r/Collatz

An underscore "_" indicates subscipt and 'a exp b" means a raised to power b

Consider the following function f : f(x_i+1) = [(x_i)3 + 1]/2 when x has a odd numerator in its simplest form (let's call this operation f1) f(x_i+1) = [(x_i)]/2 when x has an even numerator in its simplest form (let's call this operation f0) Repeated self-composition of the function f on an input x_0 results in the collatz orbit of x_0.

The domain of this function is the set R which contains all rational numbers of the form p/q such that gcd(p,q) = 1 & q is odd. (i.e. all rationals with an odd denominator in their simplest form. Those who have studied collatz conjecture in 2-adics will know that this is the most natural domain for collatz conjecture). The function f is contained in R.

Suppose we start with an input x_0 for the collatz orbit with subsequent terms being x_1, x_2, x_3, x_4 and so on.. to x_n; Let the parity vector of this sequence be P(x_0 to x_n) = p_0, p_1, p_2, p_3, p_4, p_5, till p_n-1. Here pi is parity of xi denoted as 1 for odd x_i and 0 for even x_i; the ith parity vector element pi indicates the operation f1 or f0 from x_i to x_i+1.

The diophantine reformulation is as follows: x_n = [(x_0)•(3 exp u) + S]/(2 exp n) - [let's call this equation 1 from now on]

Here u is the total number of times operation f1 is applied in the collatz orbit from x_0 to x_n; equivalently it is also the number of times 1 occurs in the parity vector string of x_0 to x_n. Here n is the total number of operations f0 and f1 combined from x_0 to x_n; equivalently it is also the total length of parity vector string of x_0 to x_n.

S = summation(u-1 to 0) OF (3 exp i) • (2 exp k_i); here k_i is the number of parity vector elements before the ith 1 value in the parity vector string such that k_i > k_(i+1). Thus a particular parity vector uniquely dictates a particular S, u and n.

(This part is maybe slightly difficult to wrap your head around, but once you try to derive this equation yourself, it becomes exceedingly obvious what I'm talking about. For proof of truth of the above equation you can look up the this paper on the internet: "On the existence of cycles of a particular length..... By Corrado Böhm & Giovanna Sontachhi" .)

Now suppose that after m steps x_0 falls into a cycle then x_m = x_(m+l) = x_(m+2l) = x_(m+cl) for any non negative integer c and some positive integer l. Let the length of P(x_0 to x_m) = m, and let the number of 1s in P(x_0 to x_m) be w Let the parity vector from x_m to x_m+l be denoted by P(x_m to x_m) which is the same thing as P[x_(m+cl) to x_{m+(c+1)l}] Here the length of P(x_m to x_m+l) is l, and let the number of 1s in P(x_m to x_m+l) be v.

Example: parity vector of 5 to 1 is 1000 m=4 and w=1 And then parity vector from 1 to 1 is 10 with l=2 and v=1 P(5 to 1) can also be written as 100010 or 10001010 and so on...

When we input the values from above example into our equation we get 1 = [(3 exp u)•5 + S]/(2 exp n) with values of u, n, S respectively which are dependent on the choice of parity vector from above options. Example: if parity vector taken as 1000 then u=w+v=1, n=m+l=4 and S accordingly. If parity vector taken as 100010 then u=w+v=2, n=m+l=4 and S accordingly. Each choice of parity vector produces different n,u,S values each of which satisfy equation 1. Now what happens if we use the parity vector 10001010101010... extending to infinity to solve equation 1? I will try to answer this

Now in the domain R, we can always find a cycle for any given parity vector; i.e we can always find a element r in the set R such that parity vector from r to r is a binary string of our choice. Thus the above example of 5 ending in 1 is example of 10 cycle but we can have any cycle 10011 or 1100011 or any binary string.

So the question more generally stated would be: what happens when we try to solve equation 1 for a rational with odd denominator x_0 which falls into a cycle (as all inputs tested yet do) while using a parity vector of x_0 whose length tends to infinity?

