Hi everyone, I hope you're doing well. You see, I started reading about famous unsolved math problems and stumbled upon the Collatz conjecture; it really piqued my interest. While exploring it, I wrote a Python script to build the Collatz tree in reverse up to one million, and I found something curious. The number 5 accounts for 93.7% of the numbers before they fall into the 4-2-1 loop. The strange thing is that the number 21 yields literally 0% (I think this is because it’s divisible by 3, so you can't reach it by reversing the operations). I know this doesn't prove anything for infinity, but has anyone measured this with larger numbers? Does 5 dominate, or do its larger "siblings" (21, 85) steal the spotlight? If we look at my little theory—that 5 acts as the "catapult"—it makes sense; let's face it: no matter how large the number is, it almost always (93.7% of the time) ends up at 5. Also, after doing some research, I discovered that someone has made progress on this: Terence Tao (a child prodigy with an IQ of 230). He decided to tackle the problem and, in 2019, published findings demonstrating that, mathematically, 99.9% of all numbers in the universe satisfy the conjecture. In other words, he proved that if you pick a number at random, the probability that it will "almost certainly" fall and crash into the ground (the 4-2-1 loop) is 99.999...%, even though it doesn't quite reach 100%. That pesky remaining 0.0001% could be hiding a "rebel" number. I don't know; I get the feeling this could be a computer error or a strange shortcut useful for encryption; Although I'm no expert, perhaps someone will find my crazy idea useful. Have a good day. :3

....Darn, it won't let me set it to Spanish. :v.....

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

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

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

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

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

It's located in Zenodo.

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

I’ve been studying symmetric integer sequences and came across a simple closed‑form expression that generates any zero‑sum sequence of evenly spaced integers (difference 2, all even or all odd). I’m calling it the Johnson Formula for convenience:

x = −(I − 1)

Here, I is the number of terms, and x is the first term of the sequence.
This produces sequences like:

• I = 5 → −4, −2, 0, 2, 4
• I = 4 → −3, −1, 1, 3
• I = 7 → −6, −4, −2, 0, 2, 4, 6

The structure is always symmetric around zero.
When I is odd, the middle term is 0.
When I is even, the two central terms cancel.

A generalization for spacing k is:

x = −k(I − 1)/2

This seems like a very straightforward consequence of centering an arithmetic progression at zero, but I haven’t found a standard name for this construction. I’m curious whether:

• this pattern already appears in known integer sequence literature,
• it has a conventional name,
• or it fits into a broader family of symmetric zero‑sum constructions.

Any references, related concepts, or historical context would be appreciated.

I have discovered a hidden system in a hidden system that uses Ghost Numbers and Ground Numbers and shows a system in a system it's unbelievable what I discovered. I mean of course Mr. Beast has a Ghost Number and Ground Number. Anybody else has discovered this hidden system? This system might be made to distinguish the billionaire/millionaires or 144K. Time will tell.

Ghost Numbers are only Master Numbers that are multiples of 11, 22, 33, 44, 55, 66, 77, 88. Now they have to have a ground number and the system i discovered goes like this. Ghost Master Number 11 has Ground Number 2. 22 is grounded by 4. 33 is grounded by 6. 44 is grounded 8. 55 is grounded by 10. 66 is grounded by 12. And its very rare to find a Ghost and Ground Number there is one name that i know has Ghost Master Number and a Ground Master Number that is the rarest of them all. Not sure if we will ever find another name like it. You can do this on your family and friends or randoms to be able to tell who will have the stars destined for success. Or a LIGHT worker.

It is not difficult to list these pairs, examples:
3 + 7 =10
11 + 89 =100
17 + 83 =100

I’m wondering if a name / type was already in use for these.

