 In chapter 83 of A Wise Man's Fear by Patrick Rothus, there's a scene in which Quoth and his friends talk about riddles, and Quoth mentions that his father used to give him incredibly difficult riddles so that he would spend hours looking for the solution only for his father to reveal that he had made up the riddle, and there was no answer. One of his friends replies that this was a cruel thing to do to a child, but Quoth says that no, that it was fun. He explains that the harder the riddle is, the more you learn on your way to the solution, and if there is no solution, well, then you could learn forever. And to me, this is the heart of mathematics, trying to solve endless riddles and seeing what we can learn from them. Getting to the solution, if it even exists, is secondary. And for that reason, I have a special love for the unsolved problems of mathematics, like the Riemann hypothesis, the Navier-Stokes equation, and of course, the Collatz conjecture, also known as the 3M plus 1 problem. This problem is famous, at least among certain nerds, because it is incredibly easy to explain, and yet no one has been able to find the solution. It goes like this. So pick a number, any number, and then what you're gonna do is that if it's even, you're gonna divide it by two. But if it's odd, you're gonna multiply by three, and then add one. And then you take whatever result you got, and you are gonna do it again, over and over. And here's the thing. Whenever we do this, we find that we eventually reach one, at which point we go up to four, and then back to one, forming an endless loop. But does it have to be that way? I mean, for example, 82 starts a sequence that grows all the way up to like 9,000 or something like that, and then it comes down. But maybe there's a sequence that never comes down. Or maybe 4 to 1 is not the only cycle. Maybe there are other cycles out there. But if they exist, no one has been able to find these sequences, and the conjecture is that they don't even exist. We've tested every number of like 3 quintillion or something like that, and indeed, every single one eventually came down to one. But this doesn't prove that it never happens. It just proves that if it happens, it happens with some really, really big numbers. Veritasio made an excellent video covering the history of this problem and why it is so hard, and one of the comments stood up to me. Everyone here, but maybe I'll be the one to find the solution. Yeah, and it's really funny, because it's true. At least for me. At one point, I thought I could solve it. And of course, that's silly, right? Because if the solution is ever found, it will be via a very skilled mathematician who has spent years learning. But that doesn't mean we should let them have all the fun. Here's a thesis of my video. You should try to solve this problem. In fact, all of us should try to solve open problems once in a while, not because any of us will be the hero that finds the solution, but because if you try, if you honestly try, you will learn so much, you have no idea. That's why for the third summer of math, I want to try something different. I want to give this problem an honest try and show you how it goes, mistakes and all, and see what we can learn in the process. Spoilers, I was able to find a connection between the collage conjecture and the salations, the branch of geometry that studies how shapes fit together. And from there, I was able to create an algorithm that lets me find any finite sequence in the collage stream. For example, if I wanted to find a sequence that increases by three over two three times, but then decreases by three over four once, and then increases twice, but then decreases four times in a row, but those times by one over four, then here it is, this is that sequence. And by changing one variable, I can find infinitely more sequences with this exact behavior. Pretty cool, huh? And the link is in the GitHub, so you are encouraged to upload your improved versions, especially if you include cool ways to visualize it. And yes, I know that someone else must have found all of these things already, but I bet you that he didn't make a YouTube video explaining it. For this challenge, I will not look up anything about the collage conjecture nor mathematics in general. I will only use what I already know, because the fun is to figure things out on your own. And if I see all the dead ends other people have already reached, it would spoil the fun of getting there myself. But this also means that if I make a mistake along the way and I don't catch it, that's it. That could make everything that comes afterwards wrong as well. And I fear that the first comment is going to point something extremely obvious I missed. And I'm going to feel so embarrassed, as if my ass had fallen off. But you shouldn't be the one to discourage yourself. You shouldn't be your own poly. So let's do this. Looking at the collage sequences, you can see that if they ever share a number, they become identical from that point on. So we can arrange the sequences into a tree like this one, whose trunk is the sequence 16, 8, 4, 2, 1. And then if the conjecture is true, all the other numbers are somewhere in this tree. Some of these branches actually stop branching, and they go on forever as a single line of numbers. And they are always made by some power of three, multiplied by higher and higher powers of two. These branches are not relevant for anything for the rest of this video, but I just wanted to mention them so that you know that I didn't forget about them. Let's establish some terminology. These three is a graph. Graphs are made of vertices connected by edges. Most numbers are only connected to two other numbers, the one before in the sequence and the one after. But some numbers are connected to three numbers, because they are the point where two sequences meet. These vertices are very special, and for that reason they need a special name. And since I've heard the word node used when talking about graphs before, that's the word I'm gonna use. Apologies to all the graph theorists out there if there was a better word I could have used. I didn't know about it. Now, something very important is that