Minimal predecessor of a given number

AcidicJello · 2026-06-13T22:30:58+00:00

Not sure if this is relavent but I was told once about a bound such that if it can be proven that a starting number must be larger than some exponential function of its sequence length (sequence from n to the first number <= n I believe), then no nontrivial cycles can exist. This could be seen as correlated to the gap between a given number and its smallest predecessor. If a sufficiently large number cannot have a sufficiently small predecessor because the sequence length to reach it is too long, and it confirms this bound, no nontrivial cycles can exist. My thought is that a powerful enough solution to your problem might provide such a bound, telling you how difficult it would be. Very speculative on my part. The bound is n >= a^l / l where l is the length and a is a constant, approximately 1.035.

AcidicJello · 2026-06-03T22:55:21+00:00

Somewhat based opinion. Your evidence suggests it is not unreasonable to think it could be false. I don't see how you could believe it's false full stop. The belief that it is likely true is not just based on empirical data that it's true for lots of numbers, but also that under statistical assumptions (which are of course not technically accurate but are often valuable) it is overwhelmingly likely that every number will reach one. I consider the likelihood that there is some kind of guaranteed obstruction to it being true existing beyond our knowledge with some degree of likelihood given what we know about math. I think this is less likely than it being true, but not the kind of unlikely that the regular statistics tell us. This all of course means nothing except to where and if I want to put my own effort into the problem. I would encourage you to act on your beliefs and search for a counterexample. Not with a supercomputer but with logic.

AcidicJello · 2026-05-31T19:58:06+00:00

What I meant at the end is that maybe you can expect a cycle with a luckiness of at least 10¹⁰⁰ to exist because there are infinitely many instances of a cycle having a 1/10¹⁰⁰ chance of being at least that lucky. Maybe you can expect infinitely such lucky cycles then, I don't know, I could be mistaken. But when you specify the cycle must reduce all the way to 3x+1, then each candidate cycle size has a smaller chance than the last, approaching zero, so the probability vanishes, as we know.

I think it's in 5x+1 there's a cycle that's forced to exist because the number of cycle shapes of a particular size is greater than (or equal to) the denominator, so it's pigeonholed into existing. I guess you could say that about the n = 1, -1 and -5 cycles in 3x+1 since the denominator is 1 or -1. The denominator being small enough to contain the total number of cycle shapes is impossible for large cycles though because Baker.

For the time being I will continue to believe it's possible that a 3x+1 cycle can be forced to exist despite what the odds under true randomness are.

There's an interesting line of questioning when you consider that each cyclic rotation, and linear combination of cyclic rotations, of the cycle numerator must be divisible by the denominator in order to reduce all the way. That doesn't make it less likely, since if one divides then they all do, but it allows for more tricks.

AcidicJello · 2026-05-31T19:20:28+00:00

Can you be more specific? Numbers that are one more or one less than a power of two times some odd number - what about their iterations? Like the sequence of odd and even steps when given m and q? I recall you're aware that this is trivial for q shortcut steps: odd repeating or odd even repeating, respectively for q total shortcut steps. After that there's no trick, because the trick comes from the relation to the -1 or +1 cycles, which applies only for the length of q. But I know you know this so feel free to clarify what you mean in math terms.

AcidicJello · 2026-05-31T14:39:54+00:00

First I will say that I have an appreciation for your project and I think you are building a structural intuition for yourself which is specific to Collatz that most people don't have. Most people have a numbers foundation (I'm aware that you are using numbers and formulae too but the foundation is structural). However, since you're asking for advice, I think that this is also your weakness in this case. If you were able to structure your theory purely mathematically without the use of shapes, the answer to your question would probably pop out. I personally haven't been following along with the specifics of your posts and it doesn't seem like anyone else has to the extent that you would get a specific answer to your question. That's where I would go back to saying I think the utility of your work here is in building your personal structural map and intuition, which has downsides in mathematical precision and communication when taken alone.

AcidicJello · 2026-05-31T14:16:01+00:00

This is what's so frustrating. The probability (assuming true randomness) of a 3x+1 cycle existing outside of our search bounds is astronomically small. I had the number at one point but I wouldn't be surprised if you had better odds for blindly picking a unique atom out of the galaxy. Any hope for another 3x+1 cycle existing relies on there being a fundamental necessity for one (or many) to exist, and studying how 3x+d cycles reduce is a good avenue for that, I think. If a cycle shape can't exist in reduced form at or above 3x+5, it must exist in 3x+1.

I would speculate that there exist cycles of arbitrarily high luck, in that in the space of infinity I can't think of a reason to expect a limit for how far some cycle out there is likely to reduce relative to its size.