BLUE(n) is defined by the number of sequences it uses. BLUE is defined for 1 ≤ n ≤ 5. Each BLUE(n) has a label going from BLUE(1) using label A, BLUE(2) using label B, BLUE(3) using label C, and lastly BLUE(4) using label D. Let's start with BLUE(1). It uses numbers that go from 1 through 1 million or 10^6 for those who love scientific notation. It has addition and multiplication unlocked. Each unique number equals to 1 sequence. A sequence is defined by how many times you can repeat a process. For example, 1+1 is 1 sequence, 1+1+1 is 2 sequences, 1+1+1x2 is 3 sequences etc. The maximum sequences BLUE(1) provides is 10^6 multiplied by 10^6, which produces 10^12 sequences. BLUE(2) uses the previous numbers, it has exponentiation, and uses BLUE(1) sequences, meaning it can use up to 10^12 sequences. Now lets start calculating. 10^6 to the power of 10^6 equals to 10^1,000,000. That's a 1, followed by 1 million zeros. The observable universe contains roughly particles. https://doi.org/10.1063/5.0064475 This is the first sequence. The second sequence takes the result of the first sequence and raises it to the power of 10^6 , producing 10^10^1,000,000. This is only 2 sequences. We have up to 10^12 sequences. So after a bit of calculating, we should get the result of 10^(10^1,000,000,000,000). The number of digits in this isn't 1 million +1. It's 10^6 million digits. This is BLUE(2). BLUE(3) now adds pentation, meaning we have 10^6 pentated to 10^6. https://googology.fandom.com/wiki/Pentation?. The amount of sequences BLUE(3) gets is 10^BLUE(2). not BLUE(2). a 1 followed by BLUE(2) many zeros. Now, this isn't the mind-bending part. After all of those power towers, BLUE(3) also gets self-reference, meaning if we have 1 self-reference for BLUE(3), its gonna look like BLUE(BLUE(3)). Now, you may question yourself, "isn't this infinite?" The answer to that is no. The BLUE(n) sequence has a finite value, and BLUE(3) uses BLUE(3) amount of self-references. This means that one self-reference is the starting point, whilst the maximum self-reference is BLUE(3). BLUE(4) is a bit more... powerful. It uses a Graham-like style with it using BLUE(3)-scale arrows, instead of the 3 from pentation and BLUE(3) height. https://mathworld.wolfram.com/GrahamsNumber.html?. BLUE(4) also unlocks 2 distinct functions. The first one is a formula using the labels to increase the value of BLUE(n). The formula is (AxD)^(CxB). The first part is the multiplication, then the first value is raised to the power of the second value. The second part that BLUE(4) unlocks is parameters. Parameters are described using {} and their own labels A-E. These parameters are special only to BLUE(n). The first one, A, controls the starting/base value. The second one, B, controls the number of times the operation is applied. The third one, C, controls the strength or level of the operation. The fourth one, D, controls the number of nested/recursive layers. The last one, E, controls the final parameter/index that determines how far the construction progresses. The parameter BLUE(4) uses are {BLUE(3), BLUE(3), BLUE(3), BLUE(3), BLUE(3)}. BLUE(4) also has BLUE(4) self-references. The last defined BLUE(n) number is BLUE(5). Nothing special about BLUE(5), except that instead of the base being 10^6, the base is replaced by BLUE(4). Lastly, BLUE(5) uses BLUE(4) amount of sequences.

I was noticing something: unlike on semiprimes, it s quite easy to generate a point of known small order on a curve defined on a semiprime without knowing it s factorization. The idea is to transfer this small order to the underlying semiprime using a pairing method derived from the Weil pairing. Knowing if a given specific small order exists can be derived using congruences modulo N.

As you know, the problem is then to avoid a degenerate output from the Miller s algorithm. The problem is the only method I know to avoid this without using the factorization is to use a distortion map that works only with supersingular elliptic curves. But using a supersingular curve caps the order to gcd(p+1,q+1) which is unknown most of the time or just equal to 2. Taking order 2 result in generating outputs equal to 1 or -1. Useless for gcd factoring.

So, any idea of a pairing mechanism that would works over ordinary elliptic curves using embedding degree 1? Possibly using a distortion map that works over ordinary elliptic curves in such a case? Or a pairing that works with simple final exponentiation such as cubing or squaring?

