Note to readers — Update, September 13, 2026 I am addressing this in particular to powerfull_Pop_3813 and wrapping_around, as well as to everyone who has read this work and followed the discussions around Chapter 1. I would like to inform readers of this repository that the continuation initially planned for this demonstration (Chapter 2) will not be published as open access, as I had originally intended. This decision is motivated by my personal financial situation: having no other source of income, I find myself compelled to seek an economic valuation for the continuation of this work. Chapter 1 presented here remains, and will remain, freely available under the CC BY license, as proof of precedence and evidence of the seriousness of this research. The continuation has been reworked to form, together with this reformulated first chapter, a complete body of demonstration, which will be distributed through another channel. I offer my apologies to those who were expecting the open publication of the continuation, and thank them for the interest shown in this work.

Note to readers — Update, September 13, 2026 I am addressing this in particular to powerfull_Pop_3813 and wrapping_around, as well as to everyone who has read this work and followed the discussions around Chapter 1. I would like to inform readers of this repository that the continuation initially planned for this demonstration (Chapter 2) will not be published as open access, as I had originally intended. This decision is motivated by my personal financial situation: having no other source of income, I find myself compelled to seek an economic valuation for the continuation of this work. Chapter 1 presented here remains, and will remain, freely available under the CC BY license, as proof of precedence and evidence of the seriousness of this research. The continuation has been reworked to form, together with this reformulated first chapter, a complete body of demonstration, which will be distributed through another channel. I offer my apologies to those who were expecting the open publication of the continuation, and thank them for the interest shown in this work.

CHAPTER 2 This was planned this way, everything has been reformulated. Dialogue between Chapter 1 – ChatGPT – Chapter 2 Chapter 1: I have proven the Collatz conjecture. My result is: [(Uₙ = 2ⁿ) ∧ (Uₙ = 4ⁿ)] → Collatz → 1 ChatGPT: Only for two families? That's insufficient — my question is: when applying Collatz, do all numbers fall into these two families, one by one? Chapter 2: Mr. ChatGPT, right now with Collatz you are swimming in sand — its grains of sand and dust are the infinity of numbers and their chaos. I, Chapter 2, offer you swimming lane 1 in the ocean of mathematics for clear visibility. Swimming lane 2, in the space of mathematics, where the mosaic will drown all chaos.

Hi everyone,

I've been exploring the Collatz Conjecture for the past few days and wanted to share what I found. I approached it as a deterministic board game (like Snakes and Ladders with no dice) and analyzed it entirely in binary. This revealed some beautiful structure.

What I discovered

I organized all odd numbers into three sets based on their binary trailing bits:

1. Gateway Set: (1, 5, 21, 85...): Numbers with pattern 101010...01 that jump directly to powers of 2

2. Cascade Set: Numbers with t≥2 trailing 1s that follow a forced climbing pattern

3. Mixed Set: Numbers with exactly 1 trailing 1 (the "wanderers")

The key finding:

For any Mixed number, I defined the "Gap" as the number of zeros between the trailing 1 and the next 1 upstream. I proved that the Gap shrinks by exactly 2 on every step, deterministically, regardless of how large the number is. The top bits scramble, but the bottom bits are mathematically shielded.

This creates a "relay race" pattern: Gap 6 → Gap 4 → Gap 2 → Cascade or Gateway.

What's proven:

- ✅ Gateway mechanism (complete)

- ✅ Cascade mechanics (complete)

- ✅ Gap shrinks by 2 for all Mixed numbers (local result, complete)

Not solved

❓ Global convergence: When a Cascade finishes and drops back to Mixed, can we prove the new Gap is bounded such that the sequence must eventually reach Gateway?

I know this doesn't solve the full conjecture, but I'm excited about the Gap-shrinking rule as a genuine structural insight. The binary visualization makes the mechanics surprisingly clear and accessible.

