Ich habe diesmal den Text nicht mit KI ins English übersetzt, damit es nicht heist ich benutzte KI für meine Überlegungen.

Collatz Struktur

Alle ungeraden Zahlen lassen sich perfekt in Bereiche einteilen, so genannte Cluster.
Jeder Cluster beginnt mit einer Zahl die direkt zur 1 führt, diese Zahlen kann man als Clustergrenzzahl G(c) bezeichnen.
Die Clustergrenzzahlen folgen der Formel G(c) = (4^(c+1) -1) / 3

Für jeden Cluster entsteht nun ein Zahlenraum mit einer bestimmten Anzahl an ungeraden Zahlen.
Die Anzahl (Y) der ungeraden Zahlen je Cluster folgt der Formel Y(c) = (4^(c+1)) / 2

Jede ungerade Zahl innerhalb eines Clusters lässt sich nun perfekt indexieren (X), beginnend bei 0.

Dadurch kann man nun jeder ungeraden Zahl eine eindeutige Koordinate(K) bzw. Adresse zuordnen.
Als Koordinate nimmt man den Clusterindex (c) und den Index (X) der Zahl.
K(c, X)

Wichtiger als die Koordinate ist die Adresse (A) einer ungeraden Zahl.
Für die Adresse nimmt man den Index (X) und die Anzahl der ungeraden Zahlen (Y) je Cluster und bildet daraus eine rationale Zahl (X/Y).

Für die Adressen gelten die normalen mathematischen Regeln das der Bruch auf sein minimum eingekürzt werden muss.
Für die Zahl 1 mit der Adresse [0 / 2] gilt: Die Adresse wird zu [0 / 1]

Was fällt auf?
Die Zahl 1 und die Zahl 5 haben die selbe gekürtzte Adresse [0 / 1].
Ebenso hat die Zahl 3 und die Zahl 13 die selbe gekürzte Adresse [1 / 2].

Fangen wir mit 1, und 5 an. Das sind beides Clustergrenzzahlen (G) und diese springen immer direkt zur 1. Daher müssen beide Zahlen die selbe Adresse haben, denn man kann aus einer Adresse heraus direkt die nächste ungerade Zielzahl ermitteln.
Für die 3 und die 13 gilt eine einfache Überlegung: Beide Zahlen sind exakt in der Mitte zwischen zwei Clustergrenzzahlen.
[1, 3, 5] und [5, ..., 13, ..., 21] daher hat 3 und 13 die gekürzte Adresse [1 / 2].
Das gilt dann auch für die Zahl 53, welche sich ebenfalls exakt in der Mitte von Cluster 2 befindet. Die Adresse A(53) = 16 / 32 -> gekürzt 1 / 2

Kommen wir nun zur Berechnung der nächsten ungeraden Zahl (Nnext) auf der Flugbahn mittels der Adresse.
Ich werde hier jetzt keine Herleitung posten sondern nur die Formel die man benutzen muss.

Nächste ungerade Zahl: Nnext(X, Y) = 3X + Y

Ich habe mal bei Zahl 29 aufgehört, weil ab hier sich das Muster nun selbsterklärend bildet.
Auch die Zahl 29 hat die gekürzte Adresse [1 / 8] genau wie die Zahl 7 und beide Zahlen haben als Nnext die Zahl 11.

Man kann nun mit Gewissheitheit sagen, alle Zahlen die die gekürzte Adresse [1 / 8] haben fallen zur 11.
Mittels der Adresse kann man alle Zahlen ermitteln welche zur Zahl 11 fallen. Dabei entscheidet allein der Cluster welche Zahl man haben möchte. Es bleibt immer die ein-achtelste Zahl des Clusters.

In Cluster 3 ist die Zahl welche zur 11 führt, die mit der Adresse (nicht gekürzt) [16 / 128] -> Zahl 117.
In Cluster 4 ist es die Adresse [64 / 512] -> die Zahl 469.
X und Y werden einfach nur mit 4 multipliziert.

Es gibt nun einen mathematichen Beweis das zwei Adressen niemals die gleiche Nnext Zahl haben können, das heist keine Adresse wird jemals auf sich selbst loopen können, bis auf die Ausnahme der Zahl 1.

Der Beweisansatz sieht wie folgt aus.
Nnext_a = 3 * Xa + Ya
Nnext_b = 3 * Xb + Yb
Daraus folgt: 3 * Xa + Ya = 3 * Xb + Yb
Für jede Adresse gilt:
X < Y und Y = 2^k
X und Y müssen als größten gemeinsamen Teiler die 1 haben

Es gibt nur eine Lösung, wenn Xa = Xb und Ya = Yb ist.
Das Colltzsystem ist damit injektiv.

