Hi, I'm an adult studying mathematics on my own outside of school, and particularly I have been doing research on the Collatz Conjecture. I fear I have made a very small mathematics error in my research, because I've ended up convincing myself that I have proven the conjecture true, as long as my understanding of theoretical mathematics and how to apply them is being done properly, and I want to be shown why I'm wrong as I've gone a bit crazy over this lately.

To rush to my point, even though I have a lot more I could talk about on this, I imagined any chosen positive integer from the original conjecture, which I'll label cₙ, that there are at least 3, but possibly 4 paths connected to it. 2 of which are obvious, 3c + 1 and c / 2. Each of these in turn lead to cₙ₊₁. However, you can also do c * 2 for cₙ₋₁. Since c is always a positive integer, you can always multiply it by 2 for another positive integer, which would represent a past step. The other possible past step, is (c - 1) / 3. This however, doesn't always give a valid positive integer answer, and this is where I started exploring.

Getting to the chase, I eventually came up with the equation a * 2^(n) = 3x+1, where a is any given odd positive integer. Instead of asking if every number leads to 1, I asked what numbers lead straight to 1, to 3, to 5, etc. And that's when I came upon this data table.

https://preview.redd.it/zbw4y6zwj8kh1.png?width=1270&format=png&auto=webp&s=3f4ee4e05aa8892d37c0150a675979c754cf3eea

Having made this data table, I immediately started trying to do math on it. I wanted to show that if a number appeared in any given cell, if it could appear in another given cell. I went back to my equation, and ended up with a * 2^(b) = 3x+1= c * 2^(d). Simplifying the middle out, a * 2^(b) = c * 2^(d) shows that there's no valid positive integer solution for a or c, which should mean that no number can appear twice.

I also noticed that in the light blue fields, each row contains 1 / 2^(n) odd numbers, and that every single odd number appears only once. This, I believe, should disprove the fact that any "loops" can occur in the Collatz conjecture, since a number would have to appear twice for a loop to exist.

I also believe that this table proves that every odd positive integer will lead to 1, as my understanding of "Busy Beaver Problems" leads me to believe. The "steps" the Busy Beaver would have, is that it would start at 1, and it would run the equation a * 2^(n) = 3x+1, putting 1 in for a. Then, it would "mark" each solution for x with a 1, and run the same equation on the lowest marked number that it has not yet run the equation on. This would mean it would run it on 1, then 5, then 3, 13, 17, 11, etc. Eventually, ever single positive integer should be "marked" as 1.

Considering the original conjecture, and that any given even number will be divided by 2 until it hits an odd number, I've convinced myself that this data table proves the original conjecture true. However, I fully admit that I must be using some of these theories and such wrong, as I've been self taught and therefore fully vulnerable to easy mistakes.

Lately, this feels close to the bottleneck we keep running into, so I went back through some earlier work to see how much of this viewpoint is already present in the literature — partly to untangle my own headache.

I keep noticing a recurring pattern in Collatz research.

We often try to make the dynamics simpler by choosing one useful coordinate:

parity words,
valuation sequences,
stopping or first-passage times,
residues mod 2^k,
2-adic or 3-adic coordinates,
affine offsets,
heights, peaks, records, etc.

This is extremely useful. But every compression also forgets something.

So I wonder whether a useful way to think about the remaining difficulty is not

“Which single coordinate compresses the Collatz dynamics best?”

but rather

“Which values in several different coordinates can actually belong to the same integer orbit?”

Here is a simple exact example.

Take the shortcut Collatz map

T(n) = n/2 if n is even
T(n) = (3n+1)/2 if n is odd.

For a block of h steps, let p_i in {0,1} be the parity bits and let

s = p_0 + … + p_(h-1).

Then exactly

2^h T^h(n) = 3^s n + Q_h,

where

Q_h = sum_(i=0)^(h-1) p_i 2^i 3^(s-s_(i+1)),

and

s_(i+1) = p_0 + … + p_i.

So (h,s) captures part of the history, but Q_h retains arithmetic information created by the +1 terms and their positions.

This already suggests two different kinds of simplification:

time/history compression
versus
arithmetic-state compression.

They are not automatically the same thing.

A coordinate can have huge fibres by itself. For example, many histories may share the same stopping time, the same odd-step count, or the same coarse residue.

But suppose we describe one orbit history H using several projections

C_1(H), C_2(H), …, C_r(H).

Then a proposed tuple (c_1,…,c_r) corresponds to a genuine orbit only if

C_1^(-1)(c_1)
∩ C_2^(-1)(c_2)
∩ …
∩ C_r^(-1)(c_r)
is nonempty.

Each individual set may be large.

The intersection may be much smaller.

I am not claiming this intersection is always small, or that this proves Collatz. The question is whether repeated compatibility across several lossy coordinates can provide rigidity that no one coordinate provides by itself.

There is substantial classical precedent for thinking this way, although usually for particular pairs of coordinates rather than as one general principle.

Terras (1976) developed the stopping-time/parity framework underlying much of the statistical study of Collatz.

Böhm–Sontacchi (1978) showed that symbolic cycle data is constrained by exact arithmetic realizability conditions.

Bernstein–Lagarias (1996) made the parity–2-adic correspondence exact via the 3x+1 conjugacy map.

Monks et al. (2012/2013) showed that back-tracing parity data with infinitely many 1s determines congruence information modulo all powers of 3, hence fixing a 3-adic state.

Tao (2019/2020) combined first-passage ideas with fine-scale arithmetic structure on 3-adic cyclic groups.

Stérin–Woods (2020) exhibited a striking dual structure where base-2 and base-3 computations coexist in a single Collatz encoding.

So perhaps one way to read part of the history of Collatz research is:

parity <-> dyadic state

symbolic path <-> affine correction

back-tracing parity <-> triadic admissibility

first passage <-> arithmetic offset

base 2 <-> base 3

This makes me wonder whether the next useful question is not simply how much more information we can remove.

Maybe it is:
What is the minimal information that must survive in each coordinate, and what compatibility conditions must all of those surviving pieces satisfy simultaneously along one actual orbit?

In other words,
compression may be only half of the problem.
The other half may be joint realizability.

I would be very interested in references where this multi-coordinate compatibility viewpoint has already been formulated explicitly, or in counterexamples showing why this framing is not useful.

References

Terras, R. (1976), A stopping time problem on the positive integers, Acta Arithmetica.

Böhm, C. & Sontacchi, G. (1978), On the existence of cycles of given length in integer sequences…

Bernstein, D. J. & Lagarias, J. C. (1996), The 3x+1 Conjugacy Map, Canadian Journal of Mathematics.

Monks, K. et al. (2012/2013), Strongly sufficient sets and the distribution of arithmetic sequences in the 3x+1 graph.

Tao, T. (2019/2020), Almost all orbits of the Collatz map attain almost bounded values.

Stérin, T. & Woods, D. (2020), The Collatz process embeds a base conversion algorithm.

Lagarias, J. C. (survey), The 3x+1 Problem: An Overview.

The ridiculous nature of a proof.

Suppose someone sees structure, another person might not see that structure, so that's a person who cannot see it will ask for a step by step proof to prove the continuity of a structure. But continuity cannot be proven by discrete steps because we have shown that infinite discreteness cannot proxy for true continuity.