I will try to answer this by rearranging equation 1 in the following manner to get equation 2 (x_0) = [(2 exp n)(x_n) - S]/(3 exp u) equivalently (x_0) = [(x_n)•(2 exp n)/(3 exp u)] - [S/(3 exp u)]

Consider the above framework with x_m, c, u,v,w and n,l,m as defined above Substituting u=cv+w and n=cl+m in equation 2 we get (x_0) = [(x_m)•(2 exp (cl+m))/(3 exp (cv+w))] - [S/(3 exp (cv+w))] :equation 3

Now the smallest parity vector which satisfies above equation is when c=0 and when we consider the larger and larger equivalent versions of our parity vector we are effectively concatenating blocks of P(x_m to x_m) to the end of original parity vector P(x_0 to x_m). The number of times we concatenate a block of P(x_m to x_m) to P(x_0 to x_m) is the variable c. Let's call the resulting parity vector the net parity vector.

Now if we want to make the length of our net effective parity vector tend to infinity we need to increase c to infinity. Let's observe what's the result of increasing c in equation 3.

Case 1 Consider the case when the ratio l/v < log3/log2 then the first term in RHS tends to 0 as c tends to infinity. So equation 3 becomes (x_0) = - [S/(3 exp (cv+w))]

Since S = summation(0 to u-1) OF (3 exp i) • (2 exp k_i) the second term S/(3 exp (cv+w)) can be rearranged as the following Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i) such that k(u-1-i) < k(u-1-(i+1)). So x_0 = - Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i)

This forms a power series in 1/3 with increasing powers of 2 in the numerator. We know that this power series definitely converges because since l/v < log3/log2 so each subsequent blocks of P(x_m to x_m) contribute a geometrically decreasing amount to the total summation.

Case 2 Consider the case when the ratio l/v > log3/log2 then (2 exp (cl+m))/(3 exp (cv+w)) in first term of RHS tends to ∞ as c tends to infinity. Since S/(3 exp (cv+w)) = Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i) such that k(u-1-i) < k(u-1-(i+1)). So we substitute S/(3 exp (cv+w)) with Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i)

So equation 3 becomes x_0 = [(x_m)•(2 exp (∞))/(3 exp (∞))] - [Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i)] Now the problem is that the RHS does not converge in the linear metric sense and so x_0 = RHS becomes absurd. But if we consider the 2-adic metric then since first term has an infinite power of 2 as it's multiple it's 2-adic value becomes 0 and similarly since l/v > log3/log2, so Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i) converges in the 2 adic sense. This again gives us the equation as derived from case 1 which is x_0 = - Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i) : equation 4

This again forms a power series in 1/3 with increasing powers of 2 in the numerator but here since l/v > log3/log2 so each subsequent blocks of P(x_m to x_m) contribute a geometrically increasing amount to the total summation as seen in the linear metric but they converge in the 2-adic metric.

Now I have verified that equation 4 works for all elements in set R whether they fall in case 1 or case 2 category.

My question: How are we able to create a system in which the notion of convergence in both the linear metric and the 2-adic metric makes sense i.e. yields correct values as per equation 4. Because in a system usually one metric of convergence is possible. Am I missing something here or Have I made a mistake in the above formulation.

Ok, suppose an amateur got really close to a proof of the collatz conjecture. To the point where they can make many determinations, like given x then y and z are true. Connections that have been so far undefined. But they lack the education to formalize a proof.

What should they do? Attend courses to achieve the education necessary to formalize a proof but risk the chance that someone smarter discovers it as well and develops a proof sooner? Reach out to a professional in hopes of co-authoring a proof? Give their head a shake because they're likely misguided?

Say their discoveries were integral to uncovering the proof but they had a co-author. Would their contributions be overshadowed by the veteran author?

Thoughts?

Here is quite interesting statement that follows from the Collatz conjecture (actually equivalent, I believe, but I did not bother to prove). I like it because it is simple, and at the first glance - not obviously related to CC at all.

