the sum of two prime numbers

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kimms0131
Posts: 2
Joined: Tue Mar 15, 2011 4:06 pm

the sum of two prime numbers

Post by kimms0131 » Tue Mar 15, 2011 4:16 pm

maybe even numbers except 2 are expressed by p+q (p,q are differnt prime numbers)
if the even number(m) are expressed only one pair of sum of two prime numbers (p<q)
let fix the set of m =A, m<10000, n(A)=? :shock:

thundre
Posts: 356
Joined: Sun Mar 27, 2011 9:01 am

Re: the sum of two prime numbers

Post by thundre » Tue May 10, 2011 1:29 pm

Goldbach's Conjecture does not have the restriction that the primes must be different. It is unknown whether there are any exceptions to it. Your restriction forces the exceptions {4,6} at least.

Image

The above image suggests A={8,10,12,14,38}.

http://en.wikipedia.org/wiki/Goldbach%27s_conjecture
http://en.wikipedia.org/wiki/File:Goldb ... ecture.gif
Image

kimms0131
Posts: 2
Joined: Tue Mar 15, 2011 4:06 pm

Re: the sum of two prime numbers

Post by kimms0131 » Fri Jul 22, 2011 2:36 pm

But Goldbach's conjecture can be possible in a huge number(very big).

In that case why only 8,10,12,14,38 exist?

I don't understand. Help me ㅠ_ㅠ :?:

thundre
Posts: 356
Joined: Sun Mar 27, 2011 9:01 am

Re: the sum of two prime numbers

Post by thundre » Fri Jul 22, 2011 6:23 pm

To actually solve the problem as stated, you would write a program that generates primes up to 10,000 and counts each m=p+q where p<q.

Code: Select all

public class Rec2632 {
    public static final int MAX_M = 10000;

    public static void traverse() {
        PrimeSieve sieve = PrimeSieve.getInstance();
        int n = sieve.primeCount(MAX_M);
        int c[] = new int[MAX_M + 1];
        
        for (int i = 1; i < n; i++) {
            int q = (int) sieve.prime[i];
            for (int j = 0; j < i; j++) {
                int p = (int) sieve.prime[j];
                int m = p + q;
                if (m <= MAX_M) {
                    c[m]++;
                }
            }
        }
        
        for (int m = 4; m <= MAX_M; m += 2) {
            if (c[m] < 2) {
                System.out.println(m + ":  " + c[m]);
            }
        }
    }
    
    public static void main(String args[]) {
        traverse();
        System.exit(0);
    }
}
The output from the above shows {4,6} have 0 solutions and {8,10,12,14,38} have 1 each.

Going back to theory, the reason that the set is limited is that as m increases, the number of ways to partition it into two different odd numbers increases linearly. n = floor((m-2)/4)

But the probability that those numbers are prime is 1/(ln m), so the expected number of partitions is greater than n/(ln m)^2, which is generally increasing when m>8.
Image

Mattwolf33
Posts: 1
Joined: Thu Feb 21, 2013 5:48 pm

Re: the sum of two prime numbers

Post by Mattwolf33 » Thu Feb 21, 2013 5:50 pm

hmm, would this work?
974267642452

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