Diagonalization proves that a continuity has more real information than the previous model. Every step-by-step proof is actually an illusion to satisfy the strange feelings. But every discreet example of a proof fails to show the actual continuity of the structure that one is claiming to exist..

What this paper proves (almost every , natural density — not the full Collatz conjecture):

T_min(n) ≤ C (log n)^A

within O(log n) shortcut steps

for every fixed A above an explicit critical exponent A_FP ≈ 9.99plus a bound on the orbit up to that same time.

Paper: https://doi.org/10.2139/ssrn.7290240

Proof walkthrough: https://shaikidris.github.io/
Lean4 formalization: https://github.com/shaikidris/FirstPassageLinearTransport

Not the same as the recent “bridges” from Tao’s logarithmic-density theorem to natural density. Those aim at every target that goes to infinity. This is a standalone natural-density argument with a fixed polylog target, a logarithmic clock, and quantitative rates.

Prior almost-all natural-density scale: Korec got powers n^θ (θ > 0.79); Inselmann later got every n^ε power . Here the landing is polylogarithmic in .

Approach: Count parity words exactly on eachdyadic shell [2^M, 2^{M+1}). Large-scale prefix bounds control the orbit; a terminal odd-step “timeout” handles small blocks that don’t cross their next threshold in time. Decreasing thresholds turn later failures into direct first passages from the original shell. That organizes long multi-landing passages in natural density without a linear time-union loss — only  O(sqrt(M log M)) cumulative passage times

All of this is weaker almost-all, not “for every n.”

Happy to answer questions / take corrections.

Hi all... Here is the proof .pdf and .lean:
https://bytes.usc.edu/~saty/Collatz

Saty

This is not a proof in itself. I don't have the chops to write the formal mathematical proof. However, I think what follows is itself the blueprint for someone with that skill to write the proof. It's in markdown+latex format. If there's a flaw, I can't find it. So I'd be happy for someone to tell me I'm being an idiot and show me why.

-----

# A Hardware-Centric Bit-Velocity Proof Architecture for the Collatz Conjecture on Domain $\mathbb{N}$


## Abstract

We present a unified, deterministic proof architecture for the Collatz $3x+1$ conjecture on the domain of natural numbers $\mathbb{N}$. By mapping the classical Collatz operation to an isomorphic non-shifting map $T(X) = 3X + 2^k$, information flow is proven to be strictly unidirectional (least significant bit to most significant bit), eliminating top-down feedback from higher bit positions. We establish an absolute physical ceiling on register head expansion ($\Delta\text{MSB} \le 2$ bits per odd step) and prove an inescapable Solid Block Exhaustion Dichotomy: any block of $b$ contiguous ones ($2^b - 1$) undergoes either top-boundary zero-injection decay under zero carry ($C=0$) or immediate multi-bit carry-wave detonation under active carry ($C \neq 0$). Because every element $x \in \mathbb{N}$ possesses a strictly finite bit length $L_0 < \infty$, infinite non-collapsing bit streams (e.g., 2-adic $-1$) are excluded by domain definition. Analyzing the minimal 4-bit 2-chunk partition ($128$ states) under bounded carries $C \in \{0, 1\}$ demonstrates an amortized tail clearance rate ($\overline{\Delta\text{LSB}} \ge 1.875\text{ bits/step}$) that strictly exceeds the average head expansion rate ($\overline{\Delta\text{MSB}} \le \log_2(3) \approx 1.585\text{ bits/step}$), guaranteeing deterministic active register contraction to the trivial attractor $x = 1$ in finite steps.


## 1. Foundational Axioms & Domain Scope


**Axiom 1: Finite Payload Length ($\mathbb{N}$)**


Every natural number $x \in \mathbb{N}^+$ is represented by a strictly finite binary word:
$$x = \sum_{i=0}^{N-1} b_i 2^i, \quad b_i \in \{0, 1\}, \quad b_0 = 1, \quad b_{N-1} = 1$$


The initial active bit length is finite:
$$L_0 = L(x) = \lfloor\log_2 x\rfloor + 1 < \infty$$


All bit positions $i \ge N$ are identically zero. Infinitely long binary representations (such as $2$-adic integers $\mathbb{Z}_2$) are excluded by domain definition.


## 2. The Non-Shifting Isomorphic Map ($3X + 2^k$)


### 2.1 Formal Operator Definition


Instead of dividing by $2$ to strip trailing zeros, we define the non-shifting feed-forward injection map:
$$X_{t+1} = 3X_t + 2^{k_t}$$
where $2^{k_t} = 2^{\text{LSB}(X_t)}$ dynamically tracks the active LSB pointer.


**Structural Isomorphism:**
 The odd sequence of standard Collatz is recovered identically by right-shifting $X_t$ by $k_t$:
$$x_t = X_t \cdot 2^{-k_t}$$


**Unidirectional Causality (Zero Top-Down Feedback):**
 Binary addition carries propagate strictly right-to-left ($i \to i+1$). Lower bit positions evolve independently of higher bit positions. High-order bits (MSB) physically cannot emit carries downward to alter lower bit blocks.


## 3. Boundary Velocity Dynamics & Physical Bounds


Define the active register length at step $t$ as:
$$L(t) = \text{MSB}(X_t) - \text{LSB}(X_t) + 1$$


The change in active register length per odd step is governed by boundary velocities:
$$\Delta L_t = \Delta\text{MSB}_t - \Delta\text{LSB}_t$$


### 3.1 Lemma 1: The Absolute MSB Expansion Ceiling


For any state $X_t$ with bit length $N$:
$$\Delta\text{MSB}_t = \text{MSB}(X_{t+1}) - \text{MSB}(X_t) \le 2\text{ bits/step}$$


**Proof:**
For $2^{N-1} \le X_t < 2^N$:
$$3X_t + 2^{k_t} < 3(2^N) + 2^N = 4(2^N) = 2^{N+2}$$


Because $Y_{\text{max}} < 2^{N+2}$, positional binary addition physically cannot spill 3 bits in a single step. Furthermore, because $3X_t + 2^{k_t} < 3 \cdot 2^N$, if a step yields a 2-bit expansion, the leading two bits are bounded by $10_2$, preventing consecutive $+2$ expansions without intermediate carry setup. $\blacksquare$


### 3.2 Lemma 2: LSB Tail Clearance Floor


The single-step tail shift is determined by the $2$-adic valuation:
$$\Delta\text{LSB}_t = v_2(3x_t + 1) \ge 1\text{ bit/step}$$


- $\Delta\text{LSB}_t = 1$ occurs if and only if $x_t \equiv 3 \pmod 4$ (...11_2).
- $\Delta\text{LSB}_t \ge 2$ occurs whenever $x_t \equiv 1 \pmod 4$ (...01_2).


## 4. The Glider Requirement for Infinite Growth


To disprove the existence of divergent trajectories, we must establish what physical conditions are required for infinite growth. 


Because the spatial distribution of the active LSB boundary over the domain $\mathbb{N}$ strictly dictates a global mean drop rate of $\overline{\Delta\text{LSB}} = 2.0\text{ bits/step}$, any arbitrary or randomized sequence of carries will mathematically pull the active register length to $1$ (since $2.0 > \log_2(3) \approx 1.585$). 


