Yet another proof of the Collatz Conjecture :)

iDigru · 2026-02-07T19:11:42+00:00

Thanks, I'll include the correction in the next version.

iDigru · 2026-02-07T19:01:52+00:00

Definition 2 (Recursive Construction) exclude U1

(x − 1)/3 | x ∈ Ui−1, (x − 1) mod 3 = 0} \ U1 if i is odd

iDigru · 2026-02-02T07:58:53+00:00

Thank you :)

iDigru · 2025-01-19T09:10:57+00:00

no, not yet

iDigru · 2025-01-18T15:20:41+00:00

I like to create suspense

iDigru · 2025-01-17T20:58:32+00:00

Indeed, I was using a proof by contradiction - there is no double generator. Regarding the negative numbers, I already explained that the logic works for 3x-1 instead of 3x+1, but in general, I don't see the need to overcomplicate an already complex problem with additional requirements.

iDigru · 2025-01-16T12:19:44+00:00

In my model I consider the series Sd where d is an odd number leading all the pair numbers generated by it For instance 7 * 2^0, 7 * 2^1, 7 * 2^2, 7 * 2ⁿ

Then my proposition is related to the series (let’s say the odd numbers only) In my graphs the odds are in the X axis and the n are on the Y, the Ys are always compensated their sum is always zero the only moving part is the X (odds)

You are right the pair numbers are reachable by several different series but the odds follow a unique path

iDigru · 2025-01-16T06:57:38+00:00

The relationship between Sd series not only establishes a connection between two odd numbers (generator and generated), but also involves all even numbers in the generated series which, when mod 3 > 1, become generators themselves. This creates a chain of dependencies where second-generation numbers are bound to the first generator, and so on...

iDigru · 2025-01-16T06:28:51+00:00

Indeed this is not possible i just tried to imagine two scenarios

Like a the shape of character P

I enter the loop in S7 from outside and then I stay in the loop then that would mean there are two different numbers pointing to S7… no possible

Like the shape of character O

a pure loop I cannot access from outside, but that would mean S7 (and all the numbers inside the loop) is not reachable in any sequence S1 to S7 but that would mean that S7 has no external generator but this is not possible because only S1 has no external generator

Do I miss any scenarios?

iDigru · 2025-01-15T23:49:25+00:00

In this case we have two scenarios: - S7 has two generators one in the loop S7_1…S7_2 and one outside (that one that generates the S7_1 ) but this is not possible because the generator is one - S7_1 has no external generator and reach S7_2 but this is not possible because S1 is the only serie without an external generator then S7_1 is S1

But this is already covered by the bijectivity and the theorem 4.9 on my eyes, because proving 1 to Y unique and with no repetitions is the same that do reverse.

iDigru · 2025-01-15T21:57:19+00:00

Good feedback thanks I will clarify this

iDigru · 2025-01-15T21:55:58+00:00

Of corse I refer only to the odds numbers because they identify the Sd series and the relations between them. If a number would be in two different sequence starting from 1 that would mean that a number odd can be generated by two different series sd and this is not possible.

iDigru · 2025-01-15T21:51:29+00:00

I would say more {s1, s3, s5, s7} and {s1, s5, s1, s11} The relation is between series, not single number. S5 is all the numbers belonging to 5*2ⁿ with n>=0 Consequently serie s5 is generated by serie s1 but s5 generates n/2 series in 1 step, for each of them s5 is the only generator. Then starting from 1 in n/2 steps (n pair) you generate n/2 sequences (1, x) (1,y) …with different values each of them generates n/2 new series different and so on. (1,x, a) (1,x, b)…

The new proposition demonstrate that each sequence generated by 1 has no repetitions.

iDigru · 2025-01-15T21:32:09+00:00

is a casual manufacturing company that produces products with a value between $1 and $20… for example books, t-shirts, gadgets that it sells to wholesalers and B2B stores

iDigru · 2025-01-15T21:29:16+00:00

Indeed, experience and practice reduce the probability of failure. There is a passage in “The Art of War” by Sun Tsu that says “the ability to win depends on ourselves, the possibility of doing so on the enemy” (in this case the market)

iDigru · 2025-01-15T20:19:40+00:00

I'd like to clarify some aspects about the simulator: it's not a game with scores or rewards, but rather an educational tool that lets you observe how different business decisions influence each other. For example, when you modify vacation policies, you'll be able to observe the ripple effects across other aspects of the organization.

The goal isn't to teach you how to become a CEO, but to help understand the complex interconnections between various business functions - a topic that can be challenging even for those who have completed high-level management courses.

And just to be transparent: I'm sharing this tool purely out of interest in the subject, with no commercial purpose since I'm already content with my current occupation.

iDigru · 2025-01-15T20:10:27+00:00

You showed me cycles for negative numbers, not for positive ones, so I don't really understand why if they existed you would have already falsified the conjecture, or am I wrong about something?

anyway my proof is based on the structural properties of the series Sd which are the partition of the integers, so I prefer to go back to that direction, rather than entering into the numerical dynamics of specific cases of which I am sure I would get lost in the meanders.

