As of my last knowledge update in September 2021, the Twin Prime Conjecture had not been proven, and it remains an open problem in mathematics. The conjecture posits that there are infinitely many pairs of twin primes, which are prime numbers that have a difference of 2 (e.g., 41 and 43, 71 and 73).

Since my information might be outdated, I recommend checking the latest mathematical research and news sources for any recent developments regarding the Twin Prime Conjecture. Keep in mind that proving such conjectures can be extremely challenging, and progress can be slow.

Arnel Yurfo, in my paper, I propose an algorithm to generate all twin primes up to a bound "n". I prove that such algorithm terminates once it reaches a critical prime, and that all pairs left are valid twin primes. Given that the boundary is of form O(n/ln^2(n)), it can be generalized that twin primes are infinite.

I use the same algorithm (with a minor variation), to prove Goldbach's conjecture.

As far as I know, attempts to address the Twin Prime Conjecture seem to critically involve:

* heuristic approximations of prime counting functions; and

* the belief that the prime divisors of n cannot be treated as mutually independent.

One promising direction yet to be researched adequately---not only for elementary (non-analytic) proofs of the Twin Prime Conjecture and its possible generalisations, but also for an elementary (non-analytic) proof of Dirichlet's Theorem and similar functions---seemingly follows from:

* non-heuristic approximations of prime counting functions; and

* the density-based proof---in the attached preprint (40_PNT)---that the prime divisors of n are, indeed, mutually independent (as is immediately evidenced by Eratosthenes sieve).

See also §22 (The significance of evidence-based reasoning for Number Theory and Integer Factorising) of my attached book (16_Anand_Dogmas).

Kind regards,

Bhup

CV: https://www.dropbox.com/s/ohrhhinjvg5zj7s/

I think there has been recent progress, considering the works of Professor James Maynard and others.

Dear @Mirzakhmet Syzdykov,

To see how we may estimate the 'extent twin primes are infinite' from their density, please see my book [An21], §21.D:

Theorem 21.45. (Twin Prime Theorem) π2(n) → inf as n → inf,

where π2(n) denotes the number of twin primes less than n.

Kind regards,

Bhup

CV: https://www.dropbox.com/s/ohrhhinjvg5zj7s/

[An21] 2021. The Significance of Evidence-based Reasoning in Mathematics, Mathematics Education, Philosophy, and the Natural Sciences. First edition commemorating UNESCO’s 2nd World Logic Day, 14 January 2020. DBA Publishing, Mumbai, India.

https://philpapers.org/rec/ANATSO-5

@ Mirzakhmet Syzdykov,

Please pardon my ignorance but, prima facie, I could not follow the language, or understand the purpose, of the code.

Does it have anything to do with the density or distribution of twin primes over the first n integers?

Please advise.

Kind regards,

Bhup

Peter Breuer, correct, the prime numbers are not infinite. The number of prime numbers is infinite. I meant that there are infinitely many prime and twin prime numbers. Their density of the latter tends to 0 (as log^2 n/ n), but there are still infinitely may of them. Just like there are infinitely many square numbers, but their density also tends to 0.

With Prof. K. Balasubramanian, we have made some study of twin primes and their extension, look at:

Article Three conjectures on extended twin primes and the existence ...

based on a previous study on the relation between primes and Mersenne numbers:

Article Mersenne Numbers, Recursive Generation of Natural Numbers, a...

Perhaps this will help or contribute to higher problem complexity

Ha! I guess with the first line of your comment is not worth a response. But just in case, my approach is not probabilistic. If you are really interested, feel free to browse through my paper. Or test my code. If you want to debate ideas on math, happy to do so. If you want to be a wise-4$$, i guess that's your problem, not mine.

Peter Breuer , this is my approach. Let’s start with the sieve of Eratosthenes. Let’s run it for the numbers up to 24. After sieving all the even numbers, we would have:

[1,3,5,7,9,11,13,15,17,19,21,23]

Sieving by 3:

[1,5,7,11,13,17,19,23]

We can stop now, and guarantee that all the numbers left in the list are primes. Why? Because the range of the numbers are less than 5^2, so any composite number below would have been necessarily sieved by either 2 or 3. If there was a composite in the list, both of its factors would have to be larger than 5. But that would make the number larger than 25.

We can calculate the number of numbers left as 25*(1-1/2)*(1-1/3)=25/3 which is 8.3333. Rounding it down, it would be 8, which is the length of the list.

So for any large number, n, we can calculate the total number of primes (sieved after the largest prime below its square root as:

\pi(n)= n \cdot \prod_{p

In the case of the s. of Eratosthenes, yes, I would add the numbers back, in order to account for the total of number of primes in the range. Let's say, we sieve numbers from 1 to 100, and in the "strict sieve", I eliminate the initial primes {2,3, 5, 7}. Then the total number of primes from 1 to 100 would be the length of the array, plus the 4 I eliminated in the sieving process. But that is just to have an accurate count. To use a lower bound, I don't need to add them back. Just say that the number of primes (or twin primes) is at least, n/log(n) or c*n/log^2(n).

As for the explanation of the 2 levels, they are 2 parallel arrays. When sieving by 3, for example, i would eliminate [3, 9, 15, ...] from the top array (and their corresponding pairs in the second [5, 11, 17, ...], and when sieving the second array, I eliminate [9, 15, 21,...] from the second, and the corresponding [7, 13, 19, ...] from the first. That is why the width changes as (1-2/p) when p>2, because you are not only removing all 0 mod 3 from the first row, but also -1 mod 3 from the second array. Then, you would sieve by 5, 7, etc.. Whenever you sieve a position in the first array, you also eliminate the corresponding one in the second array. So "vertically", all pairs remain consistent as {x, x+2}.

Peter Breuer, at each stage, I remove the "column" or pairs divisible by p, whether it is the top component, or bottom component. When sieving by 2, I am left with all odd pairs. In this case, the factor is 1/2 of the original width of the array, because they are both even, on top and bottom. Next pass, when sieving by 3, I will remove pairs like [3, 5], [9, 11], (divisible by 3 on top), but also [7, 9], [13, 15] divisible by 3 on the bottom. Thus, the width of the array is reduced by a factor of 1/3 (=1-2/3), because every 2 out of 3 pairs will match the condition of being divisible by 3; 1/3 of them on the top, and another 1/3 of them, on the bottom.

Partially, yes. Not only I remove the multiples of A_k, but also the ones of B_k. So i remove "columns" from the array. If the top part is divisible by p_k, then it gets removed, together with the corresponding bottom part. And will remove also if the bottom part is divisible by p_k with its corresponding top component.

To clarify your note above, A1 and B1 would be the resulting arrays after sieving by 2. In reality, A0 and B0 would be:

[1, 2, 3, 4, 5, ... , 99, 100]

[3, 4, 5, 6, 7, ..., 101, 102]

Sieving by 2, would give you your 2 arrays above, possibly including [1, 3] on the first column.

Remove "the multiples of p_k from A_k", I should have said...

I think what you said last is correct. And yes, I remove any multiples of p_k from the array A_k and their corresponding position from B_k. I also remove the multiples of p_k from B_k, and their corresponding position in A_k. Then, I iterate to the next prime p_k+1. You also mention nontrivial multiples. In my algorithm I remove p_k too if is it still in either array. That is why in the original equation I add them back. It may be a valid twin prime pair may be removed in the sieving process, but removing it makes the counting (and the argument) much easier.

p_k is the k-th sieving prime. So, as in the Eratosthenes sieve, you sieve by 2, then by 3, 5, 7, 11,... etc.