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-   -   Formula for prime numbers of the form (m)*(n)+1 (https://www.mersenneforum.org/showthread.php?t=26093)

Hugo1177 2020-08-15 07:47

Formula for prime numbers of the form (m)*(n)+1
 
Proof of the Twin primes Conjecture and Goldbach's conjecture We can find infinite prime numbers with the separation we want and we can express every even number as the sum of two prime numbers. [URL]http://www.academia.edu/43581083/Proof_of_the_Twin_primes_Conjecture_and_Goldbachs_conjecture[/URL]

CRGreathouse 2020-08-15 20:17

The crux of the proof is on p. 3:
[INDENT]I have tried it with high numbers and it seems that there is no problem, which would be to say that there are infinite prime numbers separated the quantity that we want.[/INDENT]
I'll note that the method requires taking large symbolic derivatives and then factoring integers far larger than the twin primes generated, so this is not an algorithmic improvement.

jnml 2020-08-16 13:35

[QUOTE=Hugo1177;553742]Proof of the Twin primes Conjecture and Goldbach's
conjecture We can find infinite prime numbers with the separation we want and we can
express every even number as the sum of two prime numbers
[URL]http://www.academia.edu/43581083/Proof_of_the_Twin_primes_Conjecture_and_Godbachs_conjecture[/URL][/QUOTE]

It's fine if you prefer to keep the PDF to just yourself.

It's fine if you prefer to sell it for money.

It's bad to "publish" it for free - but at a site that requires registration and or login. Some, if
not most people, may chose to simply go away.

sweety439 2020-08-17 00:36

[QUOTE=Hugo1177;553742]Proof of the Twin primes Conjecture and Goldbach's conjecture We can find infinite prime numbers with the separation we want and we can express every even number as the sum of two prime numbers. [URL]http://www.academia.edu/43581083/Proof_of_the_Twin_primes_Conjecture_and_Goldbachs_conjecture[/URL][/QUOTE]

It is known that there are infinite prime numbers with separation n for an even number <= 246, but[I] we don't known what this n is!!![/I] (like that, we know that at least one of zeta(5), zeta(7), zeta(9), and zeta(11) is irrational, but we don't known which number is irrational)

Also, there is no infinite (at most one pair) prime numbers with separation n when n is odd!!!

Dr Sardonicus 2020-08-17 13:18

This [i]magnum opus[/i] does not begin well.

[quote]The function that generates those two prime numbers is:

dn/dx = x^(4/p) - 3x^(2/p) + 1

Where n is the derivative of order n and p is the distance in units of the separation that we want to find.[/quote]

So, "n" is self-referencing, and the symbol p is used to denote the [i]separation[/i], or distance between two primes, rather than a prime number.

It continues into the demonstrably false:

[quote]The relationship between this function and the Lucas numbers is that in the undifferentiated function

x^(4/p) - 3x^(2/p) + 1 it's zeros are the Lucas numbers.


For example x^(4/7) - 3x^(2/7) + 1 = 0

One of its zeros is 29 which is the 7th number of Lucas[/quote]

Er, no. Substituting 29 for x in the given expression gives -0.00198416, approximately.

The zeros of y^2 - 3*y + 1 are

[tex]\frac{3\;\pm\;\sqrt{5}}{2}\;=\(\;\frac{1\;\pm\;\sqrt{5}}{2}\)^{2}[/tex]

so, taking x = y^(p/2), we have

[tex]x \; = \; \(\frac{1\;\pm\;\sqrt{5}}{2}\)^{p}[/tex]

Hugo1177 2020-10-14 23:52

Update of the Formula for prime numbers of the form (m)*(n)+1
 
[URL="https://www.researchgate.net/publication/343813220_Formula_for_prime_numbers_of_the_form_mn1"]https://www.researchgate.net/publication/343813220_Formula_for_prime_numbers_of_the_form_mn1[/URL]

Hugo1177 2020-10-16 08:47

Formulas for Prime Numbers
 
[color=red][b][size=4]MODERATOR NOTE:[/size] Moved from Lounge.[/b][/color]

[URL="https://www.researchgate.net/profile/Pedro_Garcia_Pelaez"]https://www.researchgate.net/profile/Pedro_Garcia_Pelaez[/URL]

Hugo1177 2020-10-16 22:51

Formula for prime numbers of the form (m)*(n)+1
 
Formula that returns prime numbers of the form (m)*(n)+1 how many prime numbers between 1 and 1000000 I used a polynomyal that it´s roots are the golden ratio squared and the golden ratio conjugate. When I aply fractional exponents to the x the polinomyal it´s roots returns lucas and fibonacci numbers exactly. And when I derivate this function I obtain prime numbers included in one number of the structure of the derivative, but there are more that prime numbers come in relation with the order of the derivative multiplied for the fractional exponent. Is like there a relation between fibonacci and lucas numbers and prime number where the original polynomial derivated adapted his form to returns special prime numbers.

[URL="https://www.researchgate.net/publication/343813220_Formula_for_prime_numbers_of_the_form_mn1"]https://www.researchgate.net/publication/343813220_Formula_for_prime_numbers_of_the_form_mn1[/URL]

Batalov 2020-10-16 23:09

[COLOR="Red"][SIZE="3"][B]Mod warning:[/B][/SIZE][/COLOR]

Stop spam-posting. Next duplicate posts (and crossposts) will be deleted altogether,

CRGreathouse 2020-10-17 04:43

[QUOTE=Hugo1177;560091]prime numbers of the form (m)*(n)+1[/QUOTE]

As opposed to what other kind of prime numbers? :confused:


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