Therefore, for an active register to expand infinitely ($\overline{\Delta L} > 0$), it must systematically evade the $2.0$ spatial gravity. It can only accomplish this if the active register forms a 
**Glider**
: a self-reproducing, shift-periodic bit pattern that perfectly coordinates with the $3X$ map to artificially suppress its own $\Delta\text{LSB}$ drops. If a sequence cannot form a stable glider, it is mathematically guaranteed to be annihilated by the $2.0$ global average.


## 5. The Algebraic Constraint on Glider Velocity


Let a glider exist such that an initial active register $x_0$ successfully reproduces its exact bit pattern after $S$ steps, with a total LSB drop of $K$ bits. Because the glider is a perfect cycle, $x_0 = x_S$.


The algebraic evaluation of $S$ steps of the standard Collatz map yields:
$$ x_S = \frac{3^S x_0 + C}{2^K} $$


Where $C = \sum_{j=0}^{S-1} 3^{S-1-j} 2^{K_j}$ represents the exact sequence of $+1$ carry-injections from the map. Because the operation is strictly additive, $C$ is strictly positive ($C > 0$).


Substituting $x_S = x_0$ and solving for $C$:
$$ x_0 \left(2^K - 3^S\right) = C $$


Because both $x_0 \ge 1$ and $C > 0$, the right side of the equation is positive, demanding that the left side also be strictly positive:
$$ 2^K - 3^S > 0 $$
$$ 2^K > 3^S $$
$$ K > S \log_2(3) $$
$$ \frac{K}{S} > \log_2(3) \approx 1.585\text{ bits/step} $$


**Conclusion of the Constraint:**
The algebra dictates an absolute, unbreakable constraint: 
**no glider can ever reproduce with an average velocity $\le \log_2(3)$**
. Any hypothetical sequence that attempts to maintain a rate $\le 1.585$ to cause infinite growth 
*cannot*
 be a glider; it must be a strictly non-cycling, divergent sequence.


## 6. Carry-Wake Corruption (The Three Speeds)


Because the algebra forbids a glider from moving slow enough to cause growth, any divergent trajectory must be a non-cycling, aperiodic path. However, an aperiodic path attempting to expand the register physically generates its own destruction via the 
**Carry-Wake**
.


Assume a "fugitive" bit pattern successfully maintains an average advancement rate $K/S$ to cause infinite growth. 
Because the pattern advances leftward by $K$ bits, it leaves behind a wake of bits at the LSB. At every step, the $3X+2^k$ map multiplies this wake by $3$. Over $S$ steps, the wake physically expands leftward by exactly $S \log_2(3)$ bits.


This creates an inescapable physical constraint defined by the relationship between the pattern's speed ($K$) and the wake's expansion speed ($S \log_2 3$):


### Scenario A: The Pattern Moves Slower Than the Wake ($K < S \log_2 3$)
To cause infinite active register growth, the pattern 
*must*
 move slower than the MSB expansion. However, because $S \log_2(3) > K$, the wake's expansion physically overtakes the pattern's advancement. The chaotic carries generated by the wake blast through the entire fugitive sequence, obliterating it from below.


### Scenario B: The Pattern Moves Exactly As Fast As the Wake ($K = S \log_2 3$)
This scenario is algebraically impossible because $K$ and $S$ must be integers, and $\log_2(3)$ is irrational ($3^S \neq 2^K$). Even if a sequence hovered infinitely close to this boundary, the active register length would remain strictly bounded. By the Pigeonhole Principle, a bounded active register must eventually repeat, turning the sequence into a Glider, which is explicitly forbidden by the algebraic constraint in Section 5.


### Scenario C: The Pattern Moves Faster Than the Wake ($K > S \log_2 3$)
The pattern successfully outruns the expanding wake. However, because the rate of LSB clearance ($K/S$) strictly exceeds the rate of MSB expansion ($\log_2 3$), the active register length mathematically shrinks. The sequence collapses to the $1$ attractor.


## 7. Main Theorem Conclusion


The Catch-22 of the Collatz map on $\mathbb{N}$ is absolute. There is no mathematically safe speed for infinite growth:
- 
**The Engine of Growth:**
 Infinite growth requires systematically evading the $2.0$ spatial average, which mathematically requires the formation of a periodic glider.
- 
**The Algebraic Wall:**
 The fundamental Collatz equation $x_0(2^K - 3^S) = C$ strictly forbids any glider from moving slow enough to achieve growth ($K/S > 1.585$).
- 
**The Physical Collapse:**
 A non-cycling fugitive pattern attempting to cause growth ($K < S \log_2 3$) is physically overtaken and destroyed from below by the $1.585$ expansion speed of its own carry-wake.


$$\mathbf{Q.E.D.}$$

THE COLLATZ CONJECTURE

A COMPARATIVE DENSITY PROOF OF TRAJECTORY DESCENT IN THE COLLATZ 3N+1 SYSTEM VIA 1N+1 MODULAR MODELING

Author: All mathematical ideas and constructions by Steve Tomlinson except logarithms in 2 and 3.1.

(l knew something mathematical must do this job, l didn't know what it was; logarithmic bounds.)

Essay composition by AI with many mistakes edited by Steve Tomlinson

Date: August 2026

ABSTRACT

This paper establishes a novel structural framework for analyzing the Collatz 3N+1 conjecture by introducing a perfectly descending baseline model: the 1N+1 system. While the standard 3N+1 system exhibits chaotic trajectory growth, we prove that both systems operate on base-2 modular architecture. By comparing the density pathways of the 3N+1 system against the verified, linear geometric descent of the 1N+1 model, we demonstrate that the standard Collatz mapping exhibits an absolute asymptotic density of descent equal to 1 at the infinite operational horizon.

1. THE CENTRAL BREAKTHROUGH:

THE 1N+1 STRUCTURAL BENCHMARK

To analyze the non-linear trajectories of the standard Collatz conjecture, we define a perfectly controlled model system, the 1N+1 system, governed by the following mapping for all natural numbers N:

f(N) = N/2 if N ≡ 0 (mod 2)

f(N) = 1N+1 if N ≡ 1 (mod 2)

Theorem 1.1. In the 1N+1 system, 100% of all natural numbers N > 1 are mathematically guaranteed to reach a strictly smaller value within a maximum of two operations.

Proof.

Case 1: If N is even, a single operation yields N/2, which is strictly less than N.

Case 2: If N is odd, the application of the odd rule followed by the mandatory division by 2 yields a composite operation of (1N+1)/2.

Setting up the inequality for descent:

(N+1)/2 < N => N+1 < 2N => 1 < N.

This inequality holds true for all positive odd integers greater than 1. Thus, every element shrinks locally and immediately.

By creating an arbitrary system for numbers to drop in the 1N+1 system, the entire number line is partitioned into clean, un-scrambled geometric slices:

* Step 1 (All Evens, 0+2n) accounts for exactly 1/2 (50%) of all numbers.

* Step 2 (The 1 + 4n Odds) accounts for exactly 1/4 of all numbers

*Step 3 (The 3 + 8n Odds) accounts for exactly 1/8 of all numbers.

*Step 4 (The 7 + 16n Odds) accounts for exactly 1/16 of all numbers.

Continuence of this process continues to account for exactly (2^x-1)/(2^x) of all numbers, accumulating to 100% of the number line descending within a 2-step horizon.

2. THE 3N+1 SYSTEM AS A LOG-LINEAR DISTORTION

When the odd operator is shifted to the standard Collatz rule (3N+1), the underlying base-2 modular grid is stretched. Let m represent both the family classification and the number of odd steps executed before the first downward drop below the initial value. Let a be the number of required even operations (divisions by 2).

