At least Collatz didn't write down that he had a "magnificent solution" and then died. No one can be as great a troll as Fermat.

this dude has the most soothing voice

3:54 he turns into Bob Ross

That is just the most cool professor! Love his voice!

It is always nice to start the day with a numberphile video and learn about a new undecidable problem.

Of course Brady thinks we should climb the mountain. That's Brady 101.

As a long time reader of computerphile i certainly know about the halting problem.

*" Don't Judge a b̶o̶o̶k̶ problem by its c̶o̶v̶e̶r̶ statement "* *- Collatz Conjecture*

too bad we cant get ol' Gauss on this...

1:27 he suggests getting a computer to solve the general case of: even n ---> n/2 Odd n ---> an + b. I can solve one class of cases: If a is odd and b is 0, the values diverge for almost all starting n. You’re welcome.

Trying to solve the Collatz conjecture simplifies to answering the question, "Can you get to 1 if we start from any odd number?", simply because that if we start with an even number, it will go to 1 if it is a power of 2, and it will go to an odd number otherwise. Furthermore, we know odd numbers are of the form m = 2n - 1 for all natural n. Hence 3(2n - 1) + 1 = 6n - 2 is always an even number, so we can divide by 2, obtaining 3n - 1 for all n. Then, every odd number of the form 2m - 1 = (2^p + 1)/3 is guaranteed to be go to 1 eventually since 3m - 1 = 2^p, and powers of 2 are guaranteed to go to 1. Thus, now we are interested in the question, "Starting from any odd number, can we always get to a number of the form 2(2^n + 1)/3 - 1 = 2^(n + 1)/3 + 2/3 - 1 = (2^(n + 1) - 1)/3 for some n?" An affirmative answer would prove the Collatz conjecture. Now, is this not beautiful?

3:55 close your eyes - Bob Ross is alive.

The content on this channel is so great, thanks for all these videos!

this man is my absolute favourite on any channel. he oozes intellect yet retains symmetrical humility.

Thanks, really cool and insightful part 2. Really appreciate numphile2.

This is so simple yet so hard, interest levels are going up with the passage of time!

5:40 I think it was Erdos who said that to receive the mathematical prizes he set, you would have to work below minimum wage.

This guy's like the Bob Ross of mathematics

I think you could find more about mathematics by taking the problem backwards rather than forwards. Instead of starting at a number and using the x/2 and 3x+1 rules, you could start at 1 and use a 2n and (n-1)/3 ruleset for all whole numbers. You then have a tree starting with n=1 which includes n=2 and eventually splits when the (n-1)/3 rule is available to use. Instead of checking every single number and then drawing a line to the original tree, you could create the tree from the ground up and know that you have gotten every single possible branch. This would take up as much processing power as the original way due to branches "annihilating" when they become a number that already exists in the line therefor meaning you no longer have to calculate that branch.

Love his style of teaching.

Last night after the main video I tried 5n+1 by amending a python script and got some amazing chaotic behaviour.

It's interesting that the general an+b problem is undecidable, but I was wondering if there is some list of (a; b) pairs that are known to cycle, another list for those conjectured to go at 1 and another for those conjectured to go to infinity. That's something I've been looking for a while, though it's not too hard to generate them with some code.

The 3n+1 three is build of streath lines like the powers of 2 and staircases like how 7 goes around

I would like to see more about backwards constructive approaches to this. Like, multiply by 2 and/or subtract one and divide by three (if possible) and from there on form the tree. And then analyse this. I think somewhere there lies the proof in some way like that you can prove that every integer can be constructed out of a sequence of these steps.

Love this guy

The beauty of 3n+1 is that it ends with always the same cycle: [4 → 2 → 1]... meaning that it always reaches 4. 1 is just part of the cycle that brings back to 4. The conjecture should be: "blah blah always reaches 4" but this is less sexy than 1. And 3n-1 ends-up in cycles as well but those cycles vary depending on the starting number.

this is the simplest thing to solve, in that (N/2) if a is not a factor of 2 will become odd and therefore continue looping.but when you introduce (3n+1) essentially from simple math(odd number * odd number = odd,,but if you add 1, you make it even again and therefore (N/2) checks if its a factor of two again else it keeps going. The rule is that its only factors of two that lead you straight to one with the least steps

Math is really fun and this just proves why!!!!!!!!!

