I Think I Have A "Prime Generating Formula" (without the formula)

[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

[axn]

Ho do you count the divisors without factorizing N^2 (and therefore, N as well)?

[MathDoggy]

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

[axn]

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)?

[Batalov]

Originally Posted by MathDoggy

This work can't be replicated or plagiarized

Said the man who plagiarized.

[retina]

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?

[MathDoggy]

Good point retina

[MathDoggy]

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

[MathDoggy]

Originally Posted by Batalov

Said the man who plagiarized.

What do you mean?

[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.

[retina]

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