Consider two families of infinite-dimensional vectors (or just functions from ℕ to your favorite field):

https://preview.redd.it/w68mqjd1e3bh1.png?width=1424&format=png&auto=webp&s=7c06f5ffb97772fdb730d989deabcdb6378d4f10

Then, for every Collatz cycle, there are subsets of vectors a and b, whose sum is equal:

aᵢ₁+aᵢ₂+...+aᵢₘ = bⱼ₁+bⱼ₂+...+bⱼₙ

One such subset is easy to notice with a naked eye: a₁ = b₁. As you could guess, it corresponds to the trivial cycle 1-4-2. Then, if we could prove that there are no other subsets with equal sum property, the no-cycles part of the Collatz conjecture would immediately follow.

It is quite obvious that vectors in the b list are somewhat more "sparse" than vectors in the a list, so it is tempting to try to find some reason why on sum of *a'*s could be equal to some sum of b's, however, they are not that easy.

For example, look at those two sums: a₄₁+a₆₈+a₃₄+a₁₇ and b₈+b₁₂+b₁₈+b₂₇+b₂₀+b₃₀+b₄₅ (edit: added forgotten term)

Here, I plotted them:

https://preview.redd.it/hxv2u2gff3bh1.png?width=1709&format=png&auto=webp&s=4721686bd436895916146dcbb617d30ffd31592b

Almost the same, aren't they? They would be the same, if we shift sum of b's by 2 to the right. This near-miss actually corresponds to the cycle of the 3x-1 system that starts with 17. In general, cycles of the 3x+P systems produce sums that are equal up to shift by P-1.

How was this constructed? Actually, quite simple: by mapping integers to vectors: 1→[0000...], 2→[1000...], 3→[1100...], 4→[1110...] and so on, and then mapping edges of the shortcut Collatz graph to their differences. I can write in more details if anyone is interested, though it is quite trivial.

We can now make a purely geometrical statement from which the no-cycles part of the CC follows. Consider two cones, formed by positive linear combinations of vectors (a₂, a₃, ...) and (b₂, b₃, ...). If these two cones do not intersect (except for the apex), then there are no high cycles in the Collatz system.

I think it's neat.

Proof of Collatz conjecture

Abstract

We give an elementary proof of the Collatz function T (x) = 3x + 1 [x odd], x/2 [x even],

always to take the value x = 1 after a finite number of iterations. The problem is led to

initial values 27 + 96k, and 91 + 96k, with parameters k = k(l).

Keywords: Collatz Problem, Recurrences, Sequences of Integers

2010 Mathematics Subject Classifications: 11B37; 11B83; 11-XX

https://zenodo.org/records/20709339

A^(n) + B^(n) ‡ C^(n) for all positive integers A,B,C and n, were n>2

A,B,C must be relatively prime for a non trivial solution to any such equation, if any integer solution exists. That requires that one and only one of those bases has at least one factor of 2.

For any two smaller objects (A & B) of order n to be equal in quantity to that of a larger n ordered object (C), the smaller objects when contained within the larger (on the longest line between 2 most distant vertices of the largest object and each of the smaller objects oriented in congruence with those opposite vertices of the largest object) must OVERLAP in a union, the n order quantity of that union (O^(n)) (this happens to be the minimum amount of union possible ) will equal the nth order quantity of the largest object that is disjoint from both of the smaller objects.

It becomes useful to express:

A=a+O

B=b+O

C=a+b+O O being the linear Overlap or C-(a+b)

When A^(n) shares a vertex and orientation with C^(n) and B^(n) is also located within C^(n) at the most distance vertex of C^(n) from that of A^(n) then it could be expressed that the overlap or union O^(n) would be required to equal the disjointed expressed by the binomial expansion of the(a + b)^(n) minus (a^(n)+b^(n).)