I think the availability of PCs gave people the idea that they could crank through the integers at high speed, and be the first to identify the counter-example that disproved the conjecture. We got some clever programming out of the effort, but it wasn't really aimed at *proving* the conjecture. Nobody's doing this today, since supercomputers have run the numbers vastly beyond what anybody can do with a personal rig. (I think it's been tested up to ~2^60.)

The video began with the part I was interested in (potential intractability).

What if you plot the result of "(n mod a = 0)? then (n / (a+1)), else (an+b)" repeatedly. The colour (shade) at each point being the smallest starting natural number (for "n") greater than 1 that returns to 1? The whole thing kinda reminds me on the Mandelbrot set now.

I think the reason this problem has been so famous in because it is so simple yet unsolved. Most simple unsolvable problems have been proven to be impossible to solve, such as the 5+ color map, Fermat's last theorem, the utility problem, etc. The problem is so simple a 4th grader can understand it, yet nobody has found an exception nor proven there is not an exception.

Please upload a video about this if something happens!

Pleasant man

What there is a 2nd channel?!? Nice!

The greater conjecture is that one can always find a portable hole or singularity using the collatz ( even | odd ) snakes and ladders algorithm.

Collatz have a connection to a Pillali Conjecture. Generally in 3x+b problem there is at most one cycle for selected b. We know this from 2^(x+y)-3^y=d problem (how many solutions are there, if we selected some d) and that problem is solved here: "On the generalized Pillai equation ±ax±by=c". Connetction between 2^(x+y)-3^y=d problem we can find here: "On the existence of cycles of given length in integer sequences like 𝑥_{𝑛+1}=𝑥_{𝑛}/2 if 𝑥_{𝑛} even, and 𝑥_{𝑛+1}=3𝑥_{𝑛}+1 otherwise" (I personally extended their proof for ax+b problem and it's easy to do). There is also connetction between Catalan's conjecture (or Mihăilescu's theorem) and Collatz problem. Since we knew that Catalan's conjecture is true we knew that there is no some special types of cycles in Collatz sequences (cause there are no more solutions of this problem: 2^(x+y)-3^y=1). Furthermore they are many more interesting connettions between Collataz Problem and some hard problems in math (for example Crandall Conjecture and Wieferich numbers - "On a conjecture of Crandall concerning the qx + 1 problem" or connection with a Mersenne Primes - "A Conjecture on the Collatz-Kakutani Path Length for the Mersenne Primes"). I have been analyzed this problem over the years and I love this.

I really like this guy

When he says that the an+b version of the problem is undecidable, does he mean that it cannot be proven or disproven without extra axioms? (Like the continuum hypothesis?)

The thing is: you never know whether a problem like this can discover a field of mathematics that will easily help to solve million $ problems

Why is there no video on Numberphile about Aliquot sequences? They too form graphs similar to the ones obtained from Collatz conjecture, and there's a conjecture (made by Catalan) that any sequence ends in a cycle (a perfect number, an amicable pair, or cycle of social numbers) or in 1. Several numbers though still have an undetermined sequence (like 276).

I have proven the Collatz Conjecture to be false. But alas, the character limit of this comment section is too small to contain the proof and counter example.

It would be nice to have a video on the halting problem, I'm not aware that Brady has done one on that.

People talk about arbitrary patterns, how about this/these ones: 1) Every number has one and only one thing it can become, but a number has either one or two ways to reach it (either half a number or (n-1)/3 of a number) 2) Every time you hit an even number, you eventually end up with a number whose prime factors exclude 2 - I know this is the definition of an odd number but if one could show you can get every prime combination by building the tree in reverse thats the problem (starting with one and building outward)

So, for "3n-1," it looks like it definitely loops if it hits 7 or any of its looping numbers (7, 20, 10, 5, 4, 7) or 17 or any of its looping numbers (17, 50, 25, 74, 37, 110, 55, 164, 82, 41, 122, 61, 182, 91, 272, 136, 68, 34, 17). Just running some equations through LibreOffice/Excel. Picked up on two distinct "loops" for numbers up through 15000. Most of the rest trended to 1... Not sure if it loops for any other (higher) number sets, or not? Seemed like all the loops I was seeing through 15,000 were of the 7 or 17 loop variety.