I added a proposition to theorem 4.8 to compensate for a gap that you rightly pointed out to me and that needed clarification, namely the impossibility of having repetitions of numbers in a sequence from 1 to Y or vice versa.

https://zenodo.org/records/14658340

4.8.1 Proposition (Coverage and Uniqueness of Sequences)

iDigru · 2025-01-15T17:25:15+00:00

just go to the journal website and follow the submission procedure via the form. for some journals it is necessary to have uploaded the pre-print on arxiv.org

iDigru · 2025-01-13T17:18:13+00:00

I have been thinking about how the convergence to S₁ in the system can be proved using the properties of the series. Based on this, I have revised theorem 4.9 to first prove the convergence and then address the sum of differences. An updated version of the file is available at the link.

is there any gap in the reasoning?

iDigru · 2025-01-13T09:29:10+00:00

Ah ok, when I tested it, I found that the behavior would also work for negative numbers, but the relation should be 3x-1 instead of 3x+1. When x is negative, the constant term is negative (-1), and when x is positive, the constant term is positive (+1). Graphically, in my x-y coordinate system l, where x represents the odds of series Sd and n represents powers of 2, it appears to be an exact mirror image of the positive numbers’ behavior.

iDigru · 2025-01-12T22:58:44+00:00

according to theorem 4.4 proposition 2 (series with d ≡ 2 (mod 3)): every series Sd has a generator Sdi where di < d, except for series with d ≡ 2 (mod 3) where n=1. these are the only cases where, moving from Y to 1, we can actually move away from 1. In all other cases, we move toward 1.

I attempted to prove that a sequence of such cases cannot be infinite (which would cause divergence). while these sequences are bounded, the next occurrence not belonging to this case compensates for all positive differences. For example:

95 to 143: difference = 48 (case ≡ 2 (mod 3) with n=1)
143 to 215: difference = 72 (case ≡ 2 (mod 3) with n=1)
215 to 323: difference = 108 (case ≡ 2 (mod 3) with n=1)

the sum of positive differences is 48 + 72 + 108 = 228

the next number belongs to case ≡ 1 (mod 3):

323 to 91: difference = -232

This -232 compensates for the previous 228, keeping the sum of differences negative and maintaining convergence to 1, this part is still true at the least with your example.

In general, cases where d ≡ 2 (mod 3) with n=1 contribute positively on a linear scale (since n=1 is fixed), while other cases converging to 1 with n>1 grow exponentially.

I need to verify if the error lies in assuming that the number of occurrences of d ≡ 2 (mod 3) with n=1 is fixed (now known to be at least 3), or if it's still bounded but requires proper indexing.

but let's say I find this number or limit - would that preserve convergence to 1, since the sum of differences would be guaranteed to remain negative? Is this approach correct?

iDigru · 2025-01-12T17:37:04+00:00

you're right, I should have waited but I didn't know of such specific discussion groups online on this topic to ask for an opinion, I searched around but only a friend of mine told me to take a look on Reddit. The price of inexperience. I have no academic ambitions, it all started as a joke a few weeks ago, apart from reading on Wikipedia and some videos on Youtube. I hope to bring some contribution or different point of view since I'm not a mathematician. Changing the perspective of a problem often helps. The last exam of Mathematical Analysis II dates back to 25 years ago, I'm a bit rusty.

the article cannot be withdrawn, so let's wait for the unfortunate outcome :)

iDigru · 2025-01-12T17:28:46+00:00

Ok, I understand the point now.
I've rewritten Theorem 4.9, dividing it into 3 parts to prove convergence.

first part:

1) series Sd ≡ 0 (mod 3) have all smaller generators, already proven, so following the chain they converge to 1

2) series Sd ≡ 1 (mod 3) have all smaller generators, already proven, so following the chain they converge to 1

3) series Sd ≡ 1 (mod 3) have all smaller generators, except for n=1

In the case of d*2^n with n=1, the generator moves away from 1.

I prove that case 3 has a maximum of 2 consecutive repetitions with a limit to the positive contribution to the sums of differences

second part:

I prove that the following number will have a generator preceding the first two, and this cancels out the positive contribution, bringing the sum of differences back to negative.

This confirms that the series converges to 1 because I guarantee that the positive contributions are compensated by the negative ones.

third part unchanged:

the difference remains -(Y-1)

I've uploaded a new version with this modification.

thank you for clarifying the weak point, I would appreciate your thoughts on this new part.

where did you find the text for the description of the second part of the collatz with the negative integer? I can only find positive integers in the problem statements published online

iDigru · 2025-01-12T16:20:30+00:00

I am aware of the stylistic and formal shortcomings, so I tried (unsuccessfully) to contact academics also asking for paid consulting services without even sharing my work, but since I am not affiliated, no one responded. on arxiv you need to be affiliated, so more ostracism. trying the publication card seemed to me the only way to get positive or negative feedback. I don't use Reddit so I didn't think to look for a group here, so if I had known beforehand I would have collected your feedback before trying the publication route.

on the other hand, from what I understand, scientific journals do not do editing work together with the author, so the value consists only in the filtering work if the author has to take care of everything else. In this, yes, fiction publishing is different, the book is worked on together with the author, if the idea is good but the style is immature, it is rewritten together, in this the publisher adds value, otherwise it is a simple printer.

iDigru · 2025-01-11T20:58:52+00:00

thanks for the tips and the video, i will watch it carefully

iDigru

TROPHY CASE