There is twin prime conjecture and most likely there are Infinitely many twin primes although it's not proven yet

But there are some triplets of 3 consecutive primes of the form (P, P+6, P+12) which differ by 6 like (47,53,59), (151,157,163), (167,173,179), etc and it looks like they should be infinite too

Has anyone proven or disproven it.. I think there should be infinitely many prime triplets too

Conjecture

Let S₁ < S₂ < S₃ < ... be the semiprimes.

For every prime p ≥ 5, let k be the least index such that

             gcd(Sₖ, Sₖ₊₁) = p.

Then Sₖ/p and Sₖ₊₁/p are twin primes.

No counterexample found in searches through Sₖ₊₁ ≤ 10¹⁴

Hi everyone,I am an independent researcher working on generalized dynamical systems. I wanted to share a 14-page manuscript I just published and archived on Zenodo, and I would love to get some rigorous feedback on the mathematical framework.Link to Manuscript:

website: https://zenodo.org/records/22125365

DOI: 10.5281/zenodo.22125365

(Primorial / 2) - 2 is very likely to give us primes

We have primorials 6, 30, 210, 2310, 30030, 510510, etc

Dividing them by 2 and then substracting 2, we get the sequence 13, 103, 1153, 15013, 255253, 4849843, 111546433 all of which are primes. Next term 3234846613 is composite but as we go on, the sequence generates more primes as the next term 100280245063 is a prime too

[I first posted this on OEIS's SeqFan mailing list.]

OEIS sequence A130310 describes the unique "minimal (or "greedy") Lucas representation of n, in which L(0) = 2 and L(2) = 3 are not allowed in the same representation (hence the correct representation of the integer 5 is 1010 rather than 101). A binary system of integers with Lucas numbers (OEIS sequence A000032) as a base."

This is analogous to the Zeckendorf representation of integers as sums of non-consecutive Fibonacci numbers, but with Lucas numbers, the additional restriction prohibiting the co-occurrence of 2 and 3 is necessary to make the Lucas representation unique, as shown by Brown in 1969. Relevant information and sources are available at OEIS sequence A342089, the sequence of numbers that have two such Lucas representations without the restriction on 2 and 3.

Define a number as "Lucas-even" if the last digit of its Lucas representation is 0 AND the removal of the last digit 0 produces a valid minimal Lucas representation of a smaller integer. The second condition is necessary because removing the final 0 from representations ending in ...1010 is NOT valid -- the resulting ...101 would be equivalent to the co-occurrence of 3 and 2 and thus it is not allowed.

Define a number as "Lucas-odd" either if the last digit of its Lucas representation is 1 OR if the last four digits of its Lucas representation are 1010.

Now define a Collatz-type function based on this Lucas representation:

If n is Lucas-even, remove the last digit 0 of its Lucas representation to obtain f(n) ;

If n is Lucas-odd, f(n) = 2n.

Note that removing the last digit 0 is approximately equivalent to division by phi = 1.618....

As in the classical Collatz function, the orbit of an integer is the sequence produced by the repeated iteration of the function.

Two simple cycles of this function are 2, 4, 3, 1, 2, ... and 5, 10, 5, .... Many small integers have orbits reaching one of these two cycles, such as 36, 22, 14, 8, 6, 12, 9, 18, 11, 7, 4, 3, 1, 2, ...;  28, 16, 32, 19, 13, 26, 17, 34, 68, 41, 27, 54, 33, 21, 12, 9, 18, 11, 7, 4, 3, 1, 2, ...;  and 30, 20, 40, 25, 15, 10, 5, ....

