My honest attempt at the Collatz Conjecture | Full movie  

"It is very easy to compete against dead people because they can't earn any more points" quote of the year, lmao

“Sometimes you have to do side quests to advance the campaign” - I love this!

Several years ago I dreamt I had solved the conjecture. I have never been so mad about waking up.

Whether or not your idea directly provides a proof of the Collatz Conjecture.... by God, you achieved your goal. This video was so thoroughly thought provoking, and you did an AMAZING public service by teaching proper skepticism. That alone earns this video a FANTASTIC score in my eyes for that, but the actual progress you made in digging through the Collatz tree was FANTASTIC as well!!

Regardless of any progress on the Collatz Conjecture, I'm indebted to you for introducing me to the phrase "As though my ass had fallen off."

As soon as you didn't assume my intelligence and described what a graph was, that's when I subscribed. There's nothing more frustrating than trying to follow a video that has gone too deep into whatever topic it's discussing without a proper introduction to the material first. This is a fantastically entertaining and easy to follow video, thank you.

I think were things went wrong is when going from series to Infinity. Just because an algebraic manipulation works for each step in the series doesn't mean that, that same manipulation holds when talking about the limit. You can only split a series in parts when those parts do converge and as you said Integer series don't converge so you cant just split them an expect that to still converge

I want you to know this is one of my favorite math videos I’ve ever seen. You have, more than anyone I’ve seen, encapsulated not just the beauty but the FUN of math. A+ work my dude, I hope to see more! I’d love more of that same style where you go through your own reasoning

When I was a sophomore in high school a handful of years ago, I thought I had solved the Collatz conjecture at three in the morning, called all my friends, and woke up the next morning to realize I was completely wrong.

I understand now that the Kingkiller Chronicles series has not been finished because it is a riddle with no solution! Patrick Rothfuss has been training us how to think by not releasing the third book! Bravo Rothfuss, for this 5d writing strategy.

I have been working on this problem independently for years (around 3.5 years at this time). I started in my freshman year of college, and now in my senior year I am still making discoveries. This problem is so taunting, it feels like the problem is cursed by demons to trap mathematicians into chasing after ghosts.

You indeed made a mistake. You will notice it when you try to apply this technique to some infinite sequences. For example, the arithmetic sequence a(n) = 2n+5, or a(n) = 2*(-1)^n+3. You wrongly assume that there is some "last" number a(inf) and then show that it is impossible to construct an analogous Diophantine equation for it. It does not follow from this that this sequence is not infinite.

The second part of this video shows the value of community in any kind of research: sometimes you need a fresh pair of eyes to find a mistake or inconsistency in your argument. So instead of trying to answer if you are a crackpot or not it would be better to find somebody else and ask them what they think about it. This is also the essence of supervision and peer review. One really can't approach their own writing in the way you read what other people wrote this is why you really need somebody else to go over it. I am finishing up my Japanese literature PhD disertation and I am appaled at the number of non-sequiturs and spelling mistakes I've done in it! (it's like 300 pages after all).

I like the emphasis on having intellectual humility, while still encouraging the process because it's what leads to learning. That's really the only reason I've ever fiddled with the conjecture. Here's what I came up with: If we suppose a strong induction proof might work on the Collatz conjecture, then we only need to look for starting numbers such that their sequence that follows will eventually dip below the starting number. Thus, it doesn't matter how fast we proceed through the sequence, so let's define an accelerated collatz function. If n is even, then n=(2^k)*m for odd m and some k. If n is odd, then n=(2^k)*m-1 for odd m and some k. Then the accelerated collatz function is given as follows: f(2^k*m)=m and f(2^k*m-1)=3^k*m-1. This function has some nice properties, like how k is the number of iterations of the standard collatz function (which often immediately divides 3n+1 by 2, since it is guaranteed to be even), and how f(even)=odd and f(odd)=even. This gets the same result that you got (namely, a method of generating our choice of sequence) but I think it's a bit simpler in my opinion.

