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Author : chrisdecorte@yahoo.com Page 1
Yet another Prime formula and prove
of the related open problems
Author : chrisdecorte@yahoo.com Page 2
Abstract
In this document, we derive again a new formula to calculate prime numbers...
Author : chrisdecorte@yahoo.com Page 3
Methods & Techniques
Step 1: development of another new formula to check if a numbe...
Author : chrisdecorte@yahoo.com Page 4
Step 2: splitting an even number into the sum of 2 primes
Goldbach’s conjecture sta...
Author : chrisdecorte@yahoo.com Page 5
To illustrate this with a small example for 2n=20:
The general condition for an equ...
Author : chrisdecorte@yahoo.com Page 6
Step 5: Polignac and general Twin Prime Conjecture
Polignac’s conjecture states: “F...
Author : chrisdecorte@yahoo.com Page 7
Results
It is clear from the above that our formula for differentiating primes from...
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Yet another prime formula to prove open problems

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In this document, I derive again a new formula to calculate prime numbers and use it to discuss open problems like Goldbach and Polignac or Twin prime conjectures.
The derived formula is an interesting variant of my previous one.

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Yet another prime formula to prove open problems

  1. 1. Author : chrisdecorte@yahoo.com Page 1 Yet another Prime formula and prove of the related open problems
  2. 2. Author : chrisdecorte@yahoo.com Page 2 Abstract In this document, we derive again a new formula to calculate prime numbers and use it to discuss open problems like Goldbach and Polignac or Twin prime conjectures. The derived formula is an interesting variant of my previous one. Key-words Zhang Yitang; Goldbach conjecture; twin prime conjecture; Polignac. Introduction The following document originated during our study of primes. We tried to create a schematic representation of the mechanism that lead to the creation of primes. Following, we tried to mathematically describe this graphical representation. It resulted in a first new formula to calculate if a number is prime or not. We explained about this formula in our previous document [1]. Shortly afterwards, we derived another prime formula that might be even more powerful. In the meantime, Zhang Yitang had come up with his prove with regard to prime gaps. So, we will briefly reflect on this as well.
  3. 3. Author : chrisdecorte@yahoo.com Page 3 Methods & Techniques Step 1: development of another new formula to check if a number is prime or not We will refer to the following picture as a graphical aid to derive our formula: Let’s draw an X&Y axis to start with. Draw a sinus function that starts through (0,0) and that goes through the first prime (2,0). This sinus function will be ‫ݕ‬ = sin ቀ గ௫ ଶ ቁ. The next integer where the first sinus does not go through is a prime and is 3. Draw a sinus function that starts through (0,0) and that goes through (3,0). This sinus function will be ‫ݕ‬ = sin ቀ గ௫ ଷ ቁ. The next integer where the previous sinuses do not go through is a prime and is 5. Draw a sinus function that starts through (0,0) and that goes through (5,0). This sinus function will be ‫ݕ‬ = sin ቀ గ௫ ହ ቁ. The next integer where the previous sinuses do not go through is a prime and is 7. Continue like this. Following above, we can state the formula for determining if a natural number is prime or not as (formula 1.): ݊ = Prime ( = ‫݌‬௜ାଵሻ ⇔ ܲ(݊ሻ = ෑ sin ൬ ߨ݊ ‫݌‬௜ ൰ <> 0 ∀௣೔ழ௡ ௣೔ୀଶ