Pah, formulas are for wimps! Hard graft is whats needed, so I'm going through ALL the numbers. Up to 19 so far....wish me luck :P

Its been a while since I have actually practiced any kind of college level mathematics but I was wondering if this problem might have any connection to attractive fixed points? The reason I ask is because I feel like there are some similarities.

the answer lies in the quantum of odd and even numbers in the string. And the tree can be based on prime numbers.

Think of the implication if there exists no classical logic mapping of naturals with collatz. That makes in interesting statement about nature. It may mean we will not have a unifying set of axioms, and will just have to assume or hope that different foundations map to each other.

The "modified Collatz" C' (with these operations {3n-1; n/2}) is Collatz C mirrored at 0. ( C(k) = -C'(-k) for every k ) e.g. C'(5) -> {5, 14, 7, 20, 10, 5, ...} C(-5) -> {-5, -14, -7, -20, -10, -5, ...} So instead of asking "Why are there loops when I replace + by -?" you could ask "Why are there loops for n < 0?".

Easy solve in 2atic space, every operation removes 1/4 of the area of the space. 0.75*0.75*0.75*0.75 etc converges at zero. Eventually all numbers will eventually land on a power of 2 and end.

Could we have an update with Terry Tao's almost all collatz get to almost bounded values discovery?

this man is just so likeable

there's a relation for finding the higher number, 'N', that needs 'steps' number of operations: N = 2^^steps Ex: the biggest number that take 21 operations to reach 1 is 2097152 There are no bigger numbers than that with 21 steps. Every number bigger than 2097152 has steps bigger than 21

I was playing around with the numbers and I did couple, small and big ones, and found something interesting... Every number above 10 follows almost the same sequence as the the first 10 numbers I call "base" numbers. What do I mean? I mean that an arbitrary number, say 1238, will follow a pattern, as it makes it's way down to one (we know this number reaches one at some point) that almost resembles the pattern of 8. But then the larger the number, I found that it is not only the pattern of the last digit that is Incorporated into this arbitrary number's pattern, but it is a combination of the first ten. For example, the number, as it makes it's way down to one, will have a sequence that resembles the pattern taken by 5, then a pattern taken by 8 and so on. So every number, from what I saw as I did this, will follow a pattern as it tries to get to one that is a combination of the first base numbers. It's like a collage. It was interesting for sure. I'm sure other mathematicians have noticed this too.

a little fact would also be if n = a power of 2.. then it will inevitably go down to 1

Okay I think I got this but I could be worng. This is just another "well-formed" numbering system. Treat every integer as a series of ups (3x+1) and downs (x/2). Downs will always move you more down than ups, and you will never find a loop (i.e. x/2=/=3x+1) All you done is essentially rename every number with a non-integer base. This is a binary function base.

Fun fact: the cover of Lagaria's book has an (most probably intential) error: 133 does not give 170

I recently calculated number that has more steps than the number was shown in the video my number is 10^300 and also find the weird thing about it. when I calculated the 10^245 it has the same step as the 10^300

Instead of working brute force from 1 to 2^60, is there a definition of infinity regarding its even/odd-ness?

2²-3=1, 2-3=-1, 3²-2³=1, 2⁴-3²=7, 2^6-3⁴=-17 etc.

There is an astonishing amount of complexity bias surrounding this conjecture. The Collatz Conjecture is true because the Collatz Function is a bijection between {N} and {1,1,1,…}. It’s very simple.

Hello! I have one question. Let's supose that the conjecture is true for any given natural number n. Therefore, for a particular n0 you have a particular m0 number of steps until you reach 1, so, if you have another n1 with m1 steps to reach 1 such that m0=m1, does that imply that n0=n1? I've been thinking for a few minutes about it but I can't find a nice argument or a counterexample.

Here's a nice function that gives the next number (y) from x :) y = (1/4)*(2+(7*x)-(2+5*x)*cos(pi*x))

It's pretty obvious that this will go down to one. Because: A. Multiple by 3 and adding 1 is the same as multiple by 1.5, because that makes it even. So it goes down faster than it'll go up, 1.5 up vs 2 down. B. multiple by 1.5 and rounding up will make it divisible by four every second time, so for every up you'll have at least one down, and some times more than one down. So it's pretty obvious that it'll go down fast. We could improve it by giving a similar factor for up and down, and a similar chance for up or down. An example: a=6,b=5,m=23 point=x if x modulus 23 < (m-1)*(b/(a+b)) point / b else point * a I've started it with x = 526 for a few thousands iterations and it doesn't seem to have any direction. Up and down from zero to a billion without any order.