I'm only 13 minutes into the video but I already like it heartily beacuse it seizes what most other math RU-vid videos, however comprehensible and well-made, fail to portray: the feeling of *actually* doing math. It's really all about the exploration, testing and trying out various ideas that may or may not eventually work out at all, about giving things silly and/or beautiful names in the process just for aesthetics and simply to have fun, about looking at the examples of certain behaviours first and only then trying to formalize them, as to find out whether the patterns you think are true do indeed hold and if they tell you something about the problem you're working on... or a completely different one. This is what I'm doing math for, this is why I love it. Huge props to you for being able to share this feeling with the general audience instead of trying to make yourself look perfect and your reasoning polished to avoid criticism. As the community of mathematicians, we should generally accept that making mistakes is a natural part of what we do, and we shall not hesitate to share our attempts, to correct and to be corrected. And you also helped me to look at the great unsolved problems of mathematics form a different angle. I never really considered actually trying to work on them because I'd never put in words this philosophy of "being able to learn from them forever". I had that feeling inside of me, but it was contradicted by the steel logic of "if you most probably won't be able to solve them, why even bother?" Now I can see clearly that this is not about finding the actual solution, but about what you can learn while looking for one, so now I fell completely free to give it a try. Thank you for that!

pretty interesting video. I tried slapping the collatz conjecture with my very flaccid and wobbly mathematical baseball bat 3 months ago and obviously failed at solving it but the insights i developed throughout this process were actually recognizable in your method, just with a different visualisation. Math is neat

One of the best videos i've seen in my life. The sheer love for maths here is impressive. I wish all the best for you my friend.

I am in awe! What an amazing video about an amazing topic made by an amazing guy! This really is awesome, thank you for sharing this!

"But then i had an idea..." Pretty much where things always go to hell.

You are an absolute joy to watch!! So glad I found this video, it was super entertaining, super interesting, and I learned a lot too! Will definitely be watching your other videos.

Fricking loved the video. Your raw passion of this is just so entertaining to listen to. Your discoveries and ideas are incredible to hear. I hope you enjoyed making it and i really hope you make more. You are great man. Contiune the side quests!

I love your videos ❤ Thanks for making these amazing masterpieces

"That doesn't mean we should let them have all the fun" Love it!

Hands down one of the best amateur maths videos I've ever seen. Great job!

How tantalizing would it be if the one exception to the Collatz conjecture was somewhere in the Septillions and it was proven that it was the only other set to loop around itself. Insane to think that we can’t say for sure it isn’t there

You're right that an infinite sum of nonzero integers diverges, but the sum of the differences may converge, and it looked to me like you split a sum of differences into a difference of sums when you moved to the exponent notation.

Interesting video. Fer's conjecture and the super awesome tiles seem like great ideas! Fer's conjecture could be even stronger: if the distance between nodes is 2, it's only when the last number is odd (the Lannister tile), although then it's a bit wordy haha. About infinite constellations: on 30:38 you assume that the sequences alpha(m) and beta(m) have to converge to some alpha and beta. I fail to see why that has to be the case.. Also, on 32:02, you say the potential a_inf is "clearly" not a part of the Collatz tree, but why not? Could it not connect to 1 through a different path? Finally, at 32:50, there is a problem with the exponent trick, because it essentially says that an infinite product of (3/2)s, (3/4)s and (1/4)s cannot converge. This is clearly false, as a product of just (1/4)s and (3/4)s tends to zero. (This obviously doesn't produce a valid infinite constellation, this is just a counterpoint to the exponent trick). I think the problem here is as follows: once you rewrite it as an exponent, the product becomes a sum, and then you rearrange the terms in the sum, grouping all the 3s (ln3s) together and all the 1/2s (-ln2s) together. Rearranging an infinite sum only produces the same result if it converges absolutely, which this one clearly doesn't. One more thing: on 42:39, isn't GCD(2^w, -3^r) = 1? Oh, and on the problem you've encountered in the end: it seems like your algorithm doesn't find all solutions to the diophantine equation, just one family of it. Otherwise you would get all the integer solutions by substituting integer values for b. I am not well-versed in diophantine equations myself, but that seems to be the problem? Thanks for the video!

Amazing work. I still have yet to finish the video, but your presentation is captivating!

This video is amazing! You really showed the joy and exicetment math brings to the ones who study it. Also, you inspired me to give Number theory a shot (i'm writing this as i try to read further into Vinogradov's book) so thank you for that too!

So glad I found your channel, first video of many for me to dive into, thanks for the hard work!

