexample.
Collatz's conjecture is expressed by the following equation:
7 + 7² + 7³ + 7⁴ + ... <
8 + 8² + 8³ + 8⁴ + ... <
9 + 9² + 9³ + 9⁴ + ...
The relationship between prime numbers p and q, p < p² < (p+1)² < q, requires a prime number r such that when p = 1, p < q < r.
Furthermore, when p = 1, the relationship p < p² does not hold.
Legendre's conjecture does not hold.
This is because 1 is considered a prime number.
1.5 is one without a loop i believe.
@CrazytoothpicksLotergone0:37, 7 22 is my birthday
@ggll4045Gimme a sec….I can get ChatGPT to convince me I’m the math messiah and I solved this problem in 5 minutes.
@rowyerboat1Hey @Veritasium, at 11:26 you said, "If you can show that every sequence contains a number less than the original seed, you have proven the conjecture."
Are you sure that statement is true? Because that would make it trivial. All even-numbered seeds follow this rule, so that would mean the conjecture is true for all even numbers. If it's true for all evens, it's necessarily true for all odds, since starting with an odd seed must contain an even number in the sequence. If a sequence contains a number that obeys the conjecture, then the seed must obey the conjecture (which you can prove by induction).
So I'm wondering, what is the basis for the statement you made at 11:26?
This video sent me down a rabbit hole. I built a small program to test Collatz and a few nearby variations of the rule.
Changing 3n+1 to 3n−1 produced a stable cycle (5 → 14 → 7 → 20 → 10 → 5). Other variants either exploded upward or collapsed toward zero.
this doesn’t prove anything, and I’m sure mathematicians have explored these variations before. I just found it interesting
Fun exercise either way.
I know the video was before ai popularity but can modern ai help in this problem?
@Baraa-u2q5hChat I solved it so basically just do 3x-1 instead of 3x+1
@Potatoman-r5tIt must be proven that for every starting number, you'll hit a power of 2. Have people been working on bit shifting?
@hurz506Has anyone asked Terrance Howard to take a crack at it!? 💀💀💀💀
@papastinkyThe typa stuff i do on my calculator in math class:
@ReaINoobsterMe going to AI to solve it.
Be right back.
Wouldn't the answer be undefined
@pinklion63Cses - eird algorithm
@vecthericThat type of rule is proof that we just live in a simulation.
@AJ_realHow about we trace it backwards? 1,2,4,8,16 and then we start diverging as the next number could be 32 or just 5. Let's take 5, next will be 10 and we now dont know if next will be 20 or 3 same if we take 32, next will be 64 but we won't know if next is 128 or 21. I think there's some possibility of probability theory answering this...
@its_prince4realWhy is it a Problem, what should it prove?
@dengeleng7825it's obvious, its a bomb algorithm for the timelords of dr. who
@Birch55Hey @veritasium can have ur email i have theory to prove
@ramilvaleriano8739When going into negative numbers schouldn`t the Algorithm be adapted? Example 3x -1
@ralfflar7562I dont understand where the problem is?
@m1j9s79I solved it, because it would always end up with a even number after 3x+1 and we all know that all even numbers can be divided by 2 so no matter how big the number is, it would always end up in 1 with enough steps
@jeffrey-t5rHonestly, who cares? It means nothing.
@johndomutz1052Your question is does it actually need to be solved? There's just some kind of rule or law? I think it's about it could you be used to civil different teams from algorithms generators computation processing analyzing etc.
@willtews955Solved it.
Rules are flawed.
The divide by 2 on even number causes this.
Because it means that ANY equation where X can equal 1, that hits a number divisible by by 2 will immediately hit this loop.
For example, if you replace the first part with 7x+1. And this ruleset eventually hits a number divisible by 2 (which it eventually will). It will go down to 1.
1x7+1=8
8÷2=4
4÷2=2
2÷1=1
Repeats
Other examples this works on:
3X+5 (because 3(1)+5=8 so 8 will be loop
15X+1 (because 15(1)+1=16 so 16 will be loop
That being said, the higher numbers you use, the longer it will take for equation to hit breakpoint (divisible by 2) number.
Your welcome
nice game, what is it for?
@schiacciatrolloThis is a good video.
@JanseeniumMe personally,im not here because i like maths. Im here because im tryna study for exams and continue studying the concept of space and blackholes but got interested by this.
@Glitch_Eyes2the answer has been solved by gpt and has highlighted by terence tao.
this is the answer:
Yes — for any 0 < C_1 < C_2, there are infinitely many a, b, n with
b = n/2,
a = n/2 + O(\log n),
C_1 \log n < a + b - n < C_2 \log n,
such that a! \, b! divides n! \, (a + b - n)! .