For n=2, O^(2)=2ab

For n=3, O^(3)= 3(a^(2)b) + 3a(b^(2)

For n=5, O^(5)=5(a4)b+10(a^(3)(b^2)+10(a^(2)(b^3)+5a(b^(4)

Because of the content of any such algebraic expansion of O^(n) (of particular interest were n is prime) O will be a even quantity and require either the factor a or b to be even (but not both).

That even integer would be the only source of factors of 2 for O^(n) and be required to contain at least n factors of 2. As a result O^(n) would contain 2n factors of 2 requiring that same even integer to also contain 2n factors of 2.

In short, A^(n) + B^(n) ‡ C^(n) under the conditions stated because the union of the two smaller objects when they are contained within the larger object can not have integer equivalence (the same number of factors of 2) to that which is disjoint

I have referenced this paper from a previous post, but since the paper now folds in insights from two of my recent posts, I think it is worthy of its own post.

The key result is the recognition that n mod 5 nodes allow the construction of an overlay tree over the reverse Collatz tree where there interior nodes consists only of n mod 5 nodes that the "left" branches contain a forward path from a mutliple of 3 to the n mod 5 node that is OEEE+-free (e.g. does not contain n mod 5 node except at the end).

This structure allows one to make this strong (but presently unproven) claim about any pair. of nodes along the left branch:

e >= o. log_2(3) <=> a >= b

This is possible PRECISELY because such branches do not have 5 mod 8 nodes.

What remains unproven (though almost certainly is true) is that each 5 mod 8 reaches a 0 mod 3 node via its left branch.

update: I have had to withdraw most (but not all) of the theorems in the paper because I realised the proofs didn't stack up. I have replaced the theorems with conjectures and left some implication theorems that trace out the logical implication framework should the conjectures eventually be shown to be true. I would welcome any counter examples for conjectures/postulates as currently posed.

Here is the correct image.

I think this is true:

This regex:

(((7*3)?(1|5)))*

Matches the n mod 8 representation of all "complete" odd Collatz orbits. (An orbit is "complete" if the last term is the last odd term of a Steiner circuit - that is, it ends with either a 1 or a 5)

For example, if you consider all the odd numbers in the sequence from 27 to 1 and calculate n mod 8 for each value, then you get this sequence:

317773113173177315731317777357735115573551

That sequence is matched by the regex above.

Some notes:

- each repetition of the outer group is a complete Steiner circuit
- a string of 7's is always followed by a 3 (these correspond to the leading OE terms of a Steiner circuit)
- a 7 is always followed by a 7 or 3
- a 3 is always followed by a 1 or 5
- a 1 corresponds to an OEE term
- 1 or 5 can be followed by any other term
- a 5 corresponds to an OEEE+ term where EEE+ is at least 3 even terms
- 7 and 5 have symmetrical roles - 7's permit arbitrary number of leading OE terms in each Steiner circuit, 5's permit arbitrary number of trailing E terms in each Steiner circuit.
- 3 identifies the last OEO term in a Steiner circuit.

All of this, of course, is a consequence of action of the Collatz map on the mod 8 arithmetic of sequence elements.

Another property of this representation that relates to the recent discussions in this place about the frequency of paths that pass through 5.

All odd terms between 2 5's have the same 3.k leaf - that is, if you take closest reverse branch for each of these terms, then it is guaranteed that each 3,k leaf will be the same - the 3k leaf changes at each 5 term.

What this means is that the reverse path between a node and 0 mod 3 leaf node can be represented as an unbalanced tree where the left branch is a 3k-leaf node and the right branch is another tree of the same structure.

1
|
5
/ \
3 53
/ \
15 61
/ \
81 325
/ \
303 3077
/ \
159 2429
/ \
111 445
/
27

I've written this up in a paper

update: corrected * -> ? per comments below paper will be corrected in due course.

it's really laggy because its graphing the count of how long it takes and the max number from every number from 1-100000. it automatically goes through a random number anywhere between those with 10 going at a time and its just fun to watch i hope you all like it!

