mersenneforum.org I Think I Have A "Prime Generating Formula" (without the formula)
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 2019-03-03, 03:13 #1 MathDoggy   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
 2019-03-03, 03:26 #2 axn     Jun 2003 22×32×151 Posts Ho do you count the divisors without factorizing N^2 (and therefore, N as well)?
 2019-03-03, 03:32 #3 MathDoggy   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
2019-03-03, 03:40   #4
axn

Jun 2003

22×32×151 Posts

Quote:
 Originally Posted by MathDoggy the power of the factors of the number you are checking
And how do you get the power of the factors without knowing the factors (aka factorizing the number)?

2019-03-03, 03:52   #5
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

89·113 Posts

Quote:
 Originally Posted by MathDoggy This work can't be replicated or plagiarized
Said the man who plagiarized.

2019-03-03, 05:10   #6
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

5×7×191 Posts

Quote:
 Originally Posted by MathDoggy 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
Why even bother with the squaring step?

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 2019-03-03 at 05:11

 2019-03-03, 12:21 #7 MathDoggy   Mar 2019 5·11 Posts Good point retina
2019-03-03, 12:27   #8
MathDoggy

Mar 2019

3716 Posts

Quote:
 Originally Posted by axn And how do you get the power of the factors without knowing the factors (aka factorizing the number)?
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

2019-03-03, 12:29   #9
MathDoggy

Mar 2019

5×11 Posts

Quote:
 Originally Posted by Batalov Said the man who plagiarized.
What do you mean?

 2019-03-03, 12:59 #10 Uncwilly 6809 > 6502     """"""""""""""""""" Aug 2003 101×103 Posts 252148 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 232=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 2019-03-03 at 13:00
2019-03-03, 13:18   #11
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

5×7×191 Posts

Quote:
 Originally Posted by Uncwilly 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 232=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.
https://mersenneforum.org/showpost.p...71&postcount=6

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