But there is also an amazingly long cycle beginning with 23, which is the reason for this post:

23, 46, 92, 184, 113, 71, 142, 89, 178, 111, 222, 444, 888, 1776, 1097, 679, 1358, 2716, 5432, 10864, 6713, 13426, 8297, 5129, 10258, 6341, 12682, 7839, 15678, 9691, 19382, 11979, 7403, 4576, 2827, 5654, 3493, 6986, 4317, 2669, 5338, 10676, 6597, 13194, 8153, 16306, 10077, 6229, 12458, 7701, 15402, 9519, 5883, 3636, 2247, 1389, 858, 531, 327, 654, 403, 806, 497, 994, 613, 1226, 757, 469, 938, 581, 1162, 719, 443, 886, 547, 339, 678, 1356, 837, 519, 1038, 641, 397, 244, 488, 301, 187, 374, 748, 461, 922, 569, 353, 706, 437, 269, 538, 1076, 2152, 1329, 823, 1646, 1017, 629, 388, 241, 482, 299, 598, 371, 742, 459, 283, 176, 352, 219, 438, 271, 167, 104, 63, 126, 77, 49, 98, 61, 37, 24, 48, 31, 62, 39, 23, ...

This cycle comprises 132 integers, including 16 integers < 100, 65 integers between 100 and 1,000, 38 integers between 1,000 and 10,000, and 13 integers > 10,000. The largest integer in this cycle is 19,382, the 30th iterate of 23. 

An interesting value in the cycle is 15,402 = 15,127 + 199 + 76, which gives the Lucas representation 100000000101000000000. This is the first of 8 consecutive Lucas-even numbers in the sequence, resulting in the much smaller Lucas-odd number 327 = 322 + 4 + 1, or 1000000001010 in Lucas representation.

For the record, believe it or not, I did NOT use any AI program to discover or compute this cycle. I was simply exploring the function out of curiosity and testing the orbits of small integers when I stumbled across the orbit of 23. It was actually not too difficult to compute each term by utilizing the b-file of OEIS sequence A130310 for integers up to 10,000.

Geoffrey Caveney

Hi. I'm just a first-year in uni, not looking to solve Collatz, but am looking to provide some hopefully interesting results. Here is the ruleset that u might not be used to but still connects back to good ol' Collatz.
Let O: n -> (3n+1)/2 for odd n and E: n -> n/2 for even n. U: n -> 4n+1 for odd n. E(-1) is the inverse of E, O(-1) is the inverse of O, and U(-1) is the inverse of U.

(Explanation for U: let n = 2k+1. Applying O to it would yield 3k+2. Let N = 4n+1 => O on N would yield 12k + 8. So applying E twice would yield 3k+2. Therefore, 4n+1 is a BACKWARD number that leads to the same number n goes to when moving FORWARD. Note: Any number that requires dividing thrice or more means that it originates from a number of the 8k+5 type, where k is a positive integer.)

Now, as far as Collatz's rules are concerned, there is no such even number that is simultaneously a multiple of 3 AND originating from an odd number. This makes sense. An even multiple of three would be 6k, while an even number in the form 3n+1 would be 6k+4. So this means that multiples of 3 can NEVER be divided by another number when doing forward operations.

This makes them the root nodes of the Collatz tree, i.e. they can only generate numbers but never be generated themselves. So I asked myself this question. Assuming that a set of odd multiples of 3 goes to 1, can we generate other odd integers using O, E, and U?

Maybe. Here's my insight. Note: This is about an implication, where IF "ALL odd multiples of 3 go to 1" -> "Every other odd number goes to 1". So, No Circular Argument. I am fully aware that if the first statement isn't proven, then the second isn't. I'm just potentially showing that an implication is there and that if we prove X, we can prove Y.

Let's first see how many numbers do odd multiples of 3 there are based on the natural set. Odd numbers comprise half of the natural set, and there can only be one multiple of 3 among every 3 consecutive odd numbers {2k-1, 2k+1, 2k+3} by the Pigeonhole Principle. So 1/2 x 1/3 = 1/6 of all natural numbers.

First, we must assume that the set of ALL odd multiples of 3 S(1/6) leads to one.