after another number, we always have a node. That's because if we do 3n plus 1, where n is all, we always get an even number. But that means we could have also gotten to it by halving another number, like how 16 is 32 over 2, but also 3 times 5 plus 1. But I don't like this condition of saying where n is all. Our running theme in this video is going to be that we want formulas and equations that work whenever we plug any integer number into them. For example, all the odd numbers are always one more than an even number. So they can all be written as 2n plus 1. And so when we multiply by 3 and add 1, we get 6n plus 4. This means that all the nodes in the collapse tree can always be written as 6n plus 4. This fact will be extremely important to the point that I would not call this the 3n plus 1 problem, but the 6n plus 4. Since I was already focused on the nodes, I began looking at their separation. And I noticed something really interesting. Sometimes there was a single vertex between them. But sometimes there were two vertices. And I couldn't find a single instance where this didn't happen. And from that little observation, I made my own conjecture. The distance between nodes is 1 or 2. And I set out to prove it. Notice that by this point, I have completely lost sight of the collapse conjecture. But that's fine. Sometimes you have to do side quests to advance the campaign. Okay, so if you are in a node, there are two possible ways you could reach another node. You could divide by two some number of times until you reach the next node. Or you could divide by two some number of times until you reach an odd number. At which point, you know for a fact that the next number in the sequence will be a node. Now we can write my conjecture more formally by saying that these three equations always have solutions. This one represents when you divide by two and you reach an odd number. This one represents when you divide by four and you reach an odd number. And this one represents when you divide by four and you reach another node. These are called theophantine equations named after Diophantus of Alexandria, an ancient mathematician who wrote many books about how much he loved integers and how much he hated fractions. To the point that mathematicians said like, okay, dude, we get it. If you like integers so much, then whenever we are looking for integer solutions to an equation, we are going to call it theophantine. Just chill. Simplifying these equations we get n is equal to 2k minus 1 over 3. n is equal to 4k over 3. And finally n is equal to 4k plus 2. If my conjecture was false, there should exist some number n for which no integer k makes any of these three equations true. A node made using such a number would need to be divided by 8 before reaching an odd number or reaching another number of the form 6n plus 4. Let's look at this table where I've calculated the value of n in each equation for each value of k starting in k0. As you can see, there are many fractions in this table. And that means that those values of k are not solutions for those equations. But before we jit them into the zone like Diophantus would have wanted, look, this number here starts at minus 1 but goes up by 2 every time. And this one starts at 0 and goes up by 4 every time. And that means that every third number, they will be multiples of 3. And that 3 will cancel with this 3 in the denominator, leaving us with integers. In this case, we will get all the odd numbers. And in this case, we will get all the multiples of 4. And then finally, this equation always gives us integers, but they are always 2 more than a multiple of 4. And that will give us all the even numbers that are not multiples of 4, which are the only numbers we didn't have already. And so with that, we have all the numbers. And so this means that my conjecture was true. Can we get a scoreboard? It is very easy to compete against their people because they cannot earn any more points. For every number you give me, if I make a node with it, it will be part of one of these families of nodes. So let's give them names. If n is odd, it will be a Stark. If n is a multiple of 4, it will be a Lannister. And if n is even but is not a multiple of 4, it will be a Targaryen. We can represent each kind of node as a sort of puzzle piece, and the collat streak can be made by combining them. I was editing this video and I realized that I should have explained how these styles work. Okay, so this represents a node. And then this notch represents a connection to a node number. And this knob represents a connection to an even number. That's why Stark and Lannister nodes end with a circle because they connect to other nodes through odd numbers. But Targaryen nodes have this moon-looking thing at the bottom because they connect to other nodes through even numbers. Okay, so let's go back to the video. I call these tiles the super awesome tiles, because when I was nine, I promised myself that if I ever discover something, I would call it super awesome. And I'm a man of my word. Unless someone has already discovered them and gave them some boring name. Anyway, using these tiles, we have transformed the collats conjecture from a problem in number theory to a problem in geometry, specifically in tessellations, which is a lot easier to visualize. This is how the collat streak looks when we use these tiles. Beautiful, isn't it? But we are not limited to only what we can see or what we can find in the collat streak. We can just start placing random tiles together, and we don't care about the numbers, not yet anyway. And so, for example, these tiles can represent these sequence of nodes. These tiles can represent these sequence of nodes, these tiles, these sequence of nodes, and so on. You get the idea. We need a name for these sequences of nodes that are independent of the numbers. So I'm going to call them constellations, because I like the idea of the collat streak as a huge starry sky above us. And we are just looking for patterns in the stars. This actually reminds me of Feynman diagrams. These are drawings of the different ways that particles could interact. But Feynman realized that if he gave a precise mathematical meaning to each