For a net trajectory descent to occur, the geometric growth factor must drop below 1:

(3^m) / (2^a) < 1 => 3^m < 2^a

Taking the base-2 logarithm (log₂) of both sides yields the absolute structural boundary:

a > m · log₂(3) ≈ 1.5849625m

Because log₂(3) > 1, immediate descent within a single operational cycle is impossible for odd positive integers. Instead, numbers are sorted into deterministic "m-families", where the total step horizon required to secure the necessary 'a' divisions scales linearly as a function of m:

Total Steps = m + a = ⌈2.5849625m⌉

3. THE m-FAMILY SIEVE AND EXPONENTIAL CONTRACTION

The exact proportions of the number line accounted for by these families are defined sequentially:

(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 1/16 of all numbers.

*m=4 (The 7 + 128n Odds) + (The 15 + 128n Odds) + (59 + 128n Odds) accounts for 3/128 of all numbers.

* m=5 ((The 39, 79, 95, 123, 175 and199) 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.

At this point when m reaches 6:

(6 × 2.5849626) rounded up = 16 Collatz operations accumulates to account for exactly 15/16 of all numbers shown to reach a smaller number.

* m=7 accounts for exactly 30/2048 of all numbers.

Manually proving m=8 would have taken too much paper.

3.1 The structural limits for m=3 and m=4 families in the Collatz conjecture are determined by the logarithmic boundary

a > m×log2(3), where m is the number of odd steps and a is the number of even operations. Applying this, the m=3 family requires 5 even steps for 3 odd steps, creating a 1/16 density across residues modulo 32, while m=4 requires 7 even steps for 4 odd steps, generating a 3/128 density modulo 128. This logarithmic framework accurately predicts the modular structures for specific families.

4. THE UNIFIED 2^x HORIZON INDUCTION

While the multiplier 3 introduces "bumpy" intermediate statistical fluctuations between the milestones (e.g., stabilizing around a cumulative density of ≈ 5/6 at step 6, 10/11 at step 11, 12/13 at step 13, back to exactly 7/8 at step 8 and 15/16 at 16 steps) The total system mathematically self-corrects and snaps perfectly back to the clean geometric density progression of the 1N+1 benchmark at every power-of-two operational milestone (2^x).

By mathematical induction on the operational horizon x, the cumulative density of numbers proven to have reached a smaller value satisfies:

Cumulative Density(2^x) = 1 - 1/(2^x)

As the operational step horizon scales toward the infinite limit (x → ∞):

Limit as x → ∞ of [1 / 2^x] = 0

5. CONCLUSION

By using the 1N+1 system as an absolute structural baseline, we prove that the standard 3N+1 Collatz system is not chaotic, but deterministic and rigidly bounded. The "numerical shields" created by dense clusters of binary ones (such as the 2^x - 1 Collatz steps families) only temporarily delay descent. Over an infinite horizon, the remaining density of holdout numbers converges to exactly zero.

Because 100% of all numbers must eventually reach a strictly smaller milestone, any arbitrary starting number is locked into an inescapable cascading chain of downward thresholds, forcing all trajectories to eventually collapse into the fundamental 2 → 1 trivial loop.

Q.E.D.

By shifting the analytic paradigm from stochastic modeling to comparative structural architecture, this 1N+1 baseline framework introduces a constructive element that establishes absolute structural determinism, distinguishing it from the probabilistic approach in Terence Tao’s 2019 groundbreaking density proof. While Tao’s work treats individual trajectories as non-constructive, semi-chaotic random walks, this model maps the geometric architecture of "numerical shields," demonstrating that standard Collatz mapping is rigidly constrained by a base-2 modular grid and logarithmic boundaries.

For some notational consistency:
Use the shortcut odd step: T(x) = {x/2, even; (3x+1)/2, odd}

Let the number of steps be n, and the number of odd steps be m.
If x follows a given parity sequence (example: EOEOOEE), then

• kᵢ are the indices of the odd steps (example: [1,3,4]), and
• S = ∑2^(kᵢ)3^(m-1-i) from i=0 to m-1 (example: 2^(1)3^(2) + 2^(3)3^(1) + 2^(4)3^(0))
• So, T^(n)(x) = (3^(m)x + S)/2^(n)

Per a proof in a stackexchange answer, the title claim is true, but can you prove it?

I know that this video is not directly related to Collatz but I thought I'd post a link to it anyway because it is about mod 4 and mod 8 arithmetic and mod 8 arithmetic does have a lot of relevance to Collatz (without claiming that it is the whole story - it is not)

Of relevance is the key result that numbers of the form 4^a.(8b+7) cannot be expressed as a sum of three squares. If we restrict ourselves to odd numbers, then all odd numbers that can be expressed as (8b+7) cannot be expressed as a sum of 3 squares and all other odd numbers can be.

So, these odd numbers that admit a partitioning as a sum of 3 squares can expressed generically as:

8t+r = 8t+2s+3 = a^2+b^2+c^2 , where r=3+2s and s is in {-1,0,1}

If we subtract 3 from each side we get:

8t+2s = 2(4t+s) = (a-1)(a+1) + (b-1)(b+1) + (c-1)(c+1)

The LHS is divisible by 8 iff s = 0 and also (I think) iff a,b,c are all odd

Now, I haven't found any implications of this identity that is useful for Collatz, but I thought I would throw it out there anyway, just in case it piques someone else's interest or curiosity.

I’ve been experimenting with the Collatz conjecture from a slightly different direction.

Instead of asking:

“Does every number eventually reach 1?”

I asked:

“How far backwards do we have to go to find an odd multiple of 3 whose Collatz trajectory reaches a given odd number?”

DEFINITION

For an odd integer h, define:

α(h) = the smallest odd multiple of 3 whose accelerated Collatz orbit reaches h.

Then define:

E(N) = max α(h), over odd h <= N.

Call an odd number h “hard” if it sets a new record for α(h).

So hard numbers are the targets that force us to extend the source boundary further than ever before.

AN EARLY EXAMPLE

The first hard odd after 1 is:

19

Its smallest source is:

33 -> 25 -> 19

because:

3*33 + 1 = 100 = 4*25

and:

3*25 + 1 = 76 = 4*19

So the odd-step valuation word is:

(2,2)

THE PATTERN

When hard records are computed further, one particular valuation pattern eventually appears to dominate:

(1,4)

For this branch, the source-target relationship is:

α(h) = (32h - 5)/9

for the appropriate residue class.

Therefore:

α(h)/h = 32/9 - 5/(9h)

and as h becomes large:

α(h)/h -> 32/9

where:

32/9 = 3.555555...

THE CONJECTURE

This suggests the following inverse-coverage conjecture:

E(N)/N -> 32/9 as N -> infinity.

In words:

The worst-case source needed to cover every odd target up to N appears asymptotically to be about:

3.555555... * N

A STRONGER CONJECTURE

There may also be some finite threshold H such that every hard odd h > H satisfies:

α(h) = (32h - 5)/9

In other words:

Every sufficiently large hard record may come from the same valuation word:

(1,4)

I’ll call this “eventual (1,4) dominance.”

WHAT IS ACTUALLY PROVED?

Important distinction:

I am NOT claiming a proof of the Collatz conjecture.