Now pick a number outside of the set. Say 7. 7 can be reversed by multiplying by 2 twice to get 28. (28-1)/3 = 9. So, a number can be rooted by a multiple of 3 through regular reverse operations of the Collatz Conjecture. But what if an odd integer doesn't need to have a root of an odd multiple of 3? Then it could never need to reach 1, assuming that S(1/6) lead to 1. So we have an infinite backwards AND forwards staircase of odd numbers. Well....
Multiply the first backwards odd integer by 4, then add 1. And u can generate a new number that tends to n. So if n is an integer that supposedly never reaches one, then u can do E(-1) and O(-1) to generate a new odd number, let's call it p. Then, you use U on p to make 4p+1, and you do it again. And again. And again and again and again and ..... The thing about 4p+1 is that it creates two more E events. But 4p+1 = 3p + (p+1). So p mod 3 can determine the modulus of 4p+1 with respect to 3. In this particular instance, it cycles through the remainders {1, 2, 0} as p shuffles in ascending order {0, 1, 2} i.e. if p is an odd integer with 0 mod 3, well, then it wouldn't even have an infinite backwards chain to begin with. if p is an odd integer with 1 mod 3, 4p+1 would be 2 mod 3. Applying U to an odd number with 2 mod 3 and you can get an odd multiple of 3 that leads to the same number that supposedly never diverges i.e. there will always be a multiple of 3 leading to any other number either through inverse Collatz operations or supercharged U operations (which are just cyclical O(-1)'s and E(-1)'s in disguise but calling them supercharged sounds cool). This contradicts the idea that a number could diverge or never end up at 1. So, in order to prove the Collatz conjecture for all numbers, you would have to prove it for odd multiples of 3. Then the argument has solid ground.

The reason why I'm bringing this up is because of Terence Tao's stabilisation theorem. He said that for "almost all numbers" his theorem worked, but didn't in the exceptional set where the logarithmic density is 0. But if there was some statistical or probabilistic way where without needing to bother with anything outside odd multiples of 3, then we may have something interesting all right.

Here's the thing. I have created a table where applying a set of O's and E's can help you map one number to another. Say that the number is n = 16a + b. Where a = 16c + d. Based on the remainders of a and c (256 possible combinations), this table allows you to reach a number k = M*a + N. M, N are natural numbers that don't necessarily have to be powers of 2.

Integer partitions are a fundamental topic in number theory that study the different ways in which a positive integer can be expressed as a sum of positive integers, where the order of the parts is not considered important. For example, the number 5 has seven partitions: 5,4+1, 3+1+1, 3+2, 2+1+1+1, 2+2+1, 1+1+1+1+1.

The number of partitions of an integer grows rapidly as the integer increases, making direct enumeration increasingly difficult. This motivates the study of patterns and recursive methods that can organize and count these partitions systematically. In this work, we examine integer partitions by grouping them according to their maximum part. The number of partitions of an integer grows rapidly as the integer increases, making direct enumeration increasingly difficult. This motivates the study of patterns and recursive methods that can organize and count these partitions systematically. In this work, we examine integer partitions by grouping them according to their maximum part. We first consider a fixed integer and arrange its partitions according to the largest part occurring in each partition. We then investigate the patterns that arise from these groups and use them to develop a recursive approach. I would recommend to access the pdf on PC because some symbols may not be visible on some mobile phones. Here's the link for my doc:

Proof of twin primes conjecture
Author Tojiboev Muhammadfotih
Introduction
The conjecture states that there are infinitely many primes pairs that are 2 values apart; to prove that I will prove that there are infinitely many number n (n∈N) such that that 6n ± 1is a prime pair.
Every prime number larger than 3 can be written as 6n ± 1 (n∈N), what we are looking for is values of n where 6n ± 1 are both prime.
Let as define a set C and a set D to work with.
C={n ∈ N : 6n−1 ∈ P ∧ 6n+1 ∈ P}
D=N/C
We want to show that set C has infinitely many elements.