I have a... kind of... stupidly sounding question, but give me the benefit of the doubt and believe I've spent months and maybe years thinking about it, and writing some programs trying to explore it... And try to answer from position of assuming this, thanks: What is it, that makes humans able to not infinite-loop themselves on evaluating the halting problem, but the computers and algorithms do? I mean, yes, you had video about that, but... fuzzy static evaluation? we had experiments with fuzzy AI (in games) back in 2000, where's experiments with fuzzy mathematics frameworks? which simulate the way people look at and estimate equations, etc?

I was absent when my HS math class went into all this.

Please do a video on wau

This is another unintentional asmr video

Any number ends in 1, but also in 16 for all the numbers above 16 ? This number 16 can be related to time squared (4^2) right? All iterations are related to the number 4 in math, right? It's the circumference factor of the radius of a cardioid ':)

Prof Charles Cadogan of the university of the west Indies, Cavehill campus, has proved the the Collatz conjecture. You should be able to obtain a cpy of the proof by contacting the department of mathematics at Cavehill , st. Michael , Barbados.

Lo tengo algo avanzada," La conjetura de Collatz es verdad si solo si se demuestra para los números impares de la forma 12n-1 y de la forma 12n-5."; lo puedo demostrar. Ademas tengo otra joyita fácil de demostrar " Si existiese otro bucle distinto al 1-4-2-1 entonces existen infinitos números que llevan a este bucle" ( a la fecha no se ha encontrado ningún numero de estos si existiesen son infinitos).

I have something advanced, "Collatz's conjecture is true if only if it is shown for the odd numbers of the form 12n-1 and the form 12n-5."; I can prove it. Also I have another little gem that is easy to demonstrate "If there were another loop other than 1-4-2-1 then there are infinite numbers that lead to this loop" (to date no number of these has been found if they existed are infinite).

One guy crack on the million pound problems but didn't take the money. Madness, give it to a school or hospital or university.

I was thinking about the an+b thing, maybe there is a graph/sketch on a a,b axis set that has 1 in every case that the algorithm "converges" to a power of 2 (since from that point it will go to 1), and 0 in the rest. I'm wondering if such a graph would look similar to a fractal?

I just figured this out but there is n/2 for even and 3n+3 for uneven numbers and you always get 12 - 6 - 3 sequence at the end after x steps

Are there numbers, in reduction, that are missed? Where, the only time they show in reduction, is when you initiate with them? Or numbers that present less frequently than others?

Anyone else notice the (seemingly) unrelated stuff on the board?

Thanks for your idie

If n is even then divide it by two and if it´s odd multiply it by 3 and then subtract one. I did that for 8. I ended up with a loop of 2, 1, 2, 1 forever, is that right?

I thought two Japanese guysmTaniyama-Shimura-Weil conjecture taht Wiles crack the problem.Tho I assume he meant them when he mention elliptical curves.

Idea: look into obscure properties of even and odd numbers? Probably won’t work. This is one of those problems that seems like it should be simple and easy but isn’t

So if the problem of whether such a machine can exist is undecidable, does that imply that for at least one ordered pair (a, b) it's impossible to prove whether or not the associated conjecture of whether every integer eventually reaches 1 via the sequential transformations is true? I think I've heard that some people believe the Collatz conjecture to be undecidable, but I could be thinking of the Goldbach conjecture.

Which pen you use please tell me

The collatz conjecture, only powers of two get to 1 and the conjecture's operators never get to 0. Therefore, only positive integers that are powers of 2 get to 1. So, there are positive integers that do not get to 1.

What is this wholething-problem is talking of, at 1:35. Or did't I get this right? He is a bit difficult to understand for a foreign listener, sorry.

still holding the division rule at 'if n%2 = 0 then n

What if we try 5n-1 for odd no.s ?

Big Big mind professor!

I can only find a number which is equal to the number of operations he takes to reach 1. That number is 5. 5-16-8-4-2-1 Is there any other number such Col(n) = n ; Col(n) represents the number of steps that n take to reach 1)?