And I am NOT claiming the asymptotic equality E(N)/N -> 32/9 has been proved.

What can be proved in this inverse-cylinder framework is the universal upper bound:

α(h) < (32/9)h

for every odd target h.

There is also a finite inverse-cylinder argument showing that 32/9 is the optimal worst-case slope within that finite positive-cylinder framework.

So the difficult remaining direction is:

liminf E(N)/N >= 32/9

If that could be proved, then together with the upper bound we would get:

E(N)/N -> 32/9

WHY DOES 32/9 APPEAR?

For a valuation word (a1,a2,...,ar), the nominal inverse slope is:

2^(a1+a2+...+ar) / 3^r

For the two-step word (1,4):

2^(1+4) / 3^2

= 32/9

So 32/9 is directly generated by this short valuation pattern.

COMPUTATIONAL RESULTS

I wrote an exact ascending-source scanner that records every new hard odd.

The exhaustive computation currently goes through:

10,000,000,000

Results:

Hard records found:

11,972,736

Hard records following the (1,4) formula:

11,972,706

Exceptions:

30

Largest exceptional hard odd:

87,967

New non-(1,4) hard records between 87,967 and 10,000,000,000:

NONE

THE FINAL HARD RECORD BELOW 10^10

The last hard odd found below 10 billion is:

h = 9,999,999,667

Its smallest source is:

α(h) = 35,555,554,371

And exactly:

35,555,554,371

= (32*9,999,999,667 - 5)/9

So it is another exact (1,4) record.

WHY I FIND THIS INTERESTING

Most Collatz research looks at the forward orbit:

n -> T(n) -> T^2(n) -> ...

This construction instead looks at an extremal property of the inverse graph.

What is surprising is that a complicated reverse structure appears to develop a very simple boundary:

32/9

generated by a tiny valuation pattern:

(1,4)

Early hard numbers come from several different valuation words.

Then, in the computation, those competing families disappear.

After 87,967, every single new hard record through 10 billion follows the same (1,4) formula.

THE MAIN QUESTION

Can anyone prove that every sufficiently large hard odd must come from the (1,4) family?

Or, alternatively:

Can anyone find an arbitrarily large hard odd that is NOT generated by (1,4)?

A WEAKER TARGET

Even without proving eventual (1,4) dominance, it would be enough to prove:

liminf E(N)/N >= 32/9

because the matching upper bound is already available in the finite-cover framework.

WHAT THIS DOES NOT PROVE

Even if the 32/9 conjecture is true, it does NOT automatically prove the classical Collatz conjecture.

The 32/9 statement is about reverse coverage:

“How large a source is needed to reach a target?”

The Collatz conjecture is about forward convergence:

“Does every starting value eventually reach 1?”

These are related, but they are not the same statement.

LITERATURE QUESTION

I have searched for this specific formulation:

- the inverse record function E(N)

- hard odds defined by record values of α(h)

- the 32/9 asymptotic frontier

- eventual (1,4) dominance

and I have not found an equivalent conjecture in the Collatz literature.

If anyone knows an earlier equivalent formulation, I would genuinely appreciate the reference.

WHY I’M POSTING THIS:

I’m mainly interested in three things:

1. Does someone see a theoretical reason why (1,4) should eventually dominate?
2. Can someone construct a large competing valuation family?
3. Is there an existing theorem in Collatz dynamics, p-adics, symbolic dynamics, inverse trees, or ergodic theory that naturally explains the appearance of 32/9?

I have the derivation, finite-cover argument, exact computation code, and longer write-up available if anyone wants to inspect them.

EDIT: Added an illustrative explanation of "Hard" Odds - let's say we want to construct node graph - on y axis we will keep adding all the "O" odd multiples of 3 (source branch) and then keep going forward from each odd to the next path till we reach 1 or jump to existing path of any other O. Now, we are interested to know when we increase O step by step, what are some properties that we discover - one of them is asking are all the how many odds less O are present in the graph: for example in the starting case of O=3, graph has all the odds <3 that is 1 is present; now we generalize this expression by saying when will be the next odd k be discovered that hasn't been found yet and let it's source O be E(k) - now the conjecture is about k/E(k) - attached are the graphs till O=57

Graph for O=3 | all odds <3 are discovered - no hard odd found

https://preview.redd.it/vvm8lv3hcnih1.png?width=774&format=png&auto=webp&s=51eab635a2928ea6562f189e7a6a2b1af475a030

Graph for O=9 | all odds <9 are discovered - no hard odd found

https://preview.redd.it/lr0dmlvwhnih1.png?width=1366&format=png&auto=webp&s=6865fa25a74c5d72bb811617cb741bcc26cfcba7

Graph for O=15 | all odds <15 are discovered - no hard odd found

https://preview.redd.it/zxom3xacinih1.png?width=1300&format=png&auto=webp&s=2a4b02ba093e03cb152fedf9b175493811123cd0

Graph for O=21 | all odds <21 are not discovered - hard odd found 19, with source node E(O)= 33 and valuation of (2,2)

https://preview.redd.it/kcb8xebfinih1.png?width=1332&format=png&auto=webp&s=fd77d82aa8475ddc7bedd271cfcd2b7dd99e1a8c

Now jumping to Graph for O=33 (skipped 27 as 19 continued) | all odds <33 discovered - no new hard odd found

https://preview.redd.it/5vq759mxinih1.png?width=1584&format=png&auto=webp&s=173eba6cdfeb84b945d0a64a19e789c9e11214f7

Graph for O=39 | all odds <39 are not discovered - hard odd found 37, with source node E(O) = 57 and valuation of (2,1,2,2)

https://preview.redd.it/xt26vhaqjnih1.png?width=1596&format=png&auto=webp&s=01048a186bbce20b1eb15ebfb6283c680b2ecbcf

Now jumping to Graph for O=57 (skipped 45,51 as 37 continued) | all odds <57 are not discovered - hard odd found 55, with source node E(O) = 129 and valuation of (2,2,2)

https://preview.redd.it/5f2qmcq4knih1.png?width=1416&format=png&auto=webp&s=bfbade4a009480c2e7dcdc50ceed1dfd804132ca

Now, what you are seeing is that for every E(O), let's name in Exhaustive cut-off ratio resets after every hard odd O is encountered - if we plot the graph of E(O)/O - it looks like the following

https://preview.redd.it/12k6qnknlnih1.png?width=2142&format=png&auto=webp&s=7ee42ede06ed79a05ab9562c84cba3dab49b1d4b

After 133rd odd which is 89425, all hard odds till 10 Billion have had bone path from bone parent as 1,4

https://preview.redd.it/roiaeq53mnih1.png?width=1604&format=png&auto=webp&s=2e6f7cdd07656ea3f5aa1ed8d4dbb07292a86899

This is the conjecture, as O tends to grow infinitely larger, the ratio E(O)/O starts settling down to 32/9 which is weird as so many competing paths other 1,4 could have emerged but we don't why it takes over in the long term as observed in data - logically we know the upper bound is 32/9 but it can achieved by 1,2,2 or 1,1,2,1 any other combo but 1,4 is special we don't why, secondly we don't know the lower bound and hence cannot prove convergence like is 2,2 possible for any number after 10 billion - which will disprove the conjecture.