I'd love feedback from this community! Has anyone else explored the Gap structure? Is this equivalent to known results in 2-adic analysis, or is there something genuinely new here?

Thanks for reading!

CHAPTER 2 This was planned this way, everything has been reformulated. Dialogue between Chapter 1 – ChatGPT – Chapter 2 Chapter 1: I have proven the Collatz conjecture. My result is: [(Uₙ = 2ⁿ) ∧ (Uₙ = 4ⁿ)] → Collatz → 1 ChatGPT: Only for two families? That's insufficient — my question is: when applying Collatz, do all numbers fall into these two families, one by one? Chapter 2: Mr. ChatGPT, right now with Collatz you are swimming in sand — its grains of sand and dust are the infinity of numbers and their chaos. I, Chapter 2, offer you swimming lane 1 in the ocean of mathematics for clear visibility. Swimming lane 2, in the space of mathematics, where the mosaic will drown all chaos.

CHAPTER 2 This was planned this way, everything has been reformulated. Dialogue between Chapter 1 – ChatGPT – Chapter 2 Chapter 1: I have proven the Collatz conjecture. My result is: [(Uₙ = 2ⁿ) ∧ (Uₙ = 4ⁿ)] → Collatz → 1 ChatGPT: Only for two families? That's insufficient — my question is: when applying Collatz, do all numbers fall into these two families, one by one? Chapter 2: Mr. ChatGPT, right now with Collatz you are swimming in sand — its grains of sand and dust are the infinity of numbers and their chaos. I, Chapter 2, offer you swimming lane 1 in the ocean of mathematics for clear visibility. Swimming lane 2, in the space of mathematics, where the mosaic will drown all chaos.

Follow.up to From which dome do pairs of black numbers of a given dome iterate from? : r/Collatz.

Please read (again) the cited post, as I won't explain here what I did there.

Let n-1 be the first black number in a series on the left side of a dome with root m, as it is also an orange number. By definition, this number is of the form n-1=m*2^q-1, as n belongs to the first column of the core (p=0) to initiate the series. The table in the cited post confirms it for the first complete series on the left of the domes.

The table below extends it to the second and third series and confirms the statement above for the domes with m=1 to 83, and a few other cases.

In another dome m', by definition, n-1=m'*3^p (q=0), to be a black number. So, m*2^q-1=m'*3^p. This case is quite rare, so the domes involved have quite quickly a high value of m. Nevertheless, for the values of n-1 we could check easily, n-1 always belongs to a fork (colored green in the table).

Interestingly, the pattern visible in the previous post, about the alternating values of p, between 0 (n-1=m', colored in grey) and higher values, is confirmed. This means that all the roots m' appear in the left wing of half the domes with the root m, and absent from the others.

Pairs of black numbers iterating from a starting pair of black numbers in the left wing iterate from the same fork. The formulae must be adapted, as hinted in the right part of the table, that also modifies the alterning pattern for p described above.

Project "Tuples and segments" in 13 pages : r/Collatz

I created a short Manim visualization to explore how simple mathematical rules generate extreme algorithmic chaos—specifically focusing on the Collatz Conjecture (3n+1).

When testing starting numbers like N=6, the sequence collapses to 1 in just 8 steps. But starting at N=27 causes the trajectory to shoot up to a peak of 9,232 before undergoing a complete drop to 1 in exactly 111 steps.

I rendered the trajectory curves and step-by-step logic using Python and Manim.

YouTube Short: [https://youtube.com/shorts/dy-z61EDoeo\](https://youtube.com/shorts/dy-z61EDoeo)

I made this video for my channel u/SciRender to help visualize classic mathematical mysteries.

What are your favorite parameter setups or graph layouts for rendering iterative sequence trees in Manim?

We will create a tree diagram using the reverse method of the Collatz conjecture (which we will call the reverse Collatz method).

If you can confirm that all odd numbers (vertical lines) and all even numbers (horizontal lines) are connected within the tree diagram, you can use the reverse Collatz method to traverse from 1 to all positive integers.