यह किसी भी संख्या (छोटी या बड़ी) को 9 से गुणा करने की एक बेहद अनोखी और बिल्कुल नई ट्रिक है। यह पारंपरिक तरीकों से अलग और पूरी तरह से तार्किक है।

📌 मूल सिद्धांत (Core Principle):-

गणित की भाषा में इस ट्रिक का समीकरण इस प्रकार काम करता है:

Formula = (N*9=N9) -(N+9)

📋 स्टेप-बाय-स्टेप फॉर्मेट (Step-by-Step Format):-

किसी भी संख्या को 9 से गुणा करने के लिए इन 3 आसान स्टेप्स का पालन करें:

स्टेप 1: जिस संख्या को 9 से गुणा करना है, उसके दाहिनी (Right) ओर अंक 9 लिख दें। (इससे आपको पहली नई संख्या मिलेगी)।

स्टेप 2: अपनी मूल संख्या में 9 जोड़ (Add) दें। (इससे आपको दूसरी नई संख्या मिलेगी)।

स्टेप 3 (आखिरी कदम): स्टेप 1 से मिली संख्या में से स्टेप 2 से मिली संख्या को घटा (Minus) दें। आपका सटीक उत्तर आपके सामने होगा।

🧪 लाइव उदाहरण (Examples):-

उदाहरण 1: 4-अंकों की संख्या (1516 × 9)

स्टेप 1: 1516 के आगे 9 लिखें → 15169

स्टेप 2: 1516 में 9 जोड़ें → 1516 + 9 = 1525

स्टेप 3: दोनों को घटाएं → 15169 - 1525 = 13644 (सही उत्तर: 13644)

उदाहरण 2: 2-अंकों की संख्या (25 × 9)

स्टेप 1: 25 के आगे 9 लिखें → 259

स्टेप 2: 25 में 9 जोड़ें → 25 + 9 = 34

स्टेप 3: दोनों को घटाएं → 259 - 34 = 225 (सही उत्तर: 225)

उदाहरण 3: बड़ी संख्या (8432 × 9)

स्टेप 1: 8432 के आगे 9 लिखें → 84329

स्टेप 2: 8432 में 9 जोड़ें → 8432 + 9 = 8441

स्टेप 3: दोनों को घटाएं → 84329 - 8441 = 75888 (सही उत्तर: 75888)

🌟 इस ट्रिक की सबसे बड़ी खासियतइस ट्रिक की सबसे मजेदार बात यह है कि स्टेप 3 में घटाते समय, आपको हमेशा आखिरी अंक 9 में से ही घटाना होता है (क्योंकि स्टेप 1 में अंत में हमेशा 9 रहेगा)। इससे हासिल (Borrow) लेने का झंझट बहुत कम हो जाता है और गणना तेज होती है।

Bonjour à tous,

Je ne suis pas un mathématicien universitaire, mais un passionné de logique. Aujourd'hui, le 26 juin 2026, j'ai complété la cartographie logique d'un plan de preuve complet pour la Conjecture de Collatz (3x+1).

Afin de protéger mon travail, un dépôt officiel daté par courriel a déjà été effectué auprès de mes proches avant la publication de ce post.

Mon approche ne cherche pas à tester des nombres, mais analyse la dynamique structurelle sous un angle entièrement nouveau. Mon modèle repose sur 4 piliers :

1. Une réduction condensée qui transforme le système en une symétrie probabiliste parfaite avec un facteur de contraction géométrique constant.

2. L'analyse des "nombres à forte croissance", démontrant par une formule précise que leur montée exponentielle s'autodétruit en générant une falaise arithmétique (la chute est proportionnelle à la montée).

3. Une preuve d'ergodicité : l'application répétée du système agit comme un mélangeur chaotique (modulo), détruisant la mémoire et la structure des blocs de chiffres supérieurs.

4. Un système de relais spatiaux qui ramène tout nombre géant dans une zone de sécurité numérique déjà validée par ordinateur.