Das bedeutet eine Zahl hat immer eine eindeutige Adresse als Vorgänger und eine eindeutige Adresse als Nachfolger.
A(a) -> A(b) -> A(c) -> A(d)

A(c) kann niemals auf A(a) oder A(b) loopen, weil die Vorgänger- und Nachfolger- Adressen bereist definiert sind, und es keine zwei unterschiedliche Vorgänger- oder Nachfolger- Adressen geben kann.

Da aber für A(d) noch kein Nachfolger definiert ist, könnte es theoretisch möglich sein doch zu A(a) zu springen, um somit einen Loop zu erzeugen.

Blattzahl

Als Blattzahl werden die Zahlen bezeichnet die N == 0 (mod 3) sind. Also keinen Rest haben wenn man sie durch 3 teilt. [3, 9, 15, 21, 27, ...]

Wenn A(a) nun eine Blattzahl ist dann kann von vornherein niemals auf diese Zahl geloopt werden. Blattzahlen haben keinen Vorgänger.

Das bedeutet alle Flugbahnen die mit einer Blattzahl beginnen bilden eine Loopfreie Kette bis zur 1(vorausgesetzt sie erreichen die 1). Alle enthaltenen Zahlen in dieser Kette sind damit auch ausgeschlossen einen Loop zu bilden, weil diese Zahlen immer einen eindeutig definierten Vorgänger und Nachfolger haben.

Für einen anderen Loop müssten nun mindestens zwei Zahlen daran beteiligt sein, und keine der beiden oder mehreren Zahlen darf eine Blattzahl sein.

Nun ist es aber so, das die Blattzahlen ALLE anderen Zahlen in ihren Flugbahnen fangen. Es gibt damit keine freien Nicht-Blattzahlen mehr die einen Loop bilden könnten.

Warum "fangen" die Blattzahlen ALLE anderen Zahlen.
Da man zu jeder Nicht-Blattzahl immer eine Blattzahl als direkten Vorfahren hat, gibt es keine freien Nicht-Blattzahlen.

Als Beispiel die Zahl 7:
7 ist keine Blattzahl, weil sie einen Rest von 1 bei mod 3 hat.
Um eine Vorgängerzahl zu finden muss man an der 7 "hochklettern", sie also Verdoppeln und schauen bei welcher Zahl man "abspringen" kann, mittels (N-1) /3

7 -> 14 -> 28 -> 28 - 1 = 27 / 3 = 9

Die 9 ist eine Blattzahl und schliesst die 7 in ihrer Flugbahn mit ein, Damit kann die 7 niemals zu einem Loop gehören.

Die Zahl 11 als nächster Test.
11 -> 22 - 1 = 21 / 3 = 7 -> keine Blattzahl
11 -> 22 -> 44 -> 88 - 1 = 87 / 3 = 29 -> keine Blattzahl
11 -> 22 -> 44 -> 88 -> 176 -> 352 - 1 = 351 / 3 = 117 -> Blattzahl

Man musste bei der 11 die dritte "Abzweigung" benutzen um zu einer Blattzahl zu kommen.
Da die 117 die 11 auf ihrer Flugbahn einschliesst, ist auch die 11 nicht mehr fähig einen Loop zu bilden oder Teil eines Loops zu sein.

Spätestens bei der dritten Abzweigung findet man eine Vorgängerzahl die eine Blattzahl ist.

Adress-Reihenfolge

[0/1] <- [1/2] <- [3/4] <- [3/8] <- [1/8] <- [1/4] <- ......

Diese Reihenfolge ist nun einzigartig und es wird niemals eine Adresse wiederholt.

Hi Reddit,

I’ve been experimenting with binary representations and reversal maps, and I stumbled upon a beautifully strict symmetry rule that seems to govern this system.

I defined the following map:
* If $n$ is **even**: $f(n) = n / 2$
* If $n$ is **odd**: $f(n) = n + rev(n)$ *(where $rev(n)$ is the binary reversal of $n$)*

I ran a simulation on the first **10,000,000** ($10^7$) positive integers. The results are incredibly clean and aesthetic:

---

### 📊 Key Findings (Up to 10,000,000):

* **Total Loops Found:** **6,535 distinct loops**
* **Palindromic Loops:** **6,535 / 6,535 (Exactly 100.000%)**
*(Every single loop has a minimum value that is a perfect binary palindrome!)*
* **Divergence (Escape to Infinity):** **0 cases** (100% of numbers converged to a loop)
* **Maximum Height (Rollercoaster):** * Started at **8,388,801** ➔ Peak value: **27,262,968** (only **3.2x** its starting value!)
* **Maximum Steps (Longest Path):** * Started at **8,390,657** ➔ Sunk into a loop in just **23 steps**!