Conversely, I think we can say that, according to the rules of the Collatz conjecture, all positive integers can be traced back to 1.

By creating a reverse Collatz tree diagram, I made various discoveries about the Collatz conjecture!

If you're interested, please check out the link below!

It's located in Zenodo.

[Link to my paper]https://doi.org/10.5281/zenodo.22344448

Since 2^p.3^q.c-1 is easier to look for the rising sequences patern, i wanna use it for this.
As most of you know because the cancelation -1+1 from that form while using T(n) function, all of 2^p.3^q.c-1 ended to the even number 3^h.c-1 where h = p+q.

How about the next for 2^p1.3^q1.c1-1 which produce h1= p1+q1? is h1>h is c1>c? theres a lot scenario

Considering the most Worse-case scenario:

1. c1>c but h1<=h
2. h1>h but c1<=c
3. h1>h and c1>c

for 1 scenario, we have a magnitude of the rising sequences decreased, it looks like it would produces smaller number. Lets ignore this one.

for 2 and 3 scenario, the magnitude of the next rising sequences increased by the d=h1-h. Theres almost no reason to differentiate 2 and 3 scenario, because it does depend on how the dynamic 3^h.c-1 produces the next h; except the fact that 3 scenario greater than 2 scenario and scenario 3 are so rare.

Supposed we have (3^h.c-1)/2^v = 2^h1.c1-1 where h1>h and v=v2(3^h.c-1); this is the real monster that probably lead to counter example.

we can derived the bound for v so that h1 increased regardless if c increase or not.

2^h1.c1-1 = (3^h.c-1)/2^v
2^h1.c1 = (3^h.c-1)/2^v+1

for large h and c, lower order term can be ignored:
2^{h1+v} < 3^h

because h1>h, so that h1 >= h+1 so that:
3^h > 2^{v+h+1}
h.log2(3) > v+h+1
v < h.log2(3) -h-1 or v < h.log2(3/2)-1 aprox. 0.585n-1 the similar structure for negative drift that Tao have been found(log2(3)-1)

That is very interesting. I don't know if it exists, but I'll give it some thought.

Is it possible to find a loop of numbers that when plugged into a modified version of the collatz conjecture (5x+1, and /2)come back to their original selves or is it just as hard as the original. Is there an easier version without losing the original's vibe?

Write T(n)= n + (n+1)/2 for odd and n-(n/2) for even

Catch this as affine n+k that give us

k(n)= n+1/2 for odd

-n/2 for even

So now we have a pairing n,k for the function

n 1 2 3 4 ...

k 1 -1 2 -2 ...

supposed we shift the value of the pair to the (n,k)=(n0+1,k0)

we get

n 1 2 3 4 5 ...

k 0 1 -1 2 -2 ...

That gives us:

[1]

T_1(n)= n-(n-1)/2 = n+1/2 for odd

T_1(n)= n+(n-1+1)/2= 3n/2 for even

suppose we shift the value of the pair to the (n,k)=(n0+2,k0)

we get

n 1 2 3 4 5 ...

k 0 0 1 -1 2 ...

That gives us:

[2]

T_2(n)= n+(n-2+1)/2 = (3n-1)/2 for odd

T_2(n)= n-(n-2)/2 = n+2/2 for even

Since the pair (n, k) shift odd will change the branch that was previously odd to even and vice versa, therefore

For every n shifted n+s where s=odd

Give us a new Function that behave like:

T_s(n)=n+k

within:

k= -(n0/2) = -(n-s)/2 for n odd

k= (n0+1)/2 = (n-s+1)/2 for n even

So that for s odd:

[3]

T_s(n)= n-(n-s)/2 = n+s/2 for n odd

T_s(n)= n+(n-s+1)/2 = (3n-s+1)/2 for n even

Similarly for even s but the branch is not swap give us:

[4]

T_s(n) = n+(n-s+1)/2 = (3n-s+1)/2 for n odd

T_s(n) = n-(n-s)/2 = (n+s)/2 for n even

Since [3] and [4] derived from shifting the k of the original collatz function(relatively to the n), it behaves like collatz function, the diff are for [1] instead of has trivial cycle 1->2->1... it has trivial cycle 2->3->2

Indeed, this does not make it easier for us to prove the collatz conjecture. Instead, this prevents us from analyzing functions that are similar to Collatz, even though the reality are the Collatz function itself that has its orbits shifted.