element in the drawing, he could transform these drawings into integrals. And the solution of each integral tells you how likely that process is to happen. Something similar happens with these tiles. Each tile represents a number. And the position of the tile tells you how it is related to the other numbers. So it should be possible to transform every sequence of tiles into a single deophantine equation. And the solutions to that deophantine equation would tell you where to find this pattern in the collat streak. To figure this out, we are going to have to find formulas that tell us the relationship between a node and the next node in the sequence. First, let's analyze Targaryen nodes. I know that if I divide by 4, I reach another node. So we have 6n plus 4 over 4 is equal to 6s plus 4. And so we just have to find integer solutions for this equation. And to do that, I'm just going to use Wolfram Alpha. I know I said I wasn't going to look up anything about mathematics, but Wolfram Alpha is just like a fancy calculator. It does things that I could do on my own, but they would take too long, and it makes this video shorter. Okay, so using Wolfram Alpha, the solutions are that n is equal to 4a plus 2 and s is equal to a, where a can be any number. So any number from 0 up to infinity that you put in these equations will give you a Targaryen node. So these two equations actually define the entire family of Targaryen nodes. Neat. For start nodes, the equation is a bit more complex because we have to divide by 2 and then multiply times 3 and then add 1. But when we do the solutions are n is equal to 2a plus 1 and s is equal to 3a plus 2, it's larger than n. Oh, I see. So yeah, these are the nodes that actually make the sequence increase in value. Interesting. And finally, for Lannister nodes, we have that we divide by 4 and then we multiply times 3 and then we add 1 and all of this is equal to 6s plus 4. And we find that the solutions are that n is equal to 4a and s is equal to 3a. But wait, before we use this to try to find a constellation in the tree, we need to make sure that the constellation we are looking for will actually exist somewhere in the tree because I don't know, maybe there are some mathematical restrictions like maybe there are never five start nodes in a row or maybe the sequence L, S, T, S, L never happens. I don't know, something like that. To see if this happens, we have to check if the equations of the nodes are always compatible with each other and they are, they always are. There are nine possible combinations and we check for integer solutions in the same way we already did. I'm not gonna go over them, but if you want to check them for yourself, they are on screen right now. The important thing is that there are no restrictions. There's no reason any kind of node couldn't be followed by any other kind of node, even itself. And this means that whatever constellation you make with these styles, most exist somewhere in the collage tree and we can develop an algorithm to find exactly where. But to do that, we have to tackle a slightly different problem, the infinite constellation of start nodes. This lack of restrictions is dangerous when it comes to start nodes because the other nodes always decrease in value. So there's an inherent limit to how many of them you could have in a row. But start nodes always increase in value and there's seemingly no reason why we couldn't have an infinite sequence of start nodes and such a sequence would never come down to one. Does this prove that the collapse conjecture is false? Not at all. There is a limit. Let me show you what. Let's say that we were in a start node given by a number a0 so that the node can be written as six times two a0 plus one plus four. Okay, then we know that the next node will be written as six times three times a0 plus two plus four. But then if that next node is itself a start node means that there most exist a number a1 such that the node can be written as six times two times a1 plus one plus four, which means that two times a1 plus one is equal to three times a0 plus two. And then if we solve for a for a1 we get that a1 is equal to three over two times a0 plus one over two. Okay, but then we can do the same and then we get an a2 that is this and an a3, a4, a whatever. We can do the same process any m number of times. Luckily, I had already solved many problems like this when I was learning how to solve differential equations using infinite series like the Frobenius series. Basically, all we have to do is see what is happening at each step and then generalize the pattern. For example, we can see that this number is just getting multiplied by three over two each time, but it started as three over two. So it's just three over two to the end. And then this other number is also getting multiplied by three over two, but it started as one over two. And then we have to add this other number, which was calculated in the same way, except we multiply by three over two one less times. So then we can sort of like open it up and we see that it is a series of sums. And so we cannot represent it with just this single thing. And then we can just find the solution by using Wolfram Alpha, of course. And then we get this. Am is equal to three over two to the m times a0 plus three over two to the m minus one. This is an exponential Diophantine equation. And Wolfram Alpha wouldn't give me the solutions. So I had to find them myself like a person. But once I found them, they felt pretty obvious. They are just a0 is equal to two to the m times b plus two to the m minus one. And then a m is just equal to three to the m times b plus three to the m minus one. And this number b can be any number from zero to infinity. And that might seem confusing, but it makes perfect sense. That means that this constellation made of m plus one star nodes doesn't exist in just one place. It exists in infinitely many places all over the collage tree. But here is the crucial part. The places where these constellations begin and end are always given by these formulas. For example, let's say that you have a constellation made of three star nodes. Well, then three is equal to m plus one. Because remember