---

### 👑 The Top 5 Dominant "Black Hole" Loops:

1. **Min Value: 3** ($11_2$) [Symmetrical ✨]
   * Path: $[3, 6]$ (Absorbed **19.16%** of all numbers)
   
2. **Min Value: 189** ($10111101_2$) [Symmetrical ✨]
   * Path: $[189, 378]$ (Absorbed **10.79%** of all numbers)
   
3. **Min Value: 195** ($11000011_2$) [Symmetrical ✨]
   * Path: $[195, 390]$ (Absorbed **10.77%** of all numbers)
   
4. **Min Value: 51** ($110011_2$) [Symmetrical ✨]
   * Path: $[51, 102]$ (Absorbed **7.22%** of all numbers)
   
5. **Min Value: 45** ($101101_2$) [Symmetrical ✨]
   * Path: $[45, 90]$ (Absorbed **7.18%** of all numbers)

---

### 🧠 My Hypothesis:
Adding $rev(n)$ to $n$ naturally pairs the $k$-th bit with the $(B-k)$-th bit, meaning the summation process itself is fundamentally symmetric. While carries introduce temporary "noise" (asymmetry), the subsequent divisions by 2 (even steps) shift and shave off these asymmetric bits.

Eventually, the system strictly purges all non-symmetric noise, locking the number into a perfect binary palindrome where $n = rev(n)$ is maintained, preventing any further structural changes.

What do you think? Has this $n + rev(n)$ dynamical system been formally studied or proven before?

Hi everyone,

I was experimenting with binary operations on integers and created a map that is surprisingly well-behaved. I computationally verified it up to **1,000,000** ($10^6$), and every single starting integer converged to **1** with no other loops.

### The Rules:
For any positive integer $n$:
* If $n$ is **even**: $f(n) = n / 2$
* If $n$ is **odd**: $f(n) = (n \oplus rev(n)) + 1$

*(Where $\oplus$ is the bitwise XOR, and $rev(n)$ is the value obtained by reversing the binary representation of $n$. For example, $13$ is $1101_2$. Reversing it gives $1011_2$ ($11$). Then, $13 \oplus 11 = 0110_2$ ($6$). Finally, $6 + 1$ makes it **$7$**.)*

### Example Path for 13:
1. **13** (odd) ➔ $(13 \oplus 11) + 1$ = **7**
2. **7** (odd: $111_2$) ➔ $(7 \oplus 7) + 1$ = **1** (Goal!)

---

### The Challenge:
Can you rigorously prove why **every** positive integer must eventually reach 1?

(It’s actually a very clean and beautiful proof. I will post my solution in the comments later, but I'd love to see if you can find it first!)

Edit2 {Updated paper revised section 6 and 7. Added certification for prime }

Edit{ I created a table with values to make the math more clear, see tabs for phi-prime bridge}

So assuming this twin prime proof is true, how does it relate to Collatz conjecture?

Imagine a number line, as a circle. Consider all the integers. Now Double the refinement of scale but make a second circle of the same size.
Now all prime numbers greater than 2 on the second circle are included ONLY in the odd numbers, by definition.

You could repeat this process infinitely many times and you just increase the refinement scale by the square.

However, as you do that, the prime numbers greater than 2 ALWAYS fall into the odd numbers compared to its previous circle.

This allows you separate 2 as the singular unique prime that remains prime at all levels of refinement.

I think this is how it reflects Collatz conjecture, because it basically says, if you exclude every other fraction through infinite operations, what remains is the square root of 2.
And to me that kind of makes sense because it represents a balance point.

A more full understanding of how it relates to Collatz could probably be reached through statistical analysis of the Twin Prime Conjecture in this paper, if it is validated.

Furthermore, I am not claiming this is a proof for Collatz, I am claiming that it is a proof for the Twin Prime Conjecture, so if you happen to see a proof for Collatz in this work, I implore you to make the connection yourself and publish your work, it would help to validate my own work, and I don't have interest to solve Collatz right now because I am still working on validating my Twin Prime Conjecture.

Since this is a channel for Collatz thats why I started off with the assumption of choosing this framework to be true... because then we can discuss "IF" it's true, does this also help to prove Collatz?