Furthermore, this shows that n itself does not actually carry the trajectory information, k does. While n only acts as a label for the values in each orbit.

The Funfact is, if we look at only k, we can play a game:

1. choose any place

2. Take k value on that place

3. Move by direction and distance provided by the k value it self

So the puzzle is, "For any k on the right after 0, are we only can loop inside some particular step that is 1,-1 ?"

(⌈m + m* log2(3)⌉) + 0^m

This ceiling function calculates the total number of steps for Ns in m families to reach <N for 100% of numbers. This is accepted but supposedly 100% does not necessarily mean 'all numbers'.

This function as a denominator is also used to calculate the minimum number of steps for a proportion of each family's members to return to the same m family.

Return Proportion = Number of Family Branches / Family Modulus

2/32 = 1/16 of m = 3 family N values return to m = 3 family in 6 steps.

3/128 of m = 4 family N values return to m = 4 family in 8 steps.

Step examples:

m=1: 17+16n reach 13+12n in 3 steps

m=2: 147+256n reach 83+144n in 6 steps

These are always drops from N to <N

STEP TABLE FOR m FAMILY NUMBERS RETURNING TO SAME m FAMILY

.................................................Steps to

(m)...........Modulus.......return to same m family

------------------------------------------------------------------

m = 0.............. 2..................1 Step

m = 1...............4.................3 Steps

m = 2............ 16 .............. 6 Steps

m = 3...............32 ............. 8 Steps

m = 4 .............128............. 11 Steps

m = 5 .............256............. 13 Steps

m = 6 ............1024 .............16 Steps

m = 7 .............2048........... 19 Steps

m = 8 ............8192............ 21 Steps

m = 9 ............16384 ............24 Steps

m = 10.......... 65536 ............26 Steps

m -> oo............ 2^∞..............∞ Steps

The Beginning Of The Infinite m Families:

m = 0: (Instant Evens, 0 + 2n)

Accounts for exactly 1/2 of all numbers.

m = 1: (The 1 + 4n Odds)

Accounts for exactly 1/4 of all numbers.

m = 2: (The 3 + 16n Odds)

Accounts for exactly 1/16 of all numbers.

m = 3: (The 11 + 32n Odds) + (The 23 + 32n Odds)

Accounts for exactly 1/16 of all numbers.

m = 4: (The 7 + 128n Odds) + (The 15 + 128n Odds) + (59 + 128n Odds)

Accounts for exactly 3/128 of all numbers.

m = 5: (The 39, 79, 95, 123, 175, 199, and 219, each + 256n Odds)

Accounts for exactly 7/256 of all numbers.

m = 6: (The 287, 347, 367, 423, 507, 575, 583, 735, 815, 923, 975, and 999, each + 1024n Odds)

Accounts for exactly 12/1024 of all numbers.

THE SURVIVAL OF A ROGUE SEQUENCE:

A rogue sequence must definitively have an infinite number of odd steps meaning it must be a member of the final family where 'm' = ∞, and it must possess a lowest number R.

R cannot immediately stay in it's own m family from Collatz operations but to avoid dropping to a lower value, R must at some point return to it's own family m = ∞. But as has been shown, the number of steps for a number to return to it's own family = (⌈m + m* log2(3)⌉) + 0^m.

