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Old 2019-03-03, 03:13   #1
MathDoggy
 
Mar 2019

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Default 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
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Old 2019-03-03, 03:26   #2
axn
 
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Ho do you count the divisors without factorizing N^2 (and therefore, N as well)?
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Old 2019-03-03, 03:32   #3
MathDoggy
 
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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
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Old 2019-03-03, 03:40   #4
axn
 
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Quote:
Originally Posted by MathDoggy View Post
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)?
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Old 2019-03-03, 03:52   #5
Batalov
 
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Quote:
Originally Posted by MathDoggy View Post
This work can't be replicated or plagiarized
Said the man who plagiarized.
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Old 2019-03-03, 05:10   #6
retina
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Quote:
Originally Posted by MathDoggy View Post
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
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Old 2019-03-03, 12:21   #7
MathDoggy
 
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Good point retina
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Old 2019-03-03, 12:27   #8
MathDoggy
 
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Quote:
Originally Posted by axn View Post
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
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Old 2019-03-03, 12:29   #9
MathDoggy
 
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Quote:
Originally Posted by Batalov View Post
Said the man who plagiarized.
What do you mean?
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Old 2019-03-03, 12:59   #10
Uncwilly
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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
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Old 2019-03-03, 13:18   #11
retina
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Quote:
Originally Posted by Uncwilly View Post
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.
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