  4. 4. Author : chrisdecorte@yahoo.com Page 4 Step 2: splitting an even number into the sum of 2 primes Goldbach’s conjecture states that every even number can be written as the sum of two primes. We can agree on following convention: 2n every even number q the smaller prime number greater than 2 (and q≤p) (so q starts from 3) p the larger prime number smaller than 2n-2 With: 2n=p+q So for 2n, p and q, we will be able to write: ܲ(2݊ሻ = ෑ sin ൬ ߨ2݊ ‫݌‬௜ ൰ = 0 ∀௣೔ழଶ௡ ௣೔ୀଶ This is because the first term (for ‫݌‬௜ = 2), the sinus will be 0. We can then also write: ܲ(2݊ሻ = ෑ sin ൬ ߨ2݊ ‫݌‬௜ ൰ = ∀௣೔ழଶ௡ ௣೔ୀଶ ෑ ൤sin ൬ ߨ(‫݌‬ + ‫ݍ‬ሻ ‫݌‬௜ ൰൨ ∀௣೔ழଶ௡ ௣೔ୀଶ = ෑ ൤sin ൬ ߨ‫݌‬ ‫݌‬௜ ൰ cos ൬ ߨ‫ݍ‬ ‫݌‬௜ ൰ + sin ൬ ߨ‫ݍ‬ ‫݌‬௜ ൰ cos ൬ ߨ‫݌‬ ‫݌‬௜ ൰൨ ∀௣೔ழଶ௡ ௣೔ୀଶ = 0 This means that for every 2n>=6, there will be at least one q (≥3 and ≤2n-3) that will lead to a p equal to 2n-q where the equation between brackets equals to 0 on an occasion of ‫݌‬௜. It is this occasion of ‫݌‬௜ that will make the whole product equal to.
  5. 5. Author : chrisdecorte@yahoo.com Page 5 To illustrate this with a small example for 2n=20: The general condition for an equation of type sin ߙ cos ߚ + sin ߚ cos ߙ being equal to zero is that ߙ + ߚ = ݇ ߨ. In this case, this would mean that p+q should be divisible by at least one pi where pi takes on all primes from 2 to before 2n. Already for pi=2, this condition is fulfilled as p+q=2n is an even number. The above means that we have proven Goldbach’s conjecture.
  6. 6. Author : chrisdecorte@yahoo.com Page 6 Step 5: Polignac and general Twin Prime Conjecture Polignac’s conjecture states: “For any positive even number n, there are infinitely many prime gaps of size n. In other words: There are infinitely many cases of two consecutive prime numbers with difference n.” To highlight that n is even, we will write n as 2n so that 2n=p-q. Using the previous formula’s, this would mean that we are looking for: ܲ(2݊ሻ = 0 ; ܲ(‫݌‬ሻ ≠ 0 ; ܲ(‫ݍ‬ሻ ≠ 0 We can then rewrite to: p=2n+q and get: ܲ(‫݌‬ሻ = ෑ sin ൬ ߨ‫݌‬ ‫݌‬௜ ൰ = ∀௣೔ழ௣ ௣೔ୀଶ ෑ ൤sin ൬ ߨ(2݊ + ‫ݍ‬ሻ ‫݌‬௜ ൰൨ ∀௣೔ழ௣ ௣೔ୀଶ = ෑ ൤sin ൬ ߨ2݊ ‫݌‬௜ ൰ cos ൬ ߨ‫ݍ‬ ‫݌‬௜ ൰ + sin ൬ ߨ‫ݍ‬ ‫݌‬௜ ൰ cos ൬ ߨ2݊ ‫݌‬௜ ൰൨ ∀௣೔ழ௣ ௣೔ୀଶ ≠ 0 The general condition for an equation of type sin ߙ cos ߚ + sin ߚ cos ߙ being not equal to zero is that ߙ + ߚ ≠ ݇ ߨ. In this case, this would mean that 2n+q should not be divisible by not any of the pi where pi takes on all primes from 2 to before p. This means that there is no prime in between 1 an p or that p itself is a prime.. This means that for every prime q, we can find a corresponding prime p such that p=q+2n. Since there are an infinite amount of primes to choose from at the start, it means that we have proven Polignac’s conjecture. Remark: nowhere in this prove, there is a hint that n should be less than 70,000,000.
  7. 7. Author : chrisdecorte@yahoo.com Page 7 Results It is clear from the above that our formula for differentiating primes from non primes is sufficiently strong to prove the existing open conjectures. This is done using basic trigonometric formula’s and some simple interpretations. Discussions: We should try to find the link with Yitang Zhangs work. Conclusion: 1. Discussing about primes and prime gaps is not so difficult if one uses the right formula to demonstrate prime. Acknowledgements I would like to thank this publisher, his professional staff and his volunteers for all the effort they take in reading all the papers coming to them and especially I would like to thank this reader for reading my paper till the end. I would like to thank my wife for having faith in my work. References 1. Derivation of a Prime verification formula to prove the related open problems; Chris De Corte 2. http://en.wikipedia.org/wiki/Polignac's_conjecture

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