Last image - shows how E(O)/O looks like on log scale for very high O- almost converging to 32/9

https://preview.redd.it/f2ldrh7dnnih1.png?width=968&format=png&auto=webp&s=49b7f605c5d7950cdfbae4e85a5ab7a3e054e838

Table of all hards odds upto last exceptional hard odd

hard_index	hard_odd	bone1_parent	valuation_word	family	defect_32h_minus_9parent
1	1	21	-6	exception	-157
2	19	33	(2,2)	exception	311
3	37	57	(2,1,2,2)	exception	671
4	55	129	(2,2,2)	exception	599
5	109	171	(1,2,2,2)	exception	1949
6	127	225	(2,2)	exception	2039
7	163	513	(2,2,2,2)	exception	599
8	271	759	(1,1,3,2,2,2)	exception	1841
9	379	897	(2,2,2)	exception	4055
10	487	1215	(1,1,1,1,1,2,3,4)	exception	4649
11	541	1281	(2,2,2)	exception	5783
12	649	2049	(2,2,2,2)	exception	2327
13	973	2427	(1,2,1,2,2,2,2,2)	exception	9293
14	1027	2433	(2,2,2)	exception	10967
15	1135	3585	(2,2,2,2)	exception	4055
16	1459	5187	(1,4)	(1,4)	5
17	1945	6915	(1,4)	(1,4)	5
18	2593	8193	(2,2,2,2)	exception	9239
19	2701	8535	(1,1,4,2)	exception	9617
20	2917	9711	(1,1,1,2,1,1,2,3,4)	exception	5945
21	3079	9729	(2,2,2,2)	exception	10967
22	3403	11175	(1,1,2,1,2,2,2,1,1,1,1,2,1,1,2,3,1,1,2,3,4)	exception	8321
23	4051	12801	(2,2,2,2)	exception	14423
24	4213	14979	(1,4)	(1,4)	5
25	4375	15555	(1,4)	(1,4)	5
26	4861	17283	(1,4)	(1,4)	5
27	5347	19011	(1,4)	(1,4)	5
28	5833	20739	(1,4)	(1,4)	5
29	6319	22467	(1,4)	(1,4)	5
30	7291	25569	(2,2,1,1,1,2,2,1,2,2,2,2,2,2)	exception	3191
31	7777	27651	(1,4)	(1,4)	5
32	8587	28587	(1,2,2,2,1,2,2,2,2)	exception	17501
33	8749	29127	(1,1,2,2,2,2,2,2,2)	exception	17825
34	9235	30747	(1,2,1,1,1,1,2,3,4)	exception	18797
35	9883	31233	(2,2,2,2)	exception	35159
36	10369	32769	(2,2,2,2)	exception	36887
37	10693	38019	(1,4)	(1,4)	5
38	12151	43203	(1,4)	(1,4)	5
39	13123	46659	(1,4)	(1,4)	5
40	14581	51843	(1,4)	(1,4)	5
41	16039	57027	(1,4)	(1,4)	5
42	16525	58755	(1,4)	(1,4)	5
43	17335	61635	(1,4)	(1,4)	5
44	17497	62079	(1,1,1,1,1,1,2,1,2,2,2,1,2,1,2,1,1,3,1,1,3,2,1,2,2,1,1,1,1,1,1,3,1,1,1,2,1,1,1,1,2,2,3,1,2,2,2,2,1,3,3,1,2,3,4)	exception	1193
45	17983	63939	(1,4)	(1,4)	5
46	18469	65667	(1,4)	(1,4)	5
47	18955	67395	(1,4)	(1,4)	5
48	20899	74307	(1,4)	(1,4)	5
49	21709	77187	(1,4)	(1,4)	5
50	21871	77763	(1,4)	(1,4)	5
51	22357	79491	(1,4)	(1,4)	5
52	22843	81219	(1,4)	(1,4)	5
53	23329	82947	(1,4)	(1,4)	5
54	23815	84675	(1,4)	(1,4)	5
55	25273	89859	(1,4)	(1,4)	5
56	26245	93315	(1,4)	(1,4)	5
57	27703	98499	(1,4)	(1,4)	5
58	28189	100227	(1,4)	(1,4)	5
59	29161	103683	(1,4)	(1,4)	5
60	29647	105411	(1,4)	(1,4)	5
61	30457	108291	(1,4)	(1,4)	5
62	30619	108867	(1,4)	(1,4)	5
63	31105	110595	(1,4)	(1,4)	5
64	31591	112323	(1,4)	(1,4)	5
65	32077	114051	(1,4)	(1,4)	5
66	32563	115779	(1,4)	(1,4)	5
67	33535	119235	(1,4)	(1,4)	5
68	34021	120963	(1,4)	(1,4)	5
69	34831	123843	(1,4)	(1,4)	5
70	34993	124419	(1,4)	(1,4)	5
71	35479	126147	(1,4)	(1,4)	5
72	36937	131331	(1,4)	(1,4)	5
73	37909	134787	(1,4)	(1,4)	5
74	38395	136515	(1,4)	(1,4)	5
75	39367	139971	(1,4)	(1,4)	5
76	40825	145155	(1,4)	(1,4)	5
77	41311	146883	(1,4)	(1,4)	5
78	42769	152067	(1,4)	(1,4)	5
79	43579	154947	(1,4)	(1,4)	5
80	43741	155523	(1,4)	(1,4)	5
81	44713	158979	(1,4)	(1,4)	5
82	45199	160707	(1,4)	(1,4)	5
83	45685	162435	(1,4)	(1,4)	5
84	46657	165891	(1,4)	(1,4)	5
85	47143	167619	(1,4)	(1,4)	5
86	50059	177987	(1,4)	(1,4)	5
87	50545	179715	(1,4)	(1,4)	5
88	51517	183171	(1,4)	(1,4)	5
89	52489	186627	(1,4)	(1,4)	5
90	53947	191811	(1,4)	(1,4)	5
91	54433	193539	(1,4)	(1,4)	5
92	55405	196995	(1,4)	(1,4)	5
93	55891	198723	(1,4)	(1,4)	5
94	56701	201603	(1,4)	(1,4)	5
95	56863	202179	(1,4)	(1,4)	5
96	57349	203907	(1,4)	(1,4)	5
97	57835	205635	(1,4)	(1,4)	5
98	58321	207363	(1,4)	(1,4)	5
99	58807	209091	(1,4)	(1,4)	5
100	60265	214275	(1,4)	(1,4)	5
101	61237	217731	(1,4)	(1,4)	5
102	62209	221187	(1,4)	(1,4)	5
103	62695	222915	(1,4)	(1,4)	5
104	63181	224643	(1,4)	(1,4)	5
105	64639	229827	(1,4)	(1,4)	5
106	65611	230139	(1,2,1,2,1,1,2,2,2,2,2,2,2,2)	exception	28301
107	67069	238467	(1,4)	(1,4)	5
108	67555	240195	(1,4)	(1,4)	5
109	68527	243651	(1,4)	(1,4)	5
110	69013	245379	(1,4)	(1,4)	5
111	69823	248259	(1,4)	(1,4)	5
112	69985	248835	(1,4)	(1,4)	5
113	70471	250563	(1,4)	(1,4)	5
114	70957	252291	(1,4)	(1,4)	5
115	71443	254019	(1,4)	(1,4)	5
116	71929	255747	(1,4)	(1,4)	5
117	72901	259203	(1,4)	(1,4)	5
118	73387	260931	(1,4)	(1,4)	5
119	75331	267843	(1,4)	(1,4)	5
120	75817	269571	(1,4)	(1,4)	5
121	77761	276483	(1,4)	(1,4)	5
122	78733	279939	(1,4)	(1,4)	5
123	80191	285123	(1,4)	(1,4)	5
124	80677	286851	(1,4)	(1,4)	5
125	82945	294915	(1,4)	(1,4)	5
126	83107	295491	(1,4)	(1,4)	5
127	83593	297219	(1,4)	(1,4)	5
128	84079	298947	(1,4)	(1,4)	5
129	84565	300675	(1,4)	(1,4)	5
130	85051	302403	(1,4)	(1,4)	5
131	86023	305859	(1,4)	(1,4)	5
132	87967	308559	(1,1,1,3,2,1,2,1,1,1,1,2,3,4)	exception	37913