that we begin with a0. So we need to find a1 and a2. And b could be anything. But let's make it easy and say that b is equal to zero. Then a0 is equal to three. We put that into a star node and we get 46. And indeed, we see that 46 is a star node. And so is 70. And so is 106. But 160 is a Targaryen node, because we only specified that we wanted three star nodes. Using this, we can find all other constellations made of three star nodes by simply changing the value of b. Try it for yourself. Just be a little bit careful, because these numbers can get really big, really fast. And the computer seems to struggle with numbers larger than 2 to the 64, probably because of how it stores the numbers in the memory. But I don't know what the engineers had well from Alpha did, but you can really abuse it. Like, for example, this is the number that starts a sequence of 10,000 star nodes in a row. But fair, you ask me telepathically across time and space. The formula you have for a0 always gives all numbers. But what if we make a star node using an even number? Well, boys in my head, in that case, that node won't be followed by another star node. Take 82, for example. This is a star node made with a equal to 6. But then when we put 6 in the formula for a0, we see that there are no integer solutions for b and m. Which means that this node cannot be followed by another star node. And indeed it isn't, because 124 is a Lannister node, because it is a multiple of 4. But going back to the problem of the infinite constellation of star nodes, now we can see that if m was infinite, then a0 and am would be infinite too. And this means that a finite number could very well start an extremely long sequence of star nodes, but never an infinite one. This is why there is a limit. But of course, this doesn't do anything for the Collatz conjecture, because there could very well be an infinite sequence out there. It just means that the streaks of star nodes in that sequence take a break every once in a while, before going back up. But who cares? By analyzing streaks of star nodes, we were able to get a formula to find all of them. If we get similar formulas for the streaks of the other kinds of nodes, then maybe we'll be able to find any possible constellation in the tree. The process for finding the formulas for the other kinds of nodes is pretty much the same, except the solutions for the Targaryen nodes don't need a number b, because any number can already be a solution to this equation. Because if you look at the tiles, you can see that you can always multiply by 4 to get to another Targaryen node. And this means that you can start or end the sequence anywhere. Now, the question is if we can use this to find any possible constellation. And we can. We just have to think of constellations as being made of multiple sequences of streaks. That's right, we are multi-streak drifting! To understand how we're gonna do this, we have to think about sets. At first, if you just want to find any constellation, but you don't care about which one, well then you can use any positive integer. So this is your initial set. But then if you want to find a constellation made of a streak of a single kind of node, then you can use the formula for that case. And now your set has become smaller. Now it only contains the numbers that are solutions to this equation. And then if you want to find a constellation that is made of multiple streaks of different kinds of nodes, well then you have to combine these equations in some way to get a single equation. And your final set will contain only the numbers that are solutions to that equation. For example, let's say that you wanted to find a constellation made of three stark nodes. Well then you can just calculate a2 of s. I'm using the s here to specify that it comes from a start node. And you know that the end of the last node in the sequence will be equal to this. Because remember that the last node in the sequence is always something else that we cannot control. But if you want to say that that node has to be a Lannister node, then you can just take the end of a Lannister node, which is for a of lambda and the lambda to specify that it comes from a Lannister node. And then you can just make these two things equal and then divide by four to solve for a of lambda. And then you have it. Not all numbers will satisfy this condition. And so for that reason you have reduced the set of possible solutions. Now if you want this to be a streak of four Lannister nodes in a row, you just plug this into the formula for a3 of lambda. And then you have it. The solutions to this equation tells you all the places where you can find this constellation in the collapse tree. And this is actually how my code works. Although my code goes node by node instead of using streaks, which I know makes it less efficient, but it was so much easier to write and I have so many other things to work on. This is how we can transform any constellation into a single deophantine equation. Kind of like a five-mind diagrams can transform these diagrams of particle interactions into a single integral. I call this process threading because in my mind it's like if every node was a bead in a color or a bracelet and we are threading them together, connecting them through restrictions in the set of solutions until we find the equation that links them all together. Trees, constellations, tiles, and now bracelets. My metaphors are all over the place, aren't they? But what are you gonna do about it? Subscribe? Like the video? Ring the bell? You wouldn't there? This threading is a recursive process and I think that means that the collapse conjecture must be related in some way to the Mandelbrot set and the logistic map in the sense that these are also recursive processes. And it has been known for a while that we can model chaotic systems using recursive equations and changing the initial conditions, which is pretty much the same thing we are doing here. This is probably the reason why the sequences in the collapse tree seem so chaotic. And in that sense, the super awesome tiles help us flow through the chaos. Instead of trying to predict if a number reaches one or trying to predict how many