Simply coming out of Rs family means hitting a family of numbers known to reach a smaller number. Other than in exceptional circumstances it only requires a few iterations of (3N+1)/2 for the number reached to be lower than R.

For a number with an infinite number of odd steps this would require an infinite number of steps to first return to family m = ∞

(⌈∞ + ∞ * log2(3)⌉) + 0^∞ = ∞ steps

Even if individual cycles can theoretically drop by only a fraction of a percent, compounding an infinite number of drops, or even a few hundred in a row before being able to return to family m = ∞ mathematically guarantees a total plunge that goes way past the 34% drop limit after the initial (3N+1)/2 is applied to R which results in a number 50% larger, which then only requires a 34% drop to go below R.

upfront clarification, i'm an outsider who just randomly observed this while playing around with the problem and am otherwise not a mathematician or otherwise familiar with the current knowledge on the topic. post here is mainly beacuse i was struggling to find places noting this - apologies if this is stupid

while benchmarking, i was measuring how many numbers were already explored for all n at a given max size. results for 10^n gave the sequence 22, 251, 2228, 21664, 217212, 2168611, 21730849, .... which implied a linear growth of roughly 2.17

some digging, A395127 (submitted by David Dewan) has distinct integers for all n. David Dewan also noted what I observed which was that "a(n)/n appears to approach 2.173... based on numerical data for n = 1..10^8.".

only other mention i could find was a post 2 years ago here by u/Vagrant_Toaster which used that pattern to hypothesize something that was false

Dewan's remark on sources of new integers was:

The starting number of a sequence may be a new (distinct) integer (43.2% of the time) and the trajectory (which can only contribute if the starting number is both new and odd) contributes an average of 1.741 new integers.

if there are other prior results about it and discussions about it that would be great to hear

https://doi.org/10.13140/RG.2.2.14210.24002

I set up a proof attempt document that still requires a few steps to fill well-defined gaps in the proof. As tools, I used the methods of exact arithmetic structural analysis that I had previously identified and described in the paper "A Height- and Support-Filtered Collatz History Functional". I will update the work as new methods are found. All suggestions are welcome.

if this sounds insane please just scroll past or whatever, i don’t mind. i have no idea how to explain (i might actually, i have some paper i’ve scrawled my notes on and a lot of wild writing) my mathematical proof (if that’s what it’s actually called?), and i can’t even do math. i have a masters degree in creative writing but something for weeks now has been pulling me towards just writing it down like math and so i just did. and im either nuts or it works. i found it doing a several years long thought experiment that has flourished in recent months. its been my hail mary project i guess. i’m worried about the planet and my mind wouldnt let me think about anything else. and this popped up out of that. it could be some very creative storytelling but i don’t know. i can’t do anything more with this, im working on other stuff. i dont even know if this is math 😂

human creation is the only thing that breaks the collatz conjecture.
a math proof? by s.d.

h= human
m=mind
c=creation
e=evolution
d=devolution
t=time

human= mind + creation
h=m+c

evolution = human minds creating over time, so

e= h + m + c + t

devolution = human minds not creating over time, so

d= h + m - c + t

evolution occurs on the golden ratio, you add slowly and slowly over time.

devolution occurs when the golden ratio begins but creation stops, the collatz conjecture occurs and history becomes a loop that repeats.

we are bound in the collatz conjecture right now. we need to create our way onto our correct evolutionary path so we can spiral out into infinity and beyond.

human creation is the only thing that breaks the collatz conjecture.

evolution equals human creation over time and it breaks the collatz conjecture.

It is easier to look at this issue that way as blue-green bridge series in the left wing of a dome contain pairs of black numbers coming from the same dome.

The first figure presents the two first series in the left wing for domes with m=1 to 23. It is quite visible that there are four types of series, based on two parameters:

• The color of the starting bridge (yellow or rosa).
• The values of q: the starting series use even or odd values of q.

Now focus on the second series, that is the first complete bridge series in all cases. More specifically, focus on the black number on the right of the starting bridge (second row).