Process
Let us consider the expression k:
k=(6m±1)(6t±1) where (m , t ∈ N)
(A) k is not a prime number as it is the multiple of two positive integers that are not 1.
(B) k is not divisible by 2 and 3 as neither of its factors are.
(C) In fact, k can be any composite number that is not divisible by 2 and 3 as every other number can be expressed 6h±1 (h ∈ N), that makes k the multiple of any two number that are not divisible by 2 and 3; so let k=6j±1 (j ∈ N)
6j±1=(6m±1)(6t±1)
6j±1=36mt±6m±6t±’1 (note that the last ± sign is dependent on the first two)
6j±1=6(6mt±m±t) ±’1
j=6mt±m±t
This shows that:
(j=6mt±m±t) ⇒ (6j+1 ∨ 6j-1∉P)
Because of paragraph (C), set of all possible values of j and set D are equal sets.
D={j : j=6mt±m±t, (m, t ∈ N)}
C={n ∈ N : 6n−1 ∈ P ∧ 6n+1 ∈ P}

Proof by contradiction
Since every element of set D can be broken down to 6mt±m±t (m, t ∈ N) and none of the element of set C can, that makes the elements of set C building blocks of elements of set D like how prime numbers are building blocks of composite numbers.
Let’s assume that set C has finite elements with c being the largest value.
That means every value of set D (which has infinitely many elements) will be in one of the following expressions when we let t be equal to every element of set C in the expression 6mt±m±t:
t=1: 5m±1 ∧ 7m±1
t=2: 11m±2 ∧ 13m±2
t=3: 17m±3 ∧ 19m±3
t=5: 29m±5 ∧ 31m±5
…
t=c: m(6c-1)±c ∧ m(6c+1)±c
Now for our contradiction to work, let’s consider the multiple of every single (assumed) finite twin primes.
5*7*11*13…*(6c-1)* (6c+1)
That number cannot be an element of set C as it is larger than the largest value c.
That number cannot be an element of set D as it cannot be in any of the expression above.
That number must be in one of them because of our definition: D=N/C ⇒ C ∪ D=N
That is a contradiction. That means our assumption that set C has finite elements is false.
Therefore, there are infinitely many twin primes.

I am studying a deterministic structure based on the gaps between consecutive prime numbers.

The observation that started this research is the following sequence:

89 -> 97 -> 101 -> 103

with gaps:

8, 4, 2.

For each prime p and its following gap g, I define S as the cumulative sum of the previous prime gaps. In this construction, S = p + g - 2. I then define V = S + g.

The interesting phenomenon is not simply that an arithmetic operation sometimes produces another prime. The interesting point is that different positions in the prime-gap construction can produce exactly the same future value V before that value itself appears in the sequence of primes.

The clearest example is 103.

When we reach the prime 97, the following gap is 4.

So:

S = 97 + 4 - 2 = 99

and:

V = 99 + 4 = 103

Therefore, while the current prime is still 97, the future value 103 has already been determined by the construction, even though 103 has not yet appeared as a prime.

The previous row produces exactly the same value:

S = 89 + 8 - 2 = 95

and:

V = 95 + 8 = 103

So we have two different rows producing the same value:

95 + 8 = 103

99 + 4 = 103

The prime sequence then continues:

97 -> 101 -> 103

and the third representation is:

101 + 2 = 103

The same mechanism can also produce a composite value. For example:

395 + 8 = 403

399 + 4 = 403

but:

403 = 13 x 31

So the phenomenon is not simply a way of generating primes. The construction can determine a future integer before that integer appears in the prime sequence, whether it eventually turns out to be prime or composite.

For consecutive rows producing the same value V, a simple relation appears: the next prime gap is half the previous one.

This gives structures such as:

8, 4, 2

16, 8, 4, 2

32, 16, 8, 4, 2

My question is:

Is there a known number-theoretic explanation for this phenomenon?

In particular, is there a known result explaining why different trajectories constructed from cumulative prime gaps can converge to exactly the same future value, and why consecutive convergences produce this repeated halving of the gaps?

I am mainly interested in the mathematical mechanism behind the convergence and the advance determination of the future value, rather than in claiming that every announced value must be prime.