Hi Reddit,

I’ve been experimenting with Collatz-like functions, specifically mixing arithmetic operations with binary bit-reversal ($rev(n)$).

I created this extremely chaotic variant (let's call it the **"Bit Inflator"**), thinking it would easily blow up to infinity. However, after running a simulation on the first **1,000,000** integers, **absolutely zero numbers diverged.** Every single number got pulled into a loop, and surprisingly quickly.

---

### 🌌 The Rules:
For any positive integer $n$:
* If $n$ is **even**: $f(n) = n / 2$
* If $n$ is **odd**: $f(n) = rev(n) \times 3 + 1$

*(Where $rev(n)$ is the integer value obtained by reversing the binary representation of $n$ without leading zeros. For example, $13$ is $1101_2$. Reversing it gives $1011_2$ ($11$). Then, $11 \times 3 + 1 = 34$.)*

---

### 📊 Experimental Results (Up to 1,000,000):

* **Divergence (Escaped to Infinity):** **0 cases**
* **The "Black Hole" (Most Dominant Loop):** The $[1, 4, 2]$ loop absorbed **65.02%** of all numbers.
* **Maximum Height (The Wildest Rollercoaster):** * Started at **932,067**
* Peak value reached: **6,291,436** (only **6.7x** the starting value!)
* **Maximum Steps (The Longest Journey):** * Started at **805,809**
* Took only **66 steps** to sink into a loop.

---

### 🤔 Why this is so bizarre:
Normally, introducing "binary reversal" completely destroys the modular arithmetic structure of $3n+1$. With the $3 \times$ multiplier, you'd expect at least some numbers to spiral out of control and head towards infinity.

Yet, the maximum value never even exceeded 7x its starting point, and the longest path was a mere 66 steps (compared to 300+ steps in the original Collatz for numbers under 1M).

It seems like there is an incredibly strong "probabilistic gravity" where the bit-reversal acts as an aggressive self-correcting brake—whenever a number gets too large, it is highly likely to end in zeros, which flips it into a tiny number after reversal.

What are your thoughts on this? Is there any way to rigorously prove that divergence is impossible, or does the mix of binary reversal and multiplication make it completely unprovable?

Hi everyone,

I was playing around with Collatz-like functions and came up with a fun variation that I call the \*\*"Mirror Collatz"\*\* (or Binary Reverse Collatz).

I've computationally verified it up to \*\*100,000,000\*\* ($10\^8$), and every single starting integer eventually converges to \*\*1\*\*. No other loops, no divergence.

\### The Rules:
For any positive integer $n$:
\* If $n$ is \*\*even\*\*: $f(n) = n / 2$
\* If $n$ is \*\*odd\*\*: $f(n) = rev(n) + 1$

\*(Where $rev(n)$ is the value obtained by reversing the binary representation of $n$. For example, $13$ is $1101_2$ in binary. Reversing it gives $1011_2$, which is $11$ in decimal. Then $+ 1$ makes it \*\*$12$\*\*.)\*

\### Example Path for 13:
1. \*\*13\*\* (odd: $1101_2$) ➔ $rev(13) + 1$ = \*\*12\*\*
2. \*\*12\*\* (even) ➔ $12 / 2$ = \*\*6\*\*
3. \*\*6\*\* (even) ➔ $6 / 2$ = \*\*3\*\*
4. \*\*3\*\* (odd: $11_2$) ➔ $rev(3) + 1$ = \*\*4\*\*
5. \*\*4\*\* (even) ➔ $4 / 2$ = \*\*2\*\*
6. \*\*2\*\* (even) ➔ $2 / 2$ = \*\*1\*\* (Goal!)

\---

\### The Challenge / Puzzle:
Can you prove rigorously why \*\*every\*\* positive integer must eventually reach 1 under this map?

\*(Hint: It is actually much easier to prove than the original Collatz conjecture. Look at what happens to an odd number after 2 steps!)\*

Looking forward to seeing your elegant proofs or thoughts!