So u/IntelligentTwo2175 had a post that seemed really interesting to me. Basically numbers of the form 2^(n) -1 and 2^(n+1) -1 merge for n >= 0. I think we can make it a little stronger by changing it to

m * 2^(n) -1 and m * 2^(n+1) -1 for odd m. If m was even, we can just pull out the factors of 2 into the 2^(n)

So quick proof.

m * 2^(n) -1

is a steiner circuit, so let's jump ahead to

m * 3^(n) - 1

which we know must be even, except let's back up an additional even step, so we know we are at a multiple of 4 of some number x.

2 * (m * 3^(n) - 1 ) = 4 x

Now let's follow

m * 2^(n+1) -1 for only n odd steps

m * 2 * 3^(n) -1

We can easily see this is equal to 4x + 1.

And we know x and 4x + 1 numbers must merge after O, and OEE respectively.

Recursively, we can say m * 2^(n) must be connected on the tree for a specific m and all n.

Now could this be used in a proof of collatz? Let's try.

So all even numbers can divide by 2 until an odd number.

All odd numbers can be written as the even number 1 above it - 1.

By repeatedly alternating these two steps, we create a monotonically decreasing sequence. And thus, must go to 1.

Ex. Odd number 29, write it as the even number above - 1, so 30 -1. Pull out all factors of 2

15 * 2^(1) - 1.

Using the property above, we know this must connect to

15 * 2^(0) - 1

Once we are out of factors of 2, it becomes odd - 1 which must be even. So we can divide those factors of 2 via standard collatz rules until odd again.

15 - 1 = 14

14 / 2 = 7

Once again, write it as the even above - 1

7 = 8 - 1 = 2^(3) - 1

Which using the property goes

2^(3) - 1

2^(2) - 1

2^(1) - 1 = 1

So can anyone check my logic? It's way too late and this seems way too simple for what I have lost too many nights of sleep over.

This post turned out to be too long and image-dependent to be stable on Reddit, so I've moved its contents to a Google Doc.

Link: Practical Continued Fractions for Collatz

Hello everyone,

I was recently running some tests on the Collatz's conjecture and noticed a certain pattern.

It works like this: take the set of numbers generated by the function f(n) = 2^n - 1, for n >= 3, and arrange them into pairs as follows: (3, 4), (5, 6), (7, 8) ... and so on. You will then have the values ​​(7, 15), (31, 63), (127, 255), ad infinitum.

Well then, in each of these pairs, the number of steps required for the second value in the pair to reach 1 is always the number of steps for the first value plus one. Let's take the first pair as an example to visualize this:

7 -> 22 -> 11 -> 34 -> 17 -> 52 -> 26 -> 13 -> 40 -> 20-> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 (16 steps)

15 -> 46 -> 23 -> 70 -> 35 -> 106 -> 53 -> 160 -> 80 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 (17 steps)

There is also another case: if, within this sequence, you disregard the steps of dividing by 2 and proceed directly to the odd values, then—taking the same even subset—the number of steps for the first number will equal the number of steps for the second. Using the same example, we have:

7 -> 11 -> 17 -> 13 -> 5 -> 1 (5 steps)

15 -> 23 -> 35 -> 53 -> 5 -> 1 (5 steps)

I've already tested it with pairs of considerably large numbers n, and the pattern remains. I have no idea if there are any results in the literature on this, much less if it works for infinitely many values ​​of n.

Do you know anything about this?

My investigation has focused on strings of E and O operations (i.e., E, O, EO, EEO, EEOEO, or OEOEOEEOEOEOEEEE, etc.) and their potential to be Loops (i.e., {0}, {-1/2}, {-1,-2}, {4,2,1}, {-20,-10,-5,-14,-7}, or the 18-cycle Loop, respectively).

For any given string of E and O operations... such as EEOEOEOEEEOEOEEO...
...where E=10 and O=6, and L=E+O=16
...and (2^E)-(3^O) = (2^10)-(3^6) = (1,024)-(729) = 295 > 0
...there exists one and only one rational number that can be both an input to the string, and an output.
...and since (2^E)-(3^O)>0, that rational number must be positive.

In the case of EEOEOEOEEEOEOEEO, that number is...
N = Σ / (2^E - 3^O)
N = ((2^2)*(3^5) + (2^3)*(3^4) + (2^4)*(3^3) + (2^7)*(3^2) + (2^8)*(3^1) + (2^10)*(3^0)) / ((2^10) - (3^6))
N = ((4*243)+(8*81)+(16*27)+(128*9)+(256*3)+(1,024*1)) / ((1,024) - (729))
N = (972+648+432+1,152+768+1,024) / (295)
N = 4,996/295, or approximately 16.9356...

I also understand that the first positive integer that can successfully transit that string, and beget another integer, is 732 which begets 526 through EEOEOEOEEEOEOEEO.
732->366->183->550->275->826->413->1240->620->310->155->466->233->700->350->175->526

I also understand that the next higher input is 2^E greater, while the next higher output is 3^O greater.
732->526
732+2^10=1,756
526+3^6=1,255
...and sure enough, 1,756->EEOEOEOEEEOEOEEO->1,255.
(Note: I understand that 1,255 is a Collatz-INappropriate output, but that's irrelevant to this small point I'm getting at. But if it bothers you, you can increase the increment to 2*(2^10) to get 2,780 as your next input, and 1,984 as your next output.)

Meanwhile, the next lower input is 2^E smaller, while the next lower ouput is 3^O smaller.
732->526
732-2^10= -292
526-3^6= -203
...and sure enough -292 ->EEOEOEOEEEOEOEEO-> -203
(Same note as before.)

Having fun, so far? Are all your calculations aligning with mine? Great! Here's the next part...

Let's call the linear difference between that input and output "the gap".
-3,364 -> -2,390 ___gap=974
-2,340 -> -1,661 ___gap=679
-1,316 -> -932 _____gap=384
-292 -> -203 _______gap=89
xoxoxoxoxoxoxoxoxoxoxoxox
732 -> 526 _________gap=206
1,756 -> 1,225 _____gap=501
2,780 -> 1,984 _____gap=796
3,804 -> 2,713 _____gap=1,091

Here's my question: Are there any strings where the smallest gap is more than 2^E away from the zero line?

Or, rephrased: Are there any strings where the first positive integer input to yield a positive integer ouput, has a gap that is greater than the gap of the second positive integer input to yield a positive integer output?

My intuition is telling me that there aren't... but I'm not sure why.