I absolutely love and agree with your approach. It's not the destination, it's the journey. Even if we don't solve the riddles, we still learn something, and further the combined knowledge of humanity. The geometric approach is also very intriguing

Really excellent video! I just discovered your channel and am loving your energy and ideas. Your introduction motivated me to pursue solutions and/or non math pursuits which may, at first, appear impossible

Really enjoyed video. Incredibly brave to toss your ideas out for scrutiny. Value for non-mathematians (like me) are your thought processes and strategies. ...These types of problems can be addictive even when every idea hits a dead-end. :).

Man I have been working on and off with an unsolved problem for years. Took a break but this has really inspired me. Love the king killer reference

Well you have proven that some specific sequences of tiles must start further and further away from 0 the longer they get, so those for sure cannot be present in a cycle because a0 cannot stay bounded, but the rest doesn't seem to ask the actual question (like sure you can find a number that's limit of an infinite constellation, but that's not necessarily the only possible new cycle)

The fractional values of B don't always follow the rules. You can see the sequence go 10->16->13->20.5 and such. If the rules were properly being followed you could never go from an integer to a non-integer. I think the algorithm doesn't work perfectly.

I love the way these ideas are presented

36:21 "Looks like a clear cut case of murder, stabbed in the back with a protractor while writing a 13 page blog post defending his theory."

This was magnificent! Thoroughly entertaining.....it's been a while since I've contributed to my own channel, and I would love to learn Manim and give my own pet problem an "honest attempt" youtube video for the next SoME...this comment is to put that out into the universe so I actually have to do it! Thanks for the knowledge and entertainment

Don't doubt yourself my man. This is really interesting and the jokes well timed lol

Perfect intro and theme, great video!

I love this, really enjoyed going on the journey with you

Diophantus: *in an ancient letter* “please release this letter in future proving the collatz conjecture, and disproving Fer… Diophantus: game point, Diophantus out.”

The math is way over my head but I give the video an A+ for the methodology. Mistake or not, it's a great teaching tool on working through problems.

I didn't think I could, watch a 1 hr video after not sleeping the night, at 5 in the morning. But I am invested. I surely didn't understand some stuff, but I got hooked on a lot of stuff you showcased. The thing at the end is also so great!!!

You won me in the first few seconds. Love the reference to A Wise Man's Fear.

when I tackled it I ended up going a different direction: I saw that nodes are either 4 mod 18 , 10 mod 18 , or 16 mod 18 , and they go in a cycle in the /2 path: 10->16->4->10 , 10 nodes are the solutions to 3x brahces, aka "dead" branches, 4 nodes are the solutions to when a node needs to be divided by 4 before needing to be multiplied by 3, and 16 nodes are the solutions to when a node needs to be divided by just 2 before needing to be multiplied by 3... but I had no idea what I was doing and ended up just going in circles as a layman

What freaks me out is that I made a disturbingly similar mistake at solving the same problem about a few days before this video aired. My "proof" never pretended to prove the full conjecture, but only that there wouldn't be any cycle other than those already known... and of course it was wrong. I had a result quite similar to what you get at 42:38 and for some reason believed it meant solutions would correspond to 2^a-3^b = 1 or -1. Still, I was able to prove a similar result than your "constellation", which has the consequence of proving that there can't be a periodic constellation that doesn't cycle. (so constellation that doesn't cycle must be aperiodic if they exist). Then I found out some guy already had proven this for their thesis (but I am still proud my proof can fit on two pages, is relatively easy to understand, and works for non-collatz sequences as long as they share some of its properties) I am currently trying to write a paper about it (I probably will leave it as a pre-print, but I can send you the link to it once it starts looking like something and I have an English translation)

this video Is probably the best video i've seen this month (possibly this year)

This is how research should be done. Every math undergraduate should watch this.

I remember my half attempt at this problem. I printed out the series in binary using black and white pixels and looked for patterns. It formed some very nice triangles.

Maybe I don't understand the ending, but I think that It makes sense that you can extend to the rational numbers, because for every rational number q, the numbers q/2 and 3q+1 are also rational, which means that you can kind of make what I imagine as a 'web' of rational numbers, where every number q has arrows that points to q/2 and 3q+1, and has arrows that point to it from 2q and (q-1)/3. The tree that you describe is sparser because it restricts it to only certain kinds of constellations, but I think the idea still applies. Checking the parity of rational numbers isn't really possible though.