steps before it reaches one, like a sailor desperately trying to steer a ship through a storm, we just say where we want to go and the tiles take us there. Originally, this is where I was gonna end the video. I mean, I found a connection between the collapse conjecture and the constellations and I found an algorithm that lets you find any pattern you want in the collapse tree, mission more than accomplished. But then I had an idea. A question. Could I find the starting point for an infinite constellation? Because that's one of the ways that the conjecture could be proven false, right? Okay, so we are obviously gonna find that we cannot prove that such a constellation couldn't exist, but it will still be fun to try. Okay, so how should we approach this? Okay, so first idea. If we have a constellation, we can just keep adding tiles at the end. And in that way, we could certainly have an infinite constellation. Yeah, but that doesn't work because every time we add a new tile, we change the equation for the whole thing. And it changes scoutically. So this means that every time we add a new tile, we change the places where this pattern begins and ends in the collapse tree. And that doesn't work. What we want is an infinite constellation that has a fixed starting point. Hello. At this point, I forgot to remind you that the way we calculate these equations is by using the formulas for the streaks repeatedly inside each other, like the result of one streak will be a zero in the next streak. And this will be extremely relevant by the end of this section. So just keep it in mind. Okay, so let's imagine we have an infinite constellation. In that case, we could calculate its equation from the beginning up to some point in the middle. And what we should find every time is that the solutions for a zero are always the same. Or if the equation is different, it is different in a way that always results in the same solution. But a n changes in such a way that it can always be something different. Okay, this seems really difficult, but perhaps not impossible. But then if an infinite constellation existed, we should be able to calculate its equation and not up to any point in the middle, but all the way to the end. How would that look like? Like, would we have something like a infinity is equal to infinity times a zero plus infinity? Like, this is meaningless. This tells you nothing about the sequence itself. So the only way this can exist is if these constants eventually converge to something as you calculate the equation up to infinity. Okay, and this, that seems possible. Yeah, sure, these constants could converge to something. But then how does that make sense for a infinity? How can this number go all the way to infinity if these constants are finite? Well, maybe because it doesn't go to infinity. Maybe a infinite is actually a finite number. Maybe what happens is that this infinite sequence goes up and down and up and down. And it just never happens to reach this particular number. So it's not that it rises forever, it just rises and goes down. It just never reaches this number. But that raises the question, like, what happens if you use the collapse rules with a infinity? Like, because it's not an infinite number, right? So what would happen? Well, there are two possibilities. I mean, it's clearly not part of the collapse tree. So it could either start a new infinite constellation, or it could be part of a cycle. So it seems that if we find one a zero that starts an infinite sequence, we should find infinitely more infinite sequences, infinitely more trees disjointed from the main collapse tree. But the problem with all of this are these constants, because we have to understand how is it possible for them to converge to anything. And it can be really hard to imagine what's going on here because we have all these fractions raised to different powers. But what we are going to do is that we are going to write them using Euler's number raised to these logarithms. And I know this looks very different to a fraction, but trust me, it is the exact same thing you can test it for yourself. Okay, so now what happens is that as we do the threading process, we might not be entirely sure what's going on because we don't know the order of the constellation. But we do know that every time we thread through a streak, we are adding another exponent in the sequence, right? So in an infinite constellation, we would get these infinite sums in the exponent. And that might not seem like a huge deal, because infinite sequences converge to finite values all the time. I think they even teach that in Sesame Street these days. But the problem in the here is that for an infinite sum to converge, it needs to use fractions. And these are not fractions. These are integers because they represent the length of the streaks. And an infinite sum of integers doesn't converge. It cannot converge. Even if it was just one minus one plus one minus one plus one minus one, that diverges when you take it all the way to infinity. So this sequence here, this infinite sum, whatever it is, it doesn't matter the order of the sequence. It will diverge. We will get something to e to the plus infinity minus infinity. And that just doesn't make any sense. And so that seems to prove that an infinite sequence cannot exist, because it would require an infinite sum of integers to converge. Which means I must have made a mistake somewhere. Because, okay, in the veritasium video, Derek mentions that it is relatively easy to find any number that exhibits any arbitrary behavior, which are the notes we just found in this video. So people knew about this. Like I told you this wasn't new. Someone else had found it. I was just rediscovering it on my own. But if someone else found it, they obviously didn't disprove that infinite sequences are possible in the Colette Street. So they must have seen something that I'm missing. They must have seen a reason why this doesn't work. Unless it is me, the one who is seeing something new. And at this point, I have to share with you my greatest fear, to be a science crackpot. The internet is full of these people. They are people who claim to have made amazing discoveries and