I would like to share the first chapter of an angoing work proposes a structural study of the Collatz conjecture, reformulated under an axiomatic framework, I call (PCA, PCA1= 3n+1 for n odd.

PCA2 = n/2 for n even).The preprint IS deposited on Zenodo ( DOI: https ://doi.org/10.5281/zenodo.21927860).

This chapter does not claim a full proof of the conjecture_it establishes the structural mechanism and provides convergence to 1 for two specific geometric familis of integers.

The question of whether every natural number converges toward these familis IS left open for chapter 2.

I welcome rigorous critical Feedback in particular on the sections below

Que onda creo que resolvi un contra ejemplo de la conjetura de collatz o también conocido como 3×+1 porcierto mi ortografía es horrible bueno el contra ejemplo del que hablo es el que dice si hay algún otro bucle además de 1 2 4 bueno primero vamos a ver por que sucede el bucle de 124 ocurre por que al multiplicar el 1×3 el uno crece por 3 y esta a nada de crear un de crear un número completamente divisible por 2 a lo que me refiero con esto es a un número ejemplo 4 que al dividirse por 2 una cantidad suficiente de veces 4÷2=2÷2=1 termine llegando a su número de origen osea 1 que es el número que es el numero al que multiplicamos por 3 y le sumamos 1 bueno dicho eso seguimos multiplicar 3×1 da 3 lo que significa que esta en un 75% de ser el numero divisible por 2 al que queremos llegar en este caso 4 osea que nos falta un 25% que justo el +1 termina siendo equivalente a 25% muchos pensaran este tipo me explicó algo que ya se y si esto lo saben casi todos pero la razón por la que expliqué eso es por que la única manera de que se pueda generar otro bucle es que pase exactamente igual que con el bucle 1 2 4 pero esto es imposible que pase por que al multiplicar cualquier numero impar por 3 ejemplo 3×3=9 lo dejamos a un 75% de llegar al número divisible clave para que se cree el bucle en este caso 12 nos faltaría sumar un número equivalente al 25% y el único número que podemos sumar es 1 que obviamente no equivale al 25% de hecho no equivale al 25% con absolutamente ningún número impar excepto con el 1 1×3=3+1=4÷2=2÷2=1 pero no se puede con el 3 3×3=9+1=10 ÷2=5 ni con ningun otro numero impar lo que termina pasando es que al sumar el 1 se termina creando un numero par que al dividirlo no nos lleva al número que fue multiplicado por 3 sino que nos da un número impar ejemplo 3×3=9+1=10÷2=5 como vemos aquí sumar 1 al 9 nos da dies que al dividirse da 5 osea no nos lleva al tres para que pueda ocurrir un bucle y encima con el cinco resultante ya que es un numero impar va a pasar lo mismo osea se va multiplicar por 3 le vamos a sumar uno no va a llegar al número divisible que tiene que llegar para que comience un bucle pero si va a dar otro número impar entonces esto se repite hasta llegar al bucle 1 2 4 me gustaría que opinan si ven alguna falla aunque si no entendieron algo pueden preguntarme para aclarar. Mis respetos para los que pudieron entender a la primera lo que escribí tengo sueño 😪

I'm not a researcher, but with some digging I found a couple papers posted this year that claim to solve the conjecture. They don't seem to be peer-reviewed yet, but no one has outright debunked them it seems.

So I thought I'd link them here to generate some discussion about them:

• A Structural Proof of the Collatz Conjecture

• On the Proof of the Collatz Conjecture

• Proof of the Collatz Conjecture

• Proof of the Collatz Conjecture

Many of us have independently discovered that a minimal counterexample to the Collatz conjecture would have to satisfy an increasingly stringent set of conditions, when we consider residue classes modulo 2^(k), for increasing values of k, because a larger and larger fraction of classes can be shown to "descend".

Just for example, we know that anything of the form 4k+1 "descends" to 3k+1 very quickly. Indeed, as long as k > 0. It's true that 4k+1 > 3k+1.

This kind of argument "lifts" to higher powers of 2, so for example, by the time we get to k=4, so we're looking through mod 16 lenses, the only remaining congruence classes are 7, 11, and 15. All of the others "descend". By the time we get to k=10, so we're looking through mod 1024 lenses, there are only 64 classes that aren't shown to "descend".

I saw another post about this kind of thing, and started thinking about how there are infinitely many non-descending trajectories under the Collatz map, when we apply it to rational numbers that are also 2-adic integers, i.e., rational numbers with odd denominators. Such numbers are in congruence classes mod 4, 8, 16, 32, etc., so... how does that work? Does every minimal cycle element over the rational numbers belong to classes that pass the sieves at every level?

Immediately, the thing to do seems to be to check some examples. Like, what about 19/5? It's a cycle min, so its trajectory never descends. Let's look at its 2-adic expansion:

19/5 = ....1100110011001100110111.

That four digit pattern keeps repeating to the left, and we could shorten it, using parentheses to indicate repetition, as:

19/5 = (0110)111.

Anyway, this makes its residues clear, modulo every power of 2:

19/5

≡ 1 (mod 2)
≡ 3 (mod 4)
≡ 7 (mod 8)
≡ 7 (mod 16)
≡ 23 (mod 32)

and so on. I stopped at 32 just now because that's where we have a question arise. When we test the expression 32k + 23, it seems to descend!

32k + 23
96k + 70
48k + 35
144k + 106
72k + 53
216k + 160
108k + 80
54k + 40
27k + 20

However, the number 19/5, despite being congruent to 23 (mod 32), doesn't descend! So what gives?

Let's consider the claim that the above chain is a "descent", anyway. Let n = 32k + 23, and let's compute:

32k + 23 > 27k + 20
↔ 5k > -3
↔ k > -3/5
↔ n > 32(-3/5) + 23 = (-96 + 115)/5 = 19/5

Well, how about that? If n ≡ 23 (mod 32) AND n > 19/5, then n's trajectory has to descend.

(Yes, in a way we just used the argument that 32k+23 descends as an alternative way to discover that there's a rational cycle with min element 19/5. Pretty cool, huh?)

What's your point, Gonzo?

Fair question, and I'm not sure what to do with this. What I'm saying though, with this post, is that there's a bit more going on, with these descent arguments, than we might think about at first. Each descent argument of this style looks like this:

(2^(W))k + a
...
...
...
(3^(L))k + b

where 3^(L) < 2^(W) and b < a.

(Exception for 4k+1 descending to 3k+1, in which case a = b, but we know that corresponds to descending as long as k > 0, which puts our starting number > 1.)

In fact, if we do our sieving properly, then 3^(L) is always the largest power of 3 smaller than 2^(W), or we would have caught this case for a smaller W.

So, if we now analyze the claim that this is a descent, we should obtain a condition on how large the original number has to be for that to be true:

n = (2^(W))k + a > (3^(L))k + b
↔ (2^(W) - 3^(L))k > b - a
↔ k > (b - a) / (2^(W) - 3^(L))
↔ n > (2^(W))(b - a)/(2^(W) - 3^(L)) + a = ((2^(W))b - (2^(W))a + (2^(W))a - (3^(L))a) / (2^(W) - 3^(L))
↔ n > ((2^(W))b - (3^(L))a) / (2^(W) - 3^(L))

That number there – ((2^(W))b - (3^(L))a) / (2^(W) - 3^(L)) – is the threshold where the descent argument kicks in.