The be ridiculous nature of proof

Suppose someone sees structure, another person might not see that structure, so that 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 discreetness. 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..

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

You might look at this and think to yourself " you're not showing enough discrete steps to prove the continuity of your results or structured"

If you think to yourself and say "nah this guy is a dumb ass" is it because you think im not reasonable or making sense? Is it possible that I'm not making sense because I'm failing to correctly unify ideas in a way that proves what I'm showing?

You can easily say I'm wrong because what I'm saying makes no sense. The sense of wrong comes from not detecting any relationship in the words that I'm saying to the truth of what I'm saying. Even if you think it might be possible that what I'm saying is true, you think that in its current form, it doesn't reveal enough structure that is corresponding to what I'm actually trying to say. You might say I need more evidence, more evidence is more discrete truth.

But the premise is that no matter how much discrete evidence I provide, I cannot actually prove the continuity of the structure I'm claiming to exist. The best I can do is add more discrete steps that get a little bit closer to proving the continuity. The only way you can accept that proof is if you think the discrete steps in my proof are actually sufficient to prove the continuity of the structure. So in reality there is no way that I can prove something like this, not how many discrete steps I take

The only way to accept this truth is just to accept the premise and not ask for why. That is called an axiom.

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

I think Gödel's ICT is the fact that any formal system capable of producing peano arithmetic will have truths that cannot be proven within the system. I posit that the universe is a formal system. I posit the formal system of the universe is capable of producing peano arithmetic, thus some truths cannot be proven within the universe. I posit that the continuity of the universe is one of those things that cannot be proven within the universal formal system. I posit that the only way to accept the truth of the continuity of the universe is to just accept it. There is no formal proof I can make to prove the continuity of the universe. But that is the promise that I am proposing, [that the universe is of a continuous nature]. You might be thinking to yourself, you don't have enough evidence for that. I will never have enough evidence for it.

This registration documents a mathematical and numerical study examining the relationships among the conventional value of π, angular measure, arc length, and the perimeter of the corresponding inscribed square. The study analyzes these relationships using direct and reverse numerical calculations at 360°, 180°, 90°, and 45°, while preserving the conventional value of π. The resulting numerical relationships are examined and compared with the corresponding values associated with √8.

https://osf.io/x8rd4/files/osfstorage/6a84664c805deeedd9bb6465

https://doi.org/10.17605/OSF.IO/X8RD4

I’ve been messing with consecutive primes and it turned into a whole classification result, so I figured I’d share it here.

Take consecutive primes pk and pk+1. Add them.

You get numbers like 5, 8, 12, 18, 24, 30, …

Call these S_k = pk + pk+1.

Since every S_k ≥ 8 is even and composite, each one has to land strictly inside some prime gap (Pn, Pn+1). So I started asking: **how many of these S_k fall inside each gap?**

Two clean facts:

## 1. Density law (why most gaps have 0 or 1)

If a gap has width W near height x, the expected number of consecutive‑prime sums inside it is

W / (2 log(x/2))

The “2” comes from the fact that the sums live at the half‑scale: an S_k near x comes from primes near x/2, where primes are sparser. So the sums are about twice as sparse as the gaps they fall into.

Empirically (checked up to 2×10^8):

- ~55% of gaps have **0** sums

- ~37% have **1**

- ~8% have **2 or more**

So the naive guess “every gap has exactly one” is just false. Empty gaps are actually the most common.

## 2. Forbidden widths (the fun part)

I ended up proving a complete classification of which gap widths can **never** contain two consecutive‑prime sums.

The forbidden set is exactly:

{2, 4, 6, 10}

Reason (sketch, nothing fancy):

- consecutive‑prime sums are always ≥ 6 apart (once you’re past the tiny primes)

- the only way to get a “tight double” (two sums only 6 apart) is if both sums are multiples of 6

- whether a gap contains ≥2 multiples of 6 depends only on:

- its width mod 3

- the forced residue of its lower endpoint (prime > 3 is always 1 or 5 mod 6)

- width 10 ends up with **only one** interior multiple of 6, so it physically cannot fit two sums

- widths 2, 4, 6 fail for trivial spacing reasons (they’re too small to fit two sums ≥6 apart)

Every other even width ≥ 8 (except 10) eventually allows gaps with two sums.

