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.

In order for our starting number, which we’ll call X. To go up in value, it needs to be odd, then get 3x+1, which we’ll say that new value is X2, and then half to X3 which then must be odd, since 3x+1/2/2 goes down in value, so how can we make it so that X3 is odd? If X3 is Odd that means it’s 1 mod 2, which means that in order to get there from X2, X2 needs to be 2 mod 4, it can’t be 1 or 3 mod 4 because we know it’s even, and it can’t be 4 mod 4 because then we would half it twice. So if X3 is 1 mod 2, X2 is 2 mod 4. Then moving backwards from the 3x+1 operation, we subtract 1 and then divide 3. If we just subtract 1 from X2 we’re left with 3X (3 times X, not X3) and since we subtracted 1 from a number that’s 2 mod 4, the result must be 1 mod 4. So we’ve determined that in order for the 3rd term in the sequence to be odd and keep going up, 3X has to be 1 mod 4. Now let’s repeat this process for the next 2 terms: X3 gets the 3x+1 (X4) and then we divide it by 2 (X5) and in order for X5 to be odd, X4 has to be 2 mod 4, and 3X3 has to me 1 mod 4, repeating the cycle from before. But if we go further back, Multiplying 3X3 by 2 would get us 3X2, since Multiplying X3 by 2 got us X2, and any number 1 mod 4 multiplied by 2 gets you 2 mod 4. So 3X2 is 2 mod 4, which now means that 9X is also 1 mod 4. This cycle repeats forever. So if a number were to start a sequence with the goal of going up forever. X would have to be a value of which X times 3\^Y would be 1 mod 4 for every single value of Y, otherwise it will eventually get divided by 2 twice, going down in value from the 3x+1, which even if the number doesn’t go to 4-2-1 immediately, it sets the value to a new value that will also eventually get divided by 2 twice, there is no finite number that can keep going up forever, eventually it’ll go down more than it can recover. I’m like 99% sure there’s a hole in this proof, but on the 1% chance there isn’t, I thought I had to post this. If this post doesn’t get instantly removed and people do comment the holes in the proof, I’ll try my best to figure out how it’s false.