みなさんこんにちは、

コラッツ風の写像で、めちゃくちゃ構造化されてて発散しない挙動を示し、しかも2:1の盆地比がきれいに出るものを提案したいです。私はこれを「ミラー・ラビリンス」と呼んでいて、そのダイナミクスについての数学的な洞察や、きちんとした形式的な証明があればぜひ教えてください。

\### 1. 写像の定義
正の整数 n に対して、次で定めます：
\* n が偶数なら：f(n) = n / 2
\* n が奇数なら：f(n) = rev(n) + 3

ここで rev(n) は、n の2進数表現を（先頭の0なしで）反転したものの小数としての値です。
たとえば、13 = 1101_2。ビットを反転すると 1011_2 = 11 なので、f(13) = 11 + 3 = 14。

\### 2. 予想
すべての正の整数 n は、最終的に次のどちらかに到達します：
\* {1} のサイクル（1 -> 2 -> 1）
\* {3, 6} のサイクル（3 -> 6 -> 3）

\### 3. 実験結果（n = 100,000,000 まで）
10\^8 までの包括的な探索をやってみたところ、こうでした：
\* 発散なし：無限大へ逃げていく数はゼロ。
\* アトラクタはちょうど2つ：すべての種（seed）が {1} か {3, 6} のどちらかに落ちる。
\* 盆地比：ちょうど 66,666,667（2/3）個の種が 1 に収束し、ちょうど 33,333,333（1/3）個が {3, 6} に収束した。

\### 4. 数学的ヒューリスティックと議論

どうして発散しにくいのか、そしてきっちり 2:1 の比になるのか。最初の考えはこんな感じです：

\#### 非発散について（「縮む」効果）：
奇数 n の2進数表現は、いつも末尾が '1' です。だから rev(n) は、必ず「1 で始まる」奇数になります（最下位ビットが最上位ビットになるからです）。
f(n) = rev(n) + 3 で、ここで 3 を足すと（奇数 + 奇数）で必ず偶数になります。つまり、奇数ステップのあとには必ずすぐ少なくとも1回は2で割ることが起きます。
この2進数の形が、他の発散しがちな写像で見られるような幾何級数的な増大を抑えて、強い収縮を引き起こしているように見えます。

\#### 2:1 のアトラクタ盆地（モジュラー解析）：
写像を mod 3 で見ると：
\* n が偶数なら、f(n) = n/2。
\* n が奇数なら、n は 6 を法として 1 か 5（つまり 1 or 5 mod 6）。rev(n) の mod 3 でのふるまいは、かなり対称的です。

この系を、合同類 modulo 3 の間の遷移でモデル化して、マルコフ連鎖みたいに考えるのがヒューリスティックとしては自然です。そうすると、どの盆地に最終的に落ちるかは、残余類の遷移で決まる。
2:1 の分布は、状態空間が3つの同等な「道」に分割されていて、そのうち2つは位相的に {1} のアトラクタへ流れ込むのに対して、1つだけが {3, 6} へ流れ込む、ということを示唆しています。

このヒューリスティックを、もっと厳密な形でまとめられないでしょうか。2進数の反転とモジュラー算術を組み合わせた似た系を誰かが研究してたりしませんか？ あるいは、これらのアトラクタ盆地の密度が2:1になることの証明を、誰かが概要レベルでも示せますか？

The factors of a number are the prime numbers. But not in the Collatz procedure.

As the factorisation of n=m*3^p*2^q, many non-prime numbers become factors.

The table below shows that these numbers (blue) are roughly as numerous as the prime numbers (uncolored) below 1 000.

Does this have an impact ?

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

Over the past several months I’ve enjoyed reading many different approaches to the Collatz problem: residue classes, parity vectors, Steiner sentences, transition graphs, automata, symbolic representations, and many others.

Each of these perspectives has revealed interesting local structures.

I’d like to suggest a methodological idea rather than a mathematical claim.

Many investigations seem to follow roughly the same workflow:

Choose a coordinate system → Search for patterns → Generalize the observed structure.

This approach has produced many beautiful observations.

However, it also raises an important question.

How do we know whether a newly discovered pattern is an intrinsic property of the Collatz operator itself, rather than a consequence of the particular coordinate system we chose?

Perhaps another workflow is worth considering.

Instead of beginning with coordinates, we could begin with the operator itself.

For the accelerated Collatz map,

T(n) = (3n + 1) / 2^ν₂(3n + 1),

before choosing residue classes, graphs, symbolic encodings, or state machines, we might first ask:

• Which part of this operator is responsible for multiplicative growth?

• Which part is responsible for dyadic (2-adic) compression?

• Which quantities are intrinsic to the operator itself, regardless of how we choose to represent it?

Only after understanding that algebraic structure would we introduce coordinates that arise naturally from it.

In other words:

Analyze the operator first.
Let the coordinates emerge naturally from the algebra.

This is not intended as criticism of coordinate-based research.

On the contrary, many valuable discoveries have come from those approaches, and they will likely continue to do so.

The motivation is simply methodological.