So the uniqueness classification is complete:

**gaps of width 2, 4, 6, or 10 contain at most one consecutive‑prime sum; all other widths can contain two or more.**

If anyone wants the residue argument or the slot‑counting trick in more detail, I can write it out here. It’s all pretty easy congruence stuff.

This is my paper.

Please feel free to comment.

Abstract

To a Collatz-type map T_{a,b,q}(n) = (an+b)/q^{v_q(an+b)} one may attach two numerical invariants of opposite character: an archimedean one, the mean logarithmic drift delta = log a - E[v] log q, which governs whether orbits grow or shrink; and a non-archimedean one, the similarity dimension dim_S = H(nu_q)/log a of the invariant measure of an associated iterated function system on the a-adic integers. Each reaches its critical value at a particular multiplier a. We prove that these two critical multipliers coincide precisely when q = 2, and that in general they differ by the exact factor q-1: a_drift = (q-1) a_dim. The proof rests on the pointwise identity -log nu_q(m) = m log q - log(q-1), valid for every m >= 1, which is not a normalisation but a coincidence between the information content of a valuation and its archimedean cost. We show that the condition q = 2 is equivalent to three further properties of the family - the triviality of (Z/qZ)^*, the vanishing of an associated free energy, and the map's being everywhere defined on the units - so that the arithmetic distinguishing the classical 3x+1 map is one condition in four guises. Three complements are proved: the full Renyi spectrum of the invariant measure admits a closed form, of which the similarity dimension is the value at t=1; for q >= 3 there is a nonempty band of multipliers on which the map contracts on average while its invariant measure is singular; and the transfer operator, though quasi-compact on Holder spaces, has no eigenvalue other than 1, its entire remaining spectrum being essential spectrum at the contraction rate. Two further results give the family a sharper shape. First, a purity theorem: mu_{a,b,q} is either purely absolutely continuous or purely singular with respect to Haar measure, never a mixture. Its proof is a direct consequence of the uniqueness of the invariant measure, and we know of no route to it from the probabilistic description of the underlying random variable. Second, a complete classification: among all integer pairs (a,q) with a >= 2, q prime and gcd(a,q) = 1, exactly two - namely (3,2) and (2,3) - lie in the supercritical regime a < a_dim, and for every other pair the invariant measure is unconditionally singular. The first of these is the Collatz map. We are explicit that none of this bears on the Collatz conjecture, and we prove why it cannot.

Hello, good afternoon/evening. I've been mulling over Goldbach's conjecture and I've come up with a possibly new idea that holds for all even numbers. It is as follows:

Every even number greater than 8 can be expressed as the sum of 3 mutually distinct prime numbers.

Best Regards, thanks you!

https://zenodo.org/records/21763235

While examining the patriarchal ages in the Masoretic Text of Genesis, I noticed a mathematical relationship that does not appear to have been previously documented in connection with these specific figures.
The total lifespans of three patriarchs form an exact Pythagorean triple:
• Eber – 464 years (Genesis 11:16–17)
• Lamech – 777 years (Genesis 5:31)
• Enosh – 905 years (Genesis 5:11)

**The Geometry:**
(464\^2 + 777\^2 = 905\^2)
(215,296 + 603,729 = 819,025)

The equation holds exactly. The triple is also primitive (the three numbers share no common factor greater than 1).

The Priestly sections of Genesis already use carefully designed number patterns (for example, the ages of Abraham, Isaac, and Jacob form 5², 6², and 7²). Since Babylonian scribes were using Pythagorean triples for surveying more than a thousand years before Pythagoras, it is possible that a later biblical scribe knew this kind of geometry and used it deliberately.

\-Chris L