Problems probably start happening around 3^34 because that's around 2^53. There are 53 bits of precision in a 64-bit float

Instead of the collatz conjecture, let's look at f(n) = n + 1 instead. This obviously has infinite sequences. I'll try to follow along and apply your reasoning to this. If the conclusion is reached that no infinite sequences exist then that's a problem. 12:06 If n = a, we'll call it a Unit. (So just every number) 12:55 The Unit Tile is pretty straightforward. It can just connect to another Unit Tile 13:37 We can place Unit Tiles together, for example UUU or UUUU 15:08 For the Unit it would be n + 1 = s, which has solutions n = a and s = a + 1. So that's how you could find any Unit. 20:43 For the Unit it would be a_m = a_0 + m, which has solutions a_0 = b and a_m = b + m. So that's how you could find any constellation. 24:29 We can skip this, because we only have one type of tile. We can already find any constellation. No need to do threading. 31:07 You're saying that these constants should both converge to something, in order for an infinite sequence to exists. But in our case, by looking at the equation a_m = a_0 + m we can see that Alpha does converge to 1, which is good, but Beta does not converge because it is equal to m and so will diverge as m increases. That seems like a contradiction. But maybe I'm misunderstanding. I'll admit I'm not THAT good at mathematics.

All the numbers that appear in your graphs are a node, since to be in the graph the number has to be one that appears in some Collatz sequence which passes through that number. IMO it's fine to call the nodes that have multiple connections anything you choose, but if you want to differentiate then you could refer to them as something such as Converging Node, as they are the integers where sequences converge.

Randomly, at 11:25 your graph with k = 3 and (2k-1)/3 has a typo where it says 5/5 instead of 5/3 :)

love your energy -- great video

Obviously, there are fractions that if you multiple them by 3 and add 1, enough times, you'll get an integer, but the Collatz Conjecture requires the number to be even or odd. A fraction is neither. I guess if you just checked to see if a number was even or not even, fractions would work.

your videos are so good and I love ur character, you seem like a really nice guy

11:50 "it is very easy to compete with dead people because they can't earn any more points"

Great video. By the way, Diophantus actually used lots of rational numbers, too, but for some reason whoever called the equations Diophantus equations required the solutions to be integers.

Very thought provoking video, and thank you for uploading a very nice writeup. Regarding the notion that this generalizes the Collatz in a similar way to how the gamma function generalizes the factorial - I think they are quite different, and that the differences are illuminating. This generalization is not one function, but instead a family of SuperAwesome functions indexed by a streak and a position therein, e.g. These can be viewed as rational (and indeed linear) functions, e.g. S0(a) = 2a+1, S^m0(a)=2^ma+2^m-1, while S1(a) = 3a+2 and S^m[m](a) = 3^ma + 3^m - 1. As all the tiles correspond ti rational linear functions, any (finite**) iteration of then will be linear with rational coefficients and so can be written SuperAwesome0(b) = (u/v) + (x/y) b. Integer are whenever b=(y/x)(k - v/u). Similarly there is SuperAwesome[n](b) = (t/w)+(c/d)b integer when b=(d/c)(k - t/w). So a streak that starts and ends at an integer for a specific streak are given by the intersection of the integer spectra of the first and last SuperAwesome functions for that streak. A full Collatz solution would require the intersection if the spectra of all m+1 (assuming m tiles) SuperAwesome functions. If the rational coefficients can be related to the the streak sequence, then it should be possible to bound what fraction of integral solutions are actually lost when adding one additional step to the streak. Establishing that this proportion is bounded above 0 (or even log(n)/n) could [convincingly?] show that Collatz solutions must vanish. A similar approach could be taken for cycles as well.

Just a tiny comment, but huge respect for citing the AI art and prompts in the videos. It's in keeping with my university's guidelines on using AI works, and I think it's a good precedent.

Great video. Congratulations!

34:01 Is it possible p-adics could be your solution to having integers converge to a finite value?