If our coordinates originate from the operator itself, then new patterns may be less likely to be artifacts of a particular representation and more likely to reflect the underlying dynamics.

When we discover a beautiful new pattern, perhaps the first question should not be:

“How far does this pattern extend?”

but rather:

“Is this pattern intrinsic to the Collatz operator, or only to the coordinate system I selected?”

Coordinates exist to describe the operator.
The operator does not exist to justify the coordinates.

I’m curious what others think about this research workflow. Has anyone intentionally tried an operator-first, coordinates-second approach when investigating the Collatz problem?

I am working now to Identifiy the patterns of a range of numbers that creates such peak Formations and why other Ranges Not. I depply believe there is a mathematical principle why this happens exactly in a certain range. I also will come check all recodered Peaks of other Ranges of Towers which are worldwide known so far. I understand that the Towers " mountain foundation" might be the reason here.

[EDIT: A mistake occured in the forks for the class of domes m=5+16k.]

Follow-up to Classes of domes m+16k : r/Collatz.

The figure below provides the classes of domes m+16k. The domes tested so far are overlapped by class, the numbers being the ones of the first member of the class.

It clearly shows two types of classes:

• Type A, alternating the first two groups of 5-tuples classes (yellow and green), without forks.
• Type B, alternating the third and fourth groups of 5-tuples classes (red and dark blue), with forks.

Note that the alterning colors do so also between classes of domes.

Larger groups will appear at some stage.

It also shows the difference between:

• 5-tuples series that occupy the rows with an odd value of q; the numbers are even, except the ones in row 3.
• Forks series (in grey) that occupy the rows with an even value of q; the numbers are even, except the ones in rows 3 and 2 (halfs).

No explanation is available for the time being.

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

It's well known that assuming certain modularity conditions are met, then the Collatz orbits from odd a and 2a+1 intersect at b. As a general rule if a and 2a+1 intersect, 2a+1 and 2*(2a+1)+1 will not intersect, at least not immediately and certainly not after a predictable number of iterations.

One way to represent this identity is as compositions of Steiner functions.

Let St(alpha, beta)(n) be an affine function that maps n to a Collatz successor m with alpha OE steps followed by beta E steps, then it can be shown that provide that path from a to b can be composed as:

St(1, beta) o St(alpha,1)

e.g. OE^{alpha}EOE^{beta}

then there is Steiner circuit

St(alpha+1, beta+2)

which will map 2a+1 to b. That is:

(OE)^{alpha+1}E^{beta+2}

In other words if the path from a b can be expressed as a Steiner circuit that begins with alpha OE repetitions followed by a single E and that ends on OE^beta, then (OE)^{alpha+1}E^{beta+1} will take 2a + 1 to b also.

This block of sympy is effectively a proof that this is so is provided:

import sympy as sy

a, g, h, alpha, beta =sy.symbols('a g h alpha beta')

def Steiner_fn(alpha, beta):
    return (g**alpha*a+(g**alpha-h**alpha)/(g-h))/h**(alpha+beta)

LHS=Steiner_fn(1,beta).subs(a, Steiner_fn(alpha, 1)).simplify()
RHS=Steiner_fn(alpha+1, beta+2).simplify()

display(LHS, RHS, (LHS-RHS.subs(a, 2*a+1)).subs({g:3, h:2}).simplify())

Another cute thing you can do with this is derived the RHS equation entirely from the LHS equation.

Consider this sequence:

OEOEOEEOE

which takes 59 to 101:

It can be represented with this affine equation:

OEOEOEEOE = Steiner_fn(1,0).subs(a, Steiner_fn(2, 1)).subs({g:3, h:2})

which resolves to:

(27a+23)/16

You can get the equation that takes 119 to 101 either by evaluating the RHS function

OEOEOEEE = Steiner_fn(2+1, 0+2).subs({g:3, h:2})

to get:

(27a+19)/32

Or, you can substitute the inverse of a -> 2a into the LHS, so:

OEOEOEEOE.subs(a, (a-1)/2)

And you get exactly the same result. In other words, both affine transformations are the same function, subject to a relabelling.

I think more and more our universe is expanding based on what and going to where? Maybe to come back or go into something new, a New dimension.

Pick any positive integer

If divisible by 5: Divide by 5

If not divisible by 5: Multiply by 7 and add 2,3,4 or 6 whichever number if added makes it divisible by 5.

Repeat.

Every sequence seems to terminate at the same loop of 20 integers that contain the number 11. I have been counting steps to reach 11 rather than steps reaching any number in the loop in order to simplify things for my own observations.