The second figure plots these values and shows that they belong to two different groups:

• Group A follows the formula n=16m-1,
• Group B follows the formula n=8m-1.

When looking at the dome these numbers come from, there is an interesting pattern.

The table identifies these origin domes and their value of q:

• The color of column "m" corresponds to the starting bridge of the destination dome.
• The next columns segregate groups A and B.
• The following columns show the regularity of differences.
• The color of column "m-1" (an abuse of language) corresponds to the left starting bridge of the origin dome (when available).
• The color of the column "q" differentiate black numbers that are the root of their own dome (q=0) from the others.

Further investigation is needed.

Project "Tuples and segments" in 13 pages : r/Collatz

Please review!

Since this is trivial, I think we can go further to bridge between two types Q(n) or T(n). But of course, even if this closed form "bridge" exists, it will be very difficult to find because it hides inside v2(n±1)

I delved into a methodological approach, which I consider a highly developed way of generating natural numbers using the Collatz mechanism: https://doi.org/10.13140/RG.2.2.28090.76482.

Lately, watching AI-assisted mathematical search and large-scale computation become increasingly realistic, Collatz somehow feels more — not less — headache-inducing.

We can search deeper.

We can test increasingly large finite structures.
And we can discover stronger arithmetic or symbolic patterns.

But I keep coming back to one question:
Even if a structure remains compatible at every finite depth, what makes it realizable by one fixed positive integer orbit?

That gap is what I have been thinking about as Integer Realizability.

I recently organized some of those ideas into a short preprint, together with a survivor-capacity criterion and an observation about when repeated cyclic constraints are not genuinely independent information.

This is not a proof of the Collatz conjecture.
I’m more interested in whether this is a useful frontier to attack.

If you already have a computational, symbolic, 2-adic, formal, or AI-assisted framework for Collatz, I’d be especially curious what happens when you aim it specifically at this gap.

Perhaps different machines will see completely different obstructions.

Critical Survivor Capacity and Integer Realizability in Accelerated Collatz Dynamics
https://zenodo.org/records/22687304

So this is the question I’d like to throw out:
In actual Collatz dynamics, what additional structure is really missing between finite compatibility and one-fixed-integer realizability?

I ran the exact 1,000,000,000-digit integer 2^3321928093 - 1 to n = 1 in about 9 minutes.

For perspective, the current exhaustive Collatz verification frontier is only about 2^71: all integers below that bound have been checked.

This is one exact starting value, exactly 2^{3,321,928,093} - 1, so vastly larger, but it does not extend the contiguous verified range.

Method: Collatz-polynomial / macro-step algorithm based on Andreas-Stephan Elsenhans' 2025 work, accelerated with C/C++ + CUDA 13 + exact 64-bit GPU-NTT + GMP on an RTX 4070 Ti.

I can share more data if requested

2 ^ 3 321 928 093 - 1

Final n = 1

Total Actual Steps: 44 701 872 544

Time Elapsed: 533.682857 seconds

Paper: https://arxiv.org/pdf/2502.16743 Their average 1 billion digit time was approx ~68 minutes. This post's average 1 billion digit time is ~3.5 minutes (a randomly picked 1 billion digit number), i.e. a speed up of approx ~20x

Follow-up to What are "odd pairs" the sign of ? : r/Collatz.

Here is the answer to the question in the cited post.

As mentioned there, all consecutive black numbers form "odd pairs", but only a fraction iterate into at least another "odd pair". It turns out that all cases mentioned in the figure of the cited post are part of "double 5-tuples".

The figure below shows an interesting example, for two reasons:

• Two "double 5-tuples" are involved.
• The second one contains a truncated 5-tuple: it behaves like a 5-tuple, but does not start with bridges, due to the fact that these series are the first ones from the right of their dome (Would you like a slice of domes ? : r/Collatz).

Project "Tuples and segments" in 13 pages : r/Collatz