I gave it a go and tried to approach it by finding a loop. Because even numbers just half down to some odd number, I decided to try to find some way to get to each odd and ignore the even numbers which might make such a process very lengthy as the numbers grow. To do this I invented a new operation which I called factor manipulation. When you manipulate the factor of some number (represented by fm_n) it returns the list of prime factors in descending order, with each factor labeled as f_a with a being the order that the factor is in. You can refer to these factors using the operation, and that means you can manipulate each individual factor in an equation however you want. Basically, fm_n(f_a+3) would return the a’th factor of n and add three to it. You can mention more than one factor, multiply factors, and more. The goal was to try to define what happens to the prime factors in a number when the number goes through an addition (the +1 in the conjecture) but unfortunately I don’t have any coding skills, so I’m not able to create a calculator to perform math with my operation. I had fun though! It’s really interesting to see how different people go about solving the same problem.

Hot diggity dog, dude that ending!! I didn't understand but suddenly it made sense. I never thought that anything but integers could be in the Collatz conjecture, but of course we should have instances where fractions would appear as we are talking about all real numbers up to infinity.

This is the first true extension of the Collatz conjecture I've seen (47:24) when looking on Wikipedia or other Collatz videos (including yours) they usually show fractals that are made through the continuation of the Collatz conjecture. but they have always been "artificially" extended, I don't have enough characters to explain, but the extensions aren't derived from the consequences of the Collatz conjecture. I feel like someone needs to re-do those fractals with these new values you have found... assuming it can be proven that it truly is an extension, by (probably) showing that you can't just get any number through this method. of course I'm just an amateur, so I don't know that much. I'm really excited the new are that hopefully will be created by this new discovery.

Since you mentioned Scott Aaronson, would you be willing give give a "for amateurs" explanation of his Collatz conjecture work? It sounds super interesting. Somehow it reformulates it in terms of rewriting rules on a string of characters (like your sequence letters?), and culminates in the claim that if matrices with certain properties exist, it proves the conjecture? I currently don't understand it, and its a shame there isn't a good overview of this more current research on youtube.

This is absolutely an amazing video and really inspiring. I hope the first 5 minuted of your speech about why we should try doing thr impossible on our own goed viral. Think more mathematicisns and physicist needs to think like that.

54:01 "one over eight - integers again" . . . the final number in the sequence is 7979.5, not an integer Great video! I love your enthusiasm, and your attitude toward learning and being wrong!

In the beginning of the conclusion part i imagine some enlightened monk sitting on a stone under some old tree, infront of his class of diciples, lecturing and getting sidetracked to some weird stuff and ending with "...i could be right, and i could be wrong"

A new perspective is always welcome.

I dont have enough knowledge to explain why but what you notice with the super awesome tree really strongly makes me think about 10-adic numbers, if I have the energy I should at least look for the 10-adic numbers corresponding to your rational values for b and look for patterns

It seems like when you compose diophantine equations you are only considering solutions that make both original equations integers. You also need the solutions where the first outputs a fraction but the second gets back to being an integer.

I think it's fun simply making the tree of the sequences myself and finding patterns that makes growing it slightly easier Such as the pretty trivial fact that any power of 2 will rush down to 1 as quickly as possible

Super awesome! Near the beginning, when you decided to look only at "nodes", I thought it might be interesting to look at the tree you get if you give each number two parents which can be generic rational numbers not just integers, making all numbers into "nodes". I think the results you get at the end with sequences with fractional numbers are sequences in this "completed" tree. I'm not sure what's going wrong with the equation for infinite sequences, but I think there's a couple things to note. Firstly, we could pretty easily create an equation for an arbitrarily large increasing sequence of nodes, but like you noticed, if we made it infinite the starting point would also become infinite so that's not that interesting. Secondly, the divergent form you give with two divergent infinite sums in the exponent ... just doesn't really seem to be telling us much. Conceivably we could string together tiles in some ways which cause the coefficients to converge to irrational numbers (which might mean that such "threadings" are impossible), or other ways where the coefficients diverge (also meaning such a threading is impossible), but we can't prove that all infinite arrangements of tiles map to irrational/infinite coefficients. When you recast the equation to have two infinite sums in the exponent, it becomes an indeterminate form so we can't even tell if the coefficients are finite/infinite or rational/irrational. I expect that it's really hard to find exact values for these coefficients in most cases, and the worst part is that most infinite sequences of tiles are just random and so hard to think about as the limit of finite sequences (sure every infinite sequence is the limit of its finite subsequences, but even after a sequence of arbitrary finite length it is exactly 0% of an infinite sequence). But I didn't completely follow that part or especially how the equation for all cycles was derived, I would probably have to read the paper

Finally, a near-miss that won't be named after Parker.

