20190303, 03:13  #1 
Mar 2019
5·11 Posts 
I Think I Have A "Prime Generating Formula" (without the formula)
Let N be the set of natural numbers
Let n be an element of N Then, do N^2, if n has 3 divisors do square root(n) Examples: N= 1,2,3,4,5,6,7,8,9,10... N^2= 1,4,9,16,25,36,49... 4 has three divisors so, sqrt(4)=2 which is prime This formula gives consecutive prime numbers And it works because primes are in the set of natural numbers so the square of a prime p will have always the divisors p, 1 and p^2 This work can't be replicated or plagiarize 
20190303, 03:26  #2 
Jun 2003
1001011111000_{2} Posts 
Ho do you count the divisors without factorizing N^2 (and therefore, N as well)?

20190303, 03:32  #3 
Mar 2019
5·11 Posts 
You can count the number of divisors of n by the formula: k sub 1 times k sub n+1 where k is the power of the factors of the number you are checking

20190303, 03:40  #4 
Jun 2003
2^{3}×607 Posts 

20190303, 03:52  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10010001101111_{2} Posts 

20190303, 05:10  #6  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
1011110111110_{2} Posts 
Quote:
How about this simplification: Let N be the set of natural numbers Let n be an element of N If n has 2 divisors then n is prime. Example: n= 1,2,3,4,5,6,7,8,9,10... 2 has two divisors so 2 is prime This formula gives consecutive prime numbers And it works because primes are in the set of natural numbers so a prime p will have always the divisors 1 and p This work can't be replicated or plagiarize What do I win? Last fiddled with by retina on 20190303 at 05:11 

20190303, 12:21  #7 
Mar 2019
110111_{2} Posts 
Good point retina

20190303, 12:27  #8 
Mar 2019
5·11 Posts 
I didn´t mention how would I get the factors of the numbers due to being unaware of what I wrote, you can run a factorization program to obtain the factors of a given number

20190303, 12:29  #9 
Mar 2019
5·11 Posts 

20190303, 12:59  #10 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2^{3}·7·167 Posts 
So, take a nice 'large' number. Say for example, 987654321098765432109876543210987654321.
Square that. How long might it take to find out that the number has exactly 3 factors? What about if the number was 5, 10, or 50 times as long? Hmm, 3 factors..... let's think about that! 23 is prime 23^{2}=529 If 529 has only 3 factors, they are: 1, 23, and 529 What does this tell us? First we can throw out 1 and 529, as every number has 1 and themselves as factors. So we are left with 23. Which can't be factored any more, else we would have more than 3 factors. Which means 23 is prime. And it means that 529 is the square of a prime. So, rather than just trying to sieve 23, you want us to square it, then completely factor that number, then take a costly square root of the new number, just to get back to the original 23. Go think about this for a while (take a couple of days if needed). See if you can figure out what steps in your process that can be eliminated. The see what you have left. Last fiddled with by Uncwilly on 20190303 at 13:00 
20190303, 13:18  #11  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2·3·1,013 Posts 
Quote:


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