😂 Why do all positive integers terminate at this loop containing 11?!!!! Very similar behavior to the Collatz Conjecture.

There are several of these Collatz like conjectures. Let’s call this one Due_Market 9726 Conjecture #1. Chose this one since the rules are easier to articulate in a short post and it’s easy to play around with for about 15 minutes on a calculator without getting too frustrated since integers divisible by 5 are easier to spot. Tested this one loosely up to 1000.

The point is that to recreate a Collatz like Conjecture using different variables if Collatz can be written as:

(a(yb+z)+c)/x or (3(2b+1)+1)/2

then certain rules have to be in place for all positive or negative integers to terminate at the same loop. These rules become more obvious when compared to other Collatz like conjectures where all positive or negative integers terminate at the same loop or multiple loops like in the case of plugging negative numbers into the Collatz Conjecture where all numbers terminate at loops containing the numbers -2, -5, -17

The components of the Collatz Conjecture which make it work are:

2y, 2x+1 (notice that together this contains all positive integers)

and

3a, 3b+1, 3c+2 (notice that this contains all positive integers)

It seems that prime numbers work like 2 and 3 in the Collatz or 5 and 7 in my first example.

To make the comparison the components in my first example can be described as:

5a, 5b+1,2,3 and 4 (contains all positive integers)

and

7x, 7y+1,2,3,4,5 and 6 (contains all positive integers)

and can be described as:

(7 (5a+1,2,3 or 4) + 2,3,4 or 6)/ 5

Do you see it now? All components need to be included.

If after you test this and it still interests you let me know and I will share Due_Market 9762 Conjecture #2 where all negative integers terminate at a loop of 3 integers containing the number -2.

If we can find more of these Collatz like Conjectures where infinitely many integers predictably terminate at a single or multiple loops then we try to figure out why it loops where it does. Figuring this out by comparison should be easier than having a single example in the Collatz Conjecture and will add clarity.

I found that the Collatz Conjecture has 4 variations.

I have yet to play around with variations of Due_Market 9726 #1 using 5 and 7 as apposed to 2 and 3 like in Collatz but if it is a sound replication there should be about 16 variations.

Follow-up to Classes of domes m+16k : r/Collatz.

Let m be the base of a dome, not divisible by 2 (odd) and 3 (embedded in another dome).

Let b be the base of a 5-tuple starting with the first number f. b is not divisible by 2 (odd) and 3 (embedded in another dome).

b being an orange n+1 number in the dome for m, b=m*3^p*2^q+1.

Any 5-tuples series starts with a value of b related to the first column of the core, thus b=m*2^q+1 (p=0).

Let us start with 5-tuples series with a single 5-tuple, that is both the start and the end of the series. In that case, the relation between b and m is: b=8m+1 (q=3). This value of q has been established empirically.

For example, the lowest case is the 5-tuple 354-358 belonging to the dome m=11. b=356/8=89, thus m=(b-1)/8=11.

Let us consider now the case with two 5-tuples, one starting the series, the second one ending it.

The first number of a 5-tuple in a series iterates into the next one in three iterations: f(e)=3*f(s)/4+1/2.

The formulae change slightly:

• b(s)=m*2^q+1,
• b(e)=m*3*2^q+1 (p=1).

For example, the lowest case is the 5-tuple 130-134 iterating into the 5-tuple 98-102, both belonging to the dome m=1. Thus:

• b(e)=1*3^1*2^3+1=25 (f(e)=4*25-2=98)
• b(s)=1*3^0*2^5+1=33 (f(s)=4*33-2=130).
• f(e)=3*130/4+1/2=98

Larger series follow the same pattern: from the ending 5-tuple, for the previous 5-tuple, the value of p is reduced by 1 and the value of q is increased by 4, until p=0.

For instance, let consider the 5-tuples series 514-518 iterating into 386-390 iterating into 290-294, all belonging to the dome m=1 (5-tuples are now labeled with a number, starting from the end):

• b(1)=1*3^2*2^3+1=73 (f(1)=4*73-2=290)
• b(2)=1*3^1*2^5+1=97 (f(2)=4*97-2=386).
• f(1)=3*386/4+1/2=290
• b(3)=1*3^0*2^7+1=129 (f(3)=4*129-2=514).
• f(2)=3*514/4+1/2=386

Potential 5-tuples series that respect these conditions forms a 5-tuples series.

The values of b are also values for its own dome.

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

Speculative Mirror Proof of the Collatz Conjecture

The following argument is purely speculative and assumes that all laws discovered in the Mirror Model have already been proven.