I belive the "Super Awsome Tree" is not an extension of the Collatz tree. You just generate squences with specific tiles applied, regardless if the numbers are odd or even.

You may not have solved the Collatz Conjecture, but you surely earned yourself another subscriber!

There are techniques which enable you to extract the beahviour of expansion of generating functions. Perhaphs, by reformulating the Collatz Conjecture into combinatorial problem (your structures) could give us better insights and perhaps, by approximating its generating function at infinity could give you some hints whether there is or is not infinite tiling of yours.

the way i cheered when you mentioned set theory

at 54:00 when you tried 1/8 as the input for your code, you said "integers", even though the last number was a fraction. I don't know if it changes anything, but I'm just letting you know.

Great food for thought

The number -2 is a Stark node that's followed by infinitely many Stark nodes, all equal to -2.

Naming the different types of nodes after GoT families is like naming a hockey stat you invented after like, your gamertag or something (that actually happened btw)

So I guess my next question would be is there a nice formula for all b that result in integer sequences? I don't think the super awesome tree is quite right unless you generalise the decision rule from divisible by 2 to something else, but without some thought I can't think of a generalisation that guarantees that exactly one of the rules applies for any given rational number.

Love the Phyrexian shirt!

the "X" in "TeX" is pronunced "K" since is a greek letter. Just like the statistical test "X squared"

My god, what a journey! Amazing video. Regardless of proving anything or not, your methodology and fun-factor are admirable.

I did some very brief playing around with this after being inspired by your video, one thing I found is that for each odd number that increases (4n-1), the next odd number they arrive at increases by one each time, (for example, 3 -> 10 -> 5, which is 1 odd number away, 7 -> 22 -> 11, which is 2 odd numbers and so on). Since those numbers are 2, 4, 6, 8... apart, only half of those numbers will go to another odd number that increases again (only multiples of 4 apart). Multiples of 8 will hit 2 increasing odd numbers, and so on.

The problem is that if there is no counter-example, then proving that there is none is extremely hard and requires that advanced math that only the most learned know. You cannot just "come up with a solution" if the solution is a mathematical proof that no counter-example exists. You need a huge amount of knowledge in math in order to come up with and correctly express such a proof. Plus, of course, finding the proof is extraordinarily difficult.

This raises the question 10 gives the sequence 10->5->16->8->4->2->1 in the Collatz Tree since the Collatz Tree is a subset of the Super Awesome tree, 10 should also give the same sequence in the Super Awesome Tree. However, 10 gives another sequence, namely 10->16->13->20.5->31.75->48.625->... That means in the Super Awesome Tree, a number can split into two paths, which makes the Super Awesome Tree not a tree anymore.

Interested to see if anything comes out of this.

a solution for a loop of super awesome tiles would be a solution for a infinite string of super awesome tiles but first we must define some thing that call a cfl of the loop of super awesome tiles it would be the loop but cut off when it repeat; the infinite string of super awesome tiles would be the clf repeat infinitely.

My attempt at solving it: In the case of 1, x->3x+1_>3/2 x+ 1/2_>3/4 x + 1/4 x=x Assume m is the multiplier of x, x a solution over 4, and b the addend. The equation would be mx+b=x m would have to be a fraction because a cannot be a negative number. b would have to be (1-m)x. The only way for a to increase is by using the +1 in 3x+1. The only loop in the Collatz conjecture for 10^20 is 1,4, and 2. Therefore, for b to increase to be (1-m)x (rough estimate, assuming m is 2/3) it would need 2.5*10^19 steps of 3x+1 to reach its goal. We want a number that repeats 3x+1, x/2 again and again to make b equal to 2.5x10^19. The first number that does a loop of that is 3, doing it one time (3 10 5) Then 7, doing it 2 times (7 22 11 34 17) Then 15, 31, and 63. This follows the formula 2^x+3. So, for a number to have 2.5*10^19 steps of 3x+1 in a row, it would need to be at least higher than 2.5^10^19+3, which is beyond the realm of any computer. This is an overexaggeration, extra 3x+1 steps would come after the start, but even assuming that the start only was a trillionth of the amount of 3x+1 steps, it would still require a number more than the amount of atoms. This is why I think the Collatz conjecture only works for 1, 4, and 2.