La logique globale et l'architecture de la preuve sont fluides et complètes. Cependant, je n'ai pas le jargon académique ni la maîtrise des outils matriciels ou des chaînes de Markov pour rédiger l'article rigoureux en LaTeX nécessaire pour soumettre le projet à l'Institut Clay.

Je cherche un coauteur (étudiant en master, doctorant ou chercheur en théorie des nombres) pour s'associer avec moi. Si la formalisation mathématique tient la route face à vos connaissances, nous rédigerons l'article ensemble et partagerons équitablement les crédits et le prix.

Si vous êtes ouvert d'esprit et que le défi vous intéresse, contactez-moi en DM (message privé) pour que nous puissions en discuter. Curieux et sceptiques bienvenus, tant que la discussion reste respectueuse !

Because of the 3x role in Collatz rise steps, it's useful to look at the relationship between powers of 3 and powers of 2.

Specifically, even powers of 3 (e) and odd powers of 3 (o) are always near multiples of 8 (and higher powers of 2):

3^e-1 = 8x = x*2^3 (and x*2^4, x*2^5, x*2^6, etc. for higher even powers of 3)

3^o-3 = 8x = x*2^3 (and x*2^4, x*2^5, x*2^6, etc. for higher odd powers of 3)

(Examples and proof below)

I think this may be useful when thinking about how upper terms of Collatz values change after N rise steps.

For instance, we know that a Collatz sequence starting at 2^H*C+L (where L is odd) rises to 3^H*C+Cz(L) after H steps (where Cz(L) is Collatz steps applied to lower term L).

So, any sequence that rises an even number of steps will have an upper term 3^E which by the fact given above will be (8x+1)*C. Sometimes that will be 16x, 32x, 128x.

That's more clearly interesting in binary. If you start with a number:
CCCC00000LLLL
It will rise to:
XXXCCCMMMM where MMM is Collatz expansion of L
That means the lower three binary digits (maybe more) of C are untouched after H rises.
So, even though it seems like multiple 3x+1 steps completely scramble the upper bits of the starting number, some bits are preserved.

Its easy enough to prove that these near powers of 3 are always 8x.

If 3^o-3 = 8y for any odd o (which is true for 3^3 - 3 = 24)
then: 9*(8y+3)-3 = 72y + 27 - 3 = 72y + 24 = 8*(9y+3)
so: 3^(o+2)-3 is also a multiple of 8 and by induction all 3^o-3 = 8x

If 3^e-1 = 8y for any even e (which is true for 3^2 - 1 = 8)
then: 9*(8y+1)-1 = 72y + 9 - 1 = 72y + 8 = 8*(9y+1)
so: 3^(e+2)-1 is also a multiple of 8 and by induction all 3^e-1 = 8x

what i mean is simply this consider a potential cycle that has n odd steps (3x+1) and m even steps(/2).

Have any bounds ever been proven limiting the value for m in relation to n in order for a cycle to exist?

\A= -1, 1, -1, -1, 1, 1, 1, 1, -1, 1, -1, -1, -1, -1, -1, -1, -1, 1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, -1, -1, -1, -1, -1, 1, -1, -1, 1, 1, -1, -1, -1, 1, -1, 1, -1, 1, -1, -1, -1, 1, -1, 1, 1, 1, 1, -1, 1, -1, 1, 1, 1, -1, -1, 1, -1, -1, 1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, -1, 1, 1, -1, -1, 1, -1, -1, -1, -1, 1, -1, 1, -1, -1, -1, 1, -1, 1, -1, -1, 1, 1, 1, 1, 1, 1, 1, -1, 1, -1, 1, -1, 1, -1, -1, 1, -1, 1, 1, 1, -1, -1, -1, 1, 1, 1, 1, 1, 1, 1, -1, 1, 1, -1, -1, -1, 1, 1, -1, 1, 1, 1, -1, -1, -1, 1, -1, 1, 1, -1

[ A B C D ]

[ -B A -D C ]

[ -C D A -B ]

[ -D -C B A ]

This structure

B= a backwards

C= a inverse + -

D= a both

Solution by matthew bailey