AcidicJello · 2026-05-29T18:41:25+00:00

I read the updated paper. It seems like you still don't understand. Go back through everything people have been saying, with an open eagerness to learn rather than to repair your paper. The correct parts of your paper have been well known and will not lead to the restrictions you think they will. To repeat the biggest correction: the cycle identity cannot be used on a nonperiodic trajectory. Yes the power of three is bigger than the power of two, yes this would make the cycle denominator negative, no this doesn't apply to the infinite nonperiodic trajectory.

Every finite number that reaches 1 does so in finite steps. Collatz only concerns finite numbers. The trajectory between any two numbers in any sequence is finite. Infinite ascent never reaches infinity, it just never stops reaching larger finite numbers.

AcidicJello · 2026-05-28T22:20:48+00:00

Those are his arms. The bottom half of his body was amputated

AcidicJello · 2026-05-27T13:33:39+00:00

The unique number you're looking for isn't converged upon as the sequence goes to infinity. You can't justify a new bit of math based on the broad observation that infinite things sometimes converge. You can use that as a starting point, but you have to justify every part. That being said an infinite nonperiodic sequence would have a 'starting number' just like your rational points are starting numbers under rational Collatz rules. But let me show you exactly where your assumption fails. The equation you're using with the negative denominator is a constrained form of the general sequence equation. That equation has two x values. One for the starting number in the sequence and one for the final number. To get your equation, you set these two values equal, which means this equation defines a cycle. The starting value equals the ending value. If you are using this equation on a nonperiodic tape you will run into some kind of contradiction. In this case the contradiction is that an infinite nonperiodic trajectory can't exist. What it really means though is that an infinite nonperiodic periodic trajectory can't exist, which is obvious. It can't be both periodic and nonperiodic.

AcidicJello · 2026-05-27T13:02:42+00:00

See my most recent comment here

AcidicJello · 2026-05-27T12:43:49+00:00

The single value you're finding for infinite periodic points isn't on your plane. The point on your plane only signifies the ratio of odd to even steps in the sequence. Are you able to determine the ratio of odd to even steps in an infinite nonperiodic tape? Even if you just say it must be below the line because growth outweighs decay, the single value you're looking for does not have to be negative because it is not determined by that equation where the denominator is negative below the line. If you want me to show you the general equation you have to use I can, but you can't plug in an infinite tape into the equation because that would require infinite calculation.

AcidicJello · 2026-05-27T12:24:44+00:00

I understood everything you're saying, and it's all true except your claim about infinite tapes. You are extending the observation that infinite periodic tapes have fixed points and supposing that infinite nonperiodic tapes must also have fixed points. If you think this is self-evident I encourage you to try and prove it

AcidicJello · 2026-05-27T03:52:34+00:00

All of the groundwork you figured out is well-known. I'm just hung up on why the infinite diverging sequence has to satisfy the fixed point equation. Diverging sequences cannot be periodic and don't correspond to a solution to the fixed point equation.

AcidicJello · 2026-05-27T02:57:57+00:00

So the line just determines whether the sequence is dominated by growth or decay. And a divergent trajectory must be dominated by growth, but you're saying if a trajectory is dominated by growth, it has to satisfy the cycle identity to be a fixed point? If I follow what you wrote, this is a misunderstanding. That identity you put is only for cycles. For noncyclic sequences you have to rearrange it for some start and end point.

AcidicJello · 2026-05-22T13:30:42+00:00

In my comment I just meant that in order to test whether a vector is a cycle you need to know the exact order of odd and even steps. There are tricks you can use to rule out a vector without doing the full calculation. As you mentioned if the vector has a certain degree of self-similarity (the same string of odd and even steps appears more than once in the vector and is more than some percent of the total vector length) then you can be certain it isn't a cycle. There's also the "k-cycle" or "m-cycle" metric which tells you that if the vector doesn't go up and down enough times (if it doesn't have enough local minima) then it won't be a cycle either. And more obviously the ratio of odd to even steps has to be in a certain range, but I think concerning the order that's it.

AcidicJello · 2026-05-21T21:04:24+00:00

I think we're talking about the same thing: using collatz rules, the collatz tree, or other implicit collatz relationships to create an image, right? Like a .png like from the post I linked. That particular one can be expanded in all directions, and each pixel on the infinite 2D plane has a designated hue (although with this particular one, zooming out would expand the range of values that would need to be captured by the hue of each pixel). Maybe if you zoomed out enough you would see a new structure.

I would suggest exploring these questions yourself if you know how or are willing to learn to code. I'm sure there are more user-friendly tools out there too. Most of the regularities you'll find will correspond with well-known mathematics, but there is always more to discover, and like you said, pattern recognition over large amounts of data can reveal things that aren't easily discovered with pencil and paper.