The distance between nodes: Let 2n + 1 be any odd number. It must be followed by the node 6n + 4 (you can also get there from 12n + 8). As the node is even the next number is 3n + 2, which can't be a node, but it can be even or odd. Case A) If it is odd, it must be followed by another node (9n + 7). So assume instead 3n + 2 is even, and thus n is even, say n = 2m. Then 3n + 2 is followed by 3m + 1. Case B) If m is odd, you could've arrived there from m as well, so 3m + 1 again is a node. Case C) The last case is when m is even. Then 3m + 1 is odd and thus must be followed by the node 9m + 4 (which can also be reached from 18m + 8). So in any sequence, between two nodes at most 2 other numbers can occur (and at least one other number must occur). An example with all cases: 17 52* 26 13 40*(C) 20 10*(B) 5 16*(A) 8 4*(B) 2 1. The nodes are marked with an asterisk. Note that 4 is a node because it follows 8 but also follows 1.

This problem is intriguing because it is easy to explain but so much deeper when you get into the nitty gritty. I have thought on it for about an hour with my experience in math like this being limited to a discrete math class. My first thought was to eliminate the idea of a loop from ever occurring besides x=1. If we can prove this is the only loop, that is a huge part of the problem solved. So I started simply by focusing on a loop starting with one odd number, which leads to only even numbers that will converge back to that odd number. So if x’ = 3x + 1, then x’ = 2^k * x for this to work with the simplest of loops. both x and k are integer values. I didn’t feel like making a rigorous proof for this, but you could notice that (3x + 1) / x is decreasing for natural numbers, and has a maximum of 4 when x = 1 and a minimum of 3 as x -> inf. So it’s not possible for this to be true except for the value 1 (I know I didn’t prove it all the way but it should be obvious). I then tried to introduce a second odd number into the sequence for a hypothetical number, where the sequence would go: start (odd) -> even -> odd -> even…. ….even -> start Working this out algebraically it reduces to the equation (9x + 5)/2^(k+1) = x. You can rearrange this to get (9x+5)/x = 2^(k+1), which I didn’t feel like solving algebraically but the limit is equal to 9. As you introduce more odd numbers, you can expect the numerator to be multiplied by 3 and add 2 each time, and the denominator to be doubled. This is how the equation works when you add odd numbers in a row to a sequence like start -> even -> odd -> even -> odd …. Until it gets to even numbers that reduce to the start. This made me hypothesize that maybe there is a connection between the idea of (3^k)/(2^k) = 1 being true (which it’s not due to prime factors) and there existing an X where adding an odd number anywhere would create a loop, which would mean this is impossible. I do not have the motivation to explore this idea, but I’d imagine it gets much more complicated when odd numbers are distributed throughout the sequence and not bundled at the start Has anybody else explored this idea in more detail besides pondering? I haven’t watched the video yet and it would be cool if he mentions this idea.

Oh, nice. I stumbled upon similar "number" system when certain polynoms represented in `up` and `down` notation to introduce 3n+1 and /2 operations applied to some random starting number. With random coefficitiens I tried to figure out which of the coefficients are valid solutions so i guess they are kinda isomorphic to GoT notation.

25:00 - Interesting insights so far, and maybe it does point the way towards a formal solution, but maybe we need to stretch the definition of “solution.” The Conjecture basically says that you’ll always run into a power of two and that there are no closed loops. Both have to be proved to satisfy the conjecture. Assuming that loops could be disproved, the existence of a starting number that never resolved to a power of two would imply an infinite number of starting points that never resolve to a power of two, and therefore an infinite number of independent galaxies or regimes of integers that never coincide. You’d perhaps discover a new class of number. Approaching Collatz as “purporting to seed an infinity of disparate and non-intersecting infinities” might “prove” that doubting the conjecture is absurd. On the other hand, if “any sequence” of connections is possible, one which never intersects another might be out there. It sounds like your technique to find any pattern would be unable to pick such a number, therefore that number lies outside of the entire space of the integers.