that all the other scientists are wrong, but only they are right. Only they have managed to pierce the veil and uncover the mysteries of the universe. I don't want to be like those people. I am a scientist. I am on the side of truth. I don't want to be wrong. And so that's why I insist so much that I could be wrong, because that's the one thing these people never do. They never consider that they could be wrong. And just to show you how bad it is, if you look for proofs of the Colette's Conjecture, you find many of them. Look at that. You think the Colette's Conjecture is the most proven problem in mathematics. But of course, it's not. Of course, all of these proofs are wrong. But the people who made them seem very, very confident that they are right. And they probably think that everyone else is wrong for not seeing their genius or something like that. And I never want to be like that. If I ever become like that, kill me. Okay, so how can I know if I'm wrong in this case? Okay, we have a couple of options. The first option would be like to check my reasoning and find a mistake in there. But I'm not smart enough to see any mistake in here. I just cannot see it. Okay, so another option would be to get smarter people to look at your work and find a mistake for you. But the problem is that life is too short to look at all the wrong proofs of the Colette's Conjecture. So obviously I couldn't get anyone to look at this. So then the third option I have is just to look at common elements of wrong mathematical proofs. Okay, so Scott Aronson is a scientist and he has a lot of experience reading mathematical proofs. And he has noticed some common patterns in proofs that are wrong, and he made a list of them in his blog. So let's go one by one. Number one, the authors don't use text. Okay, so text is something like a coding language, but for writing text, like all the scientific papers and textbooks are written using it. And I used a version of text known as latex. So I think I clear this point. And I also clear some other of the first ones, but spoiler alert, I don't do very well in this test. Number two, the authors don't understand the question. I hope I understand it. Like, the Colette's Conjecture is really easy to understand. If I fail this point, I'm really in trouble here, but I think I clear it. Number three, the approach seems to yield something much stronger and maybe then falls, but the authors never discuss that. Well, I think this is just pretty much limited to this one problem. So yeah, I think I also clear this one. Number four, the approach seems to conflict with a known impossible result. I'm going to say that I have cleared this. Like, as far as I know, I don't think this is happening, but maybe it is. I don't know enough mathematics. Maybe this conflicts with some other results that I am unaware of. Number five, the authors switch to whistle words by the end. Okay, so if I understand this point correctly, is that these people are not entirely sure of their result, but they present it as if they were sure of it. And I don't think I did that. Like, I'm pretty sure that an infinite sum of integers diverges. Like, I'm pretty sure these equations are correct. That's a problem. I'm pretty sure of everything I presented here. I cannot find the problem. And that's the problem. But yeah, I think point five, clear. Number six, the paper jumps into technicalities without presenting a new idea. This is where I started going wrong with this test. Yeah, I mean, my tiles are interesting and they are certainly new to me, but they are not new in general. And even if they were, they are not advanced enough to count as anything new. So yeah, point six failed. Number seven, the paper doesn't build on or even refer to any previous work. Well, I mean, that was like part of the challenge, but okay, yeah, point failed. And I don't even know what other word I could even refer to. Like, I purposely didn't research the topic. So yeah, super failed. Number eight, the paper weighs a lot of space on standard material. Okay, so this video is intended for the general public, but okay, yeah, sure. Point failed. Number nine, the paper gets poetic and philosophical. Yeah, no, I didn't do that. If I ever mistake philosophy for science, kill me. Number 10, the techniques just seem too wimpy for the problem at hand. And yeah, I agree. Absolutely point failed. That's five points out of 10. And he says that six and 10 are the most important ones and I failed those. So it seems overwhelmingly likely that I am wrong. I'm sorry, I just can't ignore it. It's right there. We've come all this way. I have to at least try. I'm sorry. Okay, so if this was correct, which it isn't, but if it was, then what about cyclical constellations? Because that's the other way the conjecture could be false, right? Aha, but this time we cannot play the same stupid game as before. We cannot argue that there's an infinite sum of exponents because a cycle, by definition, is finite. And so we cannot prove that a cycle doesn't exist, unless... At this point in the video, I gave what seemed at the time like a proof that cyclical constellations have to use negative numbers. And it was pretty cool. I used the metaphor of an auroboros to split the cycle into two constellations, which I call snakes. And I reasoned that the values for the variable A have to go up and down until they reach the same starting point, finding that this was only possible if the variable A was negative. Sadly, while editing this video, I realized that hidden in my reasoning were a series of assumptions that are not true in general. So, keep this in mind for later in the video when I mentioned that this seems to prove the colatz conjecture. That's wrong, as expected. However, what I can do is tell you that if cyclical constellations exist, we should be able to find all of them using this equation. Let me show you who I found it. Okay, if there was a constellation, you could calculate its equation, right? And since it's a cycle, the first node and the last node are the same. So A0 is equal to An. And this is great, because we actually have the solutions for A0 and An. We get them using the extended Euclidean algorithm, which