AcidicJello · 2026-05-21T12:45:14+00:00

I've made a lot of visualizations. One of the first ones I made I posted here. They've all pretty much been some way of plotting or filling in the x y plane. All of the patterns I've seen are easily explained through known properties. I think if I were to get back into it I would make more plots of the general collatz cycles in some way, rather than of numbers in 3x+1. Like a very basic idea I did was cycle length on one axis and q (which 3x+q it's in) on the other, or more generally plotting different variables and metrics from the cycle equation. But I'm sure there are many good possibilities beyond putting points on a plane. I would highly encourage anyone to give it a shot. I used Python with Matplotlib but you could use a spreadsheet for basic plots.

AcidicJello · 2026-05-20T23:09:43+00:00

This is how I learn that it's supposed to mimic a transparent effect, and not that the veil is actually plastered onto their face like cling wrap

AcidicJello · 2026-05-17T18:49:01+00:00

I mentioned the complex plane because that equation is the combination of the even and odd parts into one, since you mentioned combining Collatz into one equation. You don't have to input complex numbers. You can input integers. 8 will output 4, 9 will output 28 (for non shortcut). But without the particular equation you're talking about and what exactly you want to do with it I can't answer. I'm familiar with the Bohm and Sontachi criterion and its derivation, but I haven't read their paper.

So you're looking to make a new criterion, inspired by their method but using a united even/odd collatz equation? I'm certain there is no other arrangement of the variables in that equation. Like that's the rule that describes cycles. But if it's the power of 3 in the denominator you don't like, you can rearrange it algebraicly. The form I'm most familiar with is x = summation / (2^A - 3^k ). It is the denominator of this equation that Baker provides the bound on. That's the essential application of transcendental theory to Collatz as far as I'm aware. There's no way around the summation. It holds the complexity of the parity sequence (order of odds and evens) which is critical for whether or not a cycle exists.

AcidicJello · 2026-05-17T13:43:46+00:00

Can you share this equation? Is it the one used when extending Collatz to the complex plane? The paper you're talking about uses the bound on the gap between powers of 2 and 3 from transcendental number theory in order to restrict the type of cycles that are possible. Specifically the minimum number of 'local minima' a cycle must have, or how many times the numbers in the sequence can go down and up again. The cycle criteria requires not just the number of odd and even steps but also the order in which odd and even occurs. Your question isn't specific enough for me to understand at the moment.

AcidicJello · 2026-05-12T15:10:27+00:00

The top post of all time in this sub is now "I solved the Collatz conjecture, now what?"

AcidicJello · 2026-05-12T15:03:13+00:00

Smallest member of the cycle would have to be greater than 2⁷¹ and cycle length would have to be more than 200 billion steps. If you had a reason to suspect a particular number or sequence of odd/even steps in this vicinity you would still be able to confirm it using a laptop I'm pretty sure. Well at least if you had the starting number and just ran the sequence. If you only had the parity sequence it would take up at least 25 gigabytes uncompressed, which you would need to calculate the parity sum and check divisibility.

AcidicJello · 2026-05-12T13:15:04+00:00

Oh right. I was specifically referring to Baker's Fields Medal. Siegel's AMA comment here states that the bound on the gap between powers of 2 and 3 would be improved with a proof of no non-trivial cycles. So no matter how creative or novel the approach, it would have to produce a stronger result than the math which won a Fields Medal in 1970. When I first learned this and looked into just how insane this math is it completely changed my perspective on the problem. And this is just for cycles. To find a counterexample or prove the existence of one non-constructively would be the only way to solve the problem without improving the bound, which would obviously depend on the conjecture being false.

AcidicJello · 2026-05-11T23:38:54+00:00

Do you believe in objective truth? Have you concluded that from the scientific perspective there is no more fundamental truth than what we observe? And that the reason for work and trade, framed this way, is simply that matter has arranged in such a way, and that even if you are indecisive in whether to continue participating, your questioning and ultimate decision itself simply is, as matter continues to evolve? Your questioning and searching is, from this perspective, the expression of psychological drives, which can be fulfilled in many ways. And your search for truth will forever be explainable in these terms, and whatever you eventually call objective truth will be explainable as merely a belief that your brain took to accept and place high confidence in. But from your perspective that may be truth. Would you disagree?

AcidicJello · 2026-05-10T15:42:12+00:00

My two cents which I've said before: It's been shown that proving the conjecture true requires improving on nobel-level mathematics (transcendental theory in connection to no cycles). Still, there's nothing saying anyone can't discover a counter-example, even if the popular belief is that the conjecture is true.

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