# Upper bound for the number of representation numbers as the sum of two primes

To prove strong goldbach conjecture one can use a lower bound of number of the representations of a number as the sum of two primes. If its greater than zero, than we have conjecture.

I wonder if there is an upper bound of this number of the representations for a number $x$?

Edit: I have edited the question to tell the question in my head properly. I apologize for the people who answered this question.

• Do you ask on a bound on how many ways there are to write $n$ as the sum of two primes? Francesco understood the question differently, I believe. – Dietrich Burde Nov 21 '17 at 9:45
• @DietrichBurde exactly. – Bright Chancellor Nov 21 '17 at 9:46
• Then you should edit. This is not clear from your text, as you see. – Dietrich Burde Nov 21 '17 at 9:46
• @DietrichBurde It would makes me really happy if you can create time to edit it. – Bright Chancellor Nov 21 '17 at 9:48

By a result of Helfgott, every even number $n ≥ 4$ is the sum of at most four primes. This seems to be currently the best known result in this direction.

See the wiki page on the Goldbach's conjecture (in particular the section Rigorous results) and the references contained therein.

Let’s define
$g2(n)=$ nbr of pairs of odd integers $p,q$ such that $p+q=2n$, $p$ and $q$ are prime, $p>1$, $q>1$
$g1(n)=$ nbr of pairs of odd integers $p,q$ such that $p+q=2n$, $p$ or $q$ is prime, $p>1$, $q>1$
$g0(n)=$ nbr of pairs of odd integers $p,q$ such that $p+q=2n$, $p$ and $q$ are not prime, $p>1$, $q>1$

Now consider the set of pairs of positive odd integers $p,q$ such that $p+q=2n$.

p will go from: $$3 \to 2n-3$$ $q$ will go from: $$2n-3 \to 3$$ $p$ will be prime: $\pi(2n-3)-1$ times.
$q$ will be prime: $\pi(2n-3)-1$ times.

(the $-1$ is because we dont use $2$)

So the set of pairs will contain: $2\pi(2n-3)-2$ primes.

Therefore: $$2g2(n)+g1(n)=2\pi(2n-3)-2$$

Since $g1(n)$ cannot be $< 0$, an upper bound for $g2(n)$ is: $$\frac{2\pi(2n-3)-2}{2}= \pi(2n-3)-1$$

An obvious upper bound is $\pi(n/2)$.

EDIT: See OEIS sequence A002375 and references there, in particular the first entry in the FORMULA section.

• Yes, but I want to have a more powerful bound :) – Bright Chancellor Nov 21 '17 at 9:37