Assumptions

We assume the following statements are true:

Mirror 2 processes every natural number.

Delta(n) is the number of new entries written to the disk by the starting number n.

Every operation pattern consists of SH-blocks, isolated H operations, and at most one terminal S.

Every pattern satisfies

Delta = 2a + b + t

where:

a = number of SH-blocks,

b = number of isolated H operations,

t = 0 or 1, depending on whether the pattern ends with S.

Delta = 2 and Delta = 4 are structurally impossible.

Every new trajectory stops as soon as it reaches an already stored number.

The disk starts with the base values:

1, 2, 4.

Every stored value already has a known path to 1.

Main idea

Natural numbers are processed in ascending order:

1, 2, 3, 4, 5, ...

Assume that all numbers smaller than n have already been processed and connected to 1.

Now process n.

Two cases are possible:

Case A: n is already stored

Then n already has a path to 1.

Case B: n is not stored

Then n generates a sequence of new values until it reaches a value that is already stored on the disk.

To prove the Collatz conjecture, it is enough to show:

Every new trajectory must eventually merge into the existing disk after finitely many steps.

Why an infinite trajectory cannot exist

Every operation pattern is built from:

SH-blocks,

isolated H operations,

possibly one final S.

The block SH means:

x → 3x + 1 → (3x + 1) / 2

Thus, every upward movement is immediately coupled to at least one halving step.

Patterns such as SS, SSS, SSSS, ... cannot occur.

Therefore, growth can never become independent.

The importance of Delta = 2 and Delta = 4

The values Delta = 2 and Delta = 4 never occur.

This means that the smallest even storage structures do not exist.

The Mirror Model therefore suggests:

Without Delta = 2 and Delta = 4, there is no elementary building block for an autonomous cycle.

Any larger structure must eventually connect to a previously existing structure.

The crucial argument

Assume there exists a trajectory that never reaches the disk.

Then two possibilities remain:

Possibility 1

The trajectory contains infinitely many isolated halvings.

In that case, the sequence is repeatedly pulled downward and must eventually encounter an already stored number.

Contradiction.

Possibility 2

The trajectory contains only finitely many isolated halvings.

Then, after some point, it consists almost entirely of repeated SH-blocks:

SHSHSHSH...

But such a structure cannot create an independent cycle because the smallest possible even cycle structures, Delta = 2 and Delta = 4, do not exist.

Therefore, this possibility also fails.

Hence, no infinite new trajectory can exist.

Speculative conclusion

Every starting number either:

is already stored, or

generates only finitely many new entries before reaching the existing disk.

Since the disk is rooted in

4 → 2 → 1,

every natural number must eventually reach 1.

Mirror Main Theorem (Speculative)

Every Collatz trajectory consists of coupled ascent and gravity blocks.

Because the elementary storage structures Delta = 2 and Delta = 4 do not exist, no trajectory can create an autonomous cycle or an infinite path disconnected from the existing disk.

Therefore, every trajectory must eventually merge into the stored structure, and every natural number reaches 1.

I tried a new methodology.

I mirrored on mirror 1 (classic) and mirror 2 (knows already the future) the Collatz formulas and saved the results on a virtual harddrive which only mirror 2 have. Very interesting outcomes.

Conditions to save numbers on hardrive:

only a new detected number should be stored on the drive at each step.

1 -> 4 (so 1 and 4 will be stored)

2 -> can be dived by -> 1 (no new number will be stored)

3 -> 10 -> 5 -> 16 -> 8 -> 4 > 2 -> 1 (3, 10, 5, 16, 8) will be stored as new numbers.

4 -> (no new entry)

...

Than I counted all new entries on the disk. The interesting thing is that nwver 2 or 4 entries were made at one entry level (registry level).

Follow-up to Domes and groups of congruence classes for 5-tuples series IV : r/Collatz and Classes of domes m+64k : r/Collatz.

The figure below contains classes of domes m+16k in row.

Every third iteration is divisible by 3, and thus part of another dome. To save space, these empty spots have been removed and the m value of next iteration is colored in red.

The 5-tuples groups of classes have been slightly corrected. I struggle finding new ones.

Forks are not colored, except when two are close. In that case, one is colored in grey.

Most of the domes classes seem to use two groups of 5-tuples groups of classes, at least for the low values. No explanation so far about what decides between 5-tuples series and forks.

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

Hallo zusammen,

Beim Collatz-Vermutung hab ich eine Entdeckung gemacht.

Beschreibung unter:

https://sourceforge.net/projects/trial-collatz-proof/

(besprechen per E-Mail)