Goldbach's conjecture and number of ways in which an even number can be expressed as a sum of two primes Is there a functon that counts the number of ways in which an even number can be expressed as a sum of two primes?

Let's define g(n) as the number of decompositions of $2n$ into ordered sums of two odd primes (