3n -1 is only the symetrics of what happens in negatives numbers on 3n+1 problem

Here is the solution. The method to be applied has only two rules: - 1. For odd numbers multiply it by 3 and add 1 2. For even numbers, we keep on dividing till we get an odd number (if it's not 1 we apply rule 1 again) Now, for all powers of 2, we keep applying rule 2 we will eventually get to 1, which is 2^0. (So it has to be true for all powers of 2). All numbers which are not powers of 2, are either odd or some even number which is the product of some power of 2 and an odd number (other than 1). Now, if it's an even number we keep applying rule 2, to get to that odd number (other than 1). So, in both cases we will hit an odd number other than 1. Now, all odd numbers can be written in the form 2x + 1, where x is any whole number. So, when we apply rule 1 to it, we are actually converting it to an even number of the form 6x + 4. (3 multiplied to an odd must give an odd number, since 3 is odd. 1 added to an odd number is even. Since the number we started with was having the form 2x + 1, the even number formed by applying rule 1 will have the form 6x + 4.) But, when we apply rule 2 to this even number, the algebra tells we will never get the earlier odd number back again. This is because 3x + 2 > 2x + 1(unless x is 0, in which case 2x + 1 would have been 1). Even if, 3x + 2 is an even number when we divide it by two all the numbers that we get will be less than 2x + 1. Which means we will get different odd numbers every time. All these odd numbers, will have the form 2x + 1, but the x must have a different value for each one of them. This in turn means, the even numbers that we will get by applying rule 1 will be a different number each time. This process will continue until we hit an even number which is a power of 2, which will take us back to 1. Therefore, all numbers on which the method is applied following the rules will eventually take us back to 1. Hence proved 🙂🙂🙂

seems like all u need to prove is that every multiple of 3 plus 1 will eventually wind up on an even number after x number of iterations, and perhaps x is somewhat proportional to the log of the number or something.

One thing we know for certain, if there is a solution where n doesn't reduce to 1 then there is both an odd and even numbered solutions and therefore there would be an infinite number of solutions.

Interesting… I never realized that the generalized problem was undecidable That’s kind of insane, if you think about it… Even if we can prove the original conjecture, there will always be combinations of A/B that we will never be able to solve one way or the other I mean, given the existence of Gödel’s Incompleteness Theorems, I suppose these sorts of situations are inevitable… But even so, there’s something about the an + b situation that makes it seem like it should always be solvable, in theory

The secret to the Collatz conjecture is that you are adding halfs not integers. The Collatz conjecture is simple a geometric series where 1/2+1/4+1/8...=1.

They should put big prizes on things that are certain to matter. So if no one can figure out how to solve a problem that is preventing the improvement of solar panels, for instance, put some big money there.

Just for fun, here is what I am thinking... If n=1, stop If n is even, divide by 2 If n is odd, multiply by 3 and then add 1 So far that's the rule.. We can prove that all even numbers will eventually end up as an odd number by dividing 2. We can prove that all odd numbers multiplied by 3 will be odd, therefore all 3n+1 must result in an even number. When an even number is divided by 2, there are only two possibilities: I. Divide by 2 again. II. Ends up in an odd number. As one even number only gets divided once by 2, the next even number must get divided by 2 more than once (4, 8,16, 32...) It seems to me that the smallest set of numbers ad dividers for even numbers follow the pattern of: 2, 4, 2, 8, 2, 4, 2, 16 and grows larger as n grows. The average of this divider is 5 so since n/5 will always be less than 3n+1 for all natural numbers, doesn't that mean these rules will make any n move toward 1, the smallest number allowed based on the rules?

For what it's worth, I think that the cash prizes offered for problems like this are really quite pointless as an incentive. For one thing, many studies have found that providing a cash prize for a problem requiring even rudimentary cognitive thought usually hinders people from a solution rather than encouraging them. Secondly, anyone capable of solving a problem like this isn't doing it for the money anyway. They're doing it because they are interested in mathematics and want to know what the answer is, for themselves and for the future of mathematics. Granted, these people who make these breakthroughs to find the answers are deserving of recognition and a fitting reward for their contribution, but we shouldn't act like dangling that reward in front of them like a carrot on a stick is going to get them there any faster. If anything, it will distract them.

Just out of curiosity. What happens if someone proves this? Or at least makes some work towards it. Where should he send it to? If he sends it to an university could his/her work be stolen by one of the proffessors? Has this happened before with other problems?