I'm not going to explain because there are a trillion other videos about it. But now we can solve for B. And there you have it. The integer solutions to this equation should help us find all the cycles of the colatz 3. In fact, alpha has to be a power of 2 and beta has to be a power of 3. But I'm too tired, so I'm not going to explain why. I mean, I mean, this is left as an exercise for the viewer, as if anyone ever does any of this. It feels disappointing not being able to give you a proof, or at least something that resembles a proof. But I guess one of the oldest traditions in mathematics is saying, I give up. If you figure it out. In fact, some mathematicians became famous not for the problems they solved, but for those they couldn't solve. Like Riemann, Goldach, Poncaire, and of course, Colatz, among many, many others. And if this equation can indeed predict cycles, then I am now a small part of that tradition too. Unless someone else has already found it. But just to be clear, the question is not to find whether this equation has solutions or not. It is to find all the solutions. Because we already know a couple of them. With positive numbers, we only have the 4 to 1 cycle. But with negative numbers, there's more than one cycle. There's three actually. You are seeing them on screen right now, along with their equations and how they satisfy my formula for b. And yes, my code can find all of them. Anyway, that's all I have to say, and now me from the past has a couple of cool things to say about negative cycles. But anyway, so with the two cycles that my code can find, I can see that they have b equal to minus one. But then when I change b out of curiosity, I found many other sequences that have the same structure. They have the same kinds of nodes in the same order. But the difference is that they are not cyclical. They don't loop around. And I think I know why. We can think of cyclical constellations as one cycle, but also two cycles and three cycles and four cycles and any number of cycles. And all of these represent different constellations of different lengths. And they obviously have different equations and different solutions. But for each of these possible constellations, there most exist one number whose prime coefficients are just right to cancel all the powers of two and three and make all of these equations identical. And so these constellations are disjointed everywhere else, except here in this one point. And if we were to start the constellation in any other node, we would find another family of constellations that are also disjointed everywhere else, except here. This seems to prove that cyclical constellations are impossible. And that plus my previous proof that infinite sequences are also impossible seems to prove that the conjecture is true. Which means that this video has become just one more in this large of fake proofs of the collapse conjecture that swam through the internet. But I hope this video can have more value than all of those precisely because I am scientist enough to recognize that I am wrong. And I welcome and indeed crave to know what I did wrong. And when I know you can bet I'm going to make a video explaining what mistakes I did so that we can all learn from my mistakes together. Okay, so what did we learn? Well, in the one hand, we went through a journey through the authentic equations and infinite series and set theory and I don't know what else to find the super awesome tiles and to find this algorithm that finds any pattern in the collapse tree. And all of this is really cool. And I'm pretty sure it is right because I've tested it a lot. But in the other hand, we also have this walkie walkie walkie, I don't know, false proof of the collapse conjecture. And you know what, this feels about right, because if we had not taken the risk of being wrong, we would have never found the stuff that is true. And so I think that the trick to being a good scientist is to always be open to the possibility that you could be right so that you take the risk and maybe you find this cool stuff that is true, but that you also never forget that you could be wrong so that you don't mistake everything you find as being true. And finally, I have to thank all my patrons, but especially Carlo Fascioli, Robert Bigman, Alec Luhman, Otimo Tiozum, Azuria Salia, Mattia Serdi, Ryan Roberts, T. Heidel, Valerie Hyde, Dan Babbage, Daniel McGillibright, David Saunders, Jadon Hansel, John Stein, Marlo Vron and Matt Zweig. All my patrons, thanks a thousand times, thanks a million times. And that is where I was when I end the video, but I just have to improve my code, right? Just have to improve it a little more. Okay, so here's what happened. Do you remember how 82 starts this extremely long sequence? Well, I wanted to test my code, I improved my code and I wanted to test it, I wanted to take it to the limit. And so I made this simple code that you give it a sequence of numbers. Well, you give it a number in the collapse tree and it not only gives you the constellation, right? Like the nodes. It doesn't give you like the numbers, you already know the numbers. It tells you what kinds of nodes you find here, all the way until you reach one. And it was like, great. And I started testing it with many things. But then I eventually thought, well, this is a very long constellation, I'm going to try with this one. Okay, so we copy this constellation, I already have it here. Okay, so then this is my code. You give it a constellation and it gives you the equation. And then when I run it, there it is. It works. So here you can see that we have the, ah, let me make this a little bit bigger. Okay, so here this is the equation. It has some very large numbers, so that's to be expected. Here are the solutions. And here it gives you one example of a sequence of numbers that matches this constellation, because remember, the constellations exist in infinitely many places all over the collapse tree. And I thought, well, all of this is fine. But I want the example that starts with 82. And I was very confused, very perplexed by this. Because as you can see, I'm using the constant B equal to zero. Well, zero over one, which is actually zero. The entire code has to work with fractions to avoid issues with floating point numbers. But anyway, so I was like, okay, but how can I, what else can I do? Because if I use a negative number, if I use B is equal to minus one over one, which is minus one, well, then I get negative numbers. I was like, where is 82? Like, if I use larger numbers, if I use larger values for B, I'm gonna get larger numbers here. And if I use negative numbers, I'm gonna get like, I'm gonna use, I'm gonna get negative numbers. So what number do I have to put in here to get 82? I was very confused. I was thinking that maybe I had made a mistake that maybe my algorithm could find many constellations, but maybe it couldn't, there were constellations that I couldn't find. And so, well, I went to, well, from alpha, of course. And then I said, okay, so this is the equation for B. I mean, for a zero, as you can see, this is a zero. And I just basically, sorry, I copy it. I paste it here. And then I know that 82 has an a zero equal to six, right? And that has to be true, like, regardless of anything else, right? And so I thought, well, okay, so tell me, well, from alpha, what value of B do I need to make this equal to six? And this was Wolfram's response. Not only a fraction, but a negative fraction. And I was very confused. But I thought, well, here goes nothing. Let's try it out. Okay, so here is the thing Wolfram alpha gave me. Let's try it. And it works. It gives me the right sequence, 82, 124, 94, blah, blah, blah, and so on, on the way up to the last node, which is four. I was like, extremely confused at this point. It was like, what, well, I'm still very confused. I was thinking, like, what I'm using everything is Diofantine, right? I'm using only integers. I made everything using integers. Why, why do I need a fraction to find this? And then I thought, well, what happened? Okay, so what if this was one? What happens then? And Wolfram alpha gives me another solution. So let's try it. And we get the sequence that looks starts reasonable enough, 10, 16, oh, I've seen this before, but then it goes 13, 20, 31. And very oddly enough, finishes with two, four, one, and then some decimal points. And then I thought, well, I can just put any number in here. I can just put any integer in here. And Wolfram alpha will find a solution for B. And then I can take that value of B and put it in here. And I'm going to get a sequence. And then it gets weirder because I was like, well, what if I try some other constellation? Okay, and then I decided to try this constellation. And I'm going to try it with B equal to zero. Okay, we get something that is reasonable enough, right? Only integers like to be expected. And then we put B equal to one. Again, only integers B equal to two. Like, of course, like, what am I saying? Like, of course, you know, this is going to get integers. But the weird part is what happens when we put a fraction one over two. Look at that. Not only, obviously, we got a result. But look at that. All of these are integers. All of these are integers. If you look for these, if you want to find these sequence in the collage tree, using this equation, you have to use B equal to one over two. And that's not all. One over three, it works. One over four, it works. One over five. Okay, now it gives us things that are fractions. But what happens if we go to one over six? Integers again. One over seven. Fractions again. One over eight. Integers again. One over nine. Fractions. One over ten. Fractions. One over eleven. Fractions. One over twelve. Integers. No idea what's going on. Okay. Okay. Okay. Okay. The only conclusion I can find here is that B is not an integer number. B can be any number from zero up to infinity, but it's a continuous number. So, B is a continuous number. It has to be. We have to allow B to be a rational number in order to actually find all the constellations, because if we only use integer values of B, there are sequences we are not going to find. But then, if we can, if we are allowing us to use some values of B that are not integers, what right do we have to exclude other values of B that are also not integers? Like, we have to allow all of the values of B. But if we allow all of the values of B, then we get these other sequences, right, that are not made of integers. And the only conclusion I can find here is that these are also sequences in the tree. Like, we thought this tree was made only of integer numbers, but it isn't. Using these equations, we can generalize the tree to all the numbers from zero up to infinity in continuous, all the rational numbers from zero up to infinity, not only the integers. And it makes sense, right? Like, we are using the same equations in both cases, and we are getting results that, yeah, make sense. They follow, they certainly follow the rules. So, yeah, I guess the collapse tree was just a much smaller version of a larger tree, which I have to call the super awesome tree, just in case someone else has not found it already, so that I can fulfill my promise. So, yeah, this is the first glimpse into the branches of the super awesome tree that is made when you thread using the collapse rules, but you allow the variable B to have values, to have any value. Yeah, that's what happens. Because, yeah, and the collapse tree is only when you allow the values of B that result in integer values of a zero and a n, and I don't know, I think this is pretty cool. And the best part is that all of this, like, is independent of whether or not I'm right about the collapse conjecture, right? Because, yeah, this is true, like, the algorithm works. The algorithm certainly works. I tested it enough to be convinced of it. So, the algorithm works, and if the algorithm works, and it also gives the sequences, then the sequences also have to be right. And so, yeah, even if I'm wrong about the collapse conjecture, I think I'm right about this. I mean, I would be, oh, okay, maybe not. But I think I am. This is pretty cool. Yeah, I think, yeah, that's all I wanted to say. I just wanted to share this with you and say, like, yeah, this is pretty cool. And this is the super awesome tree. Okay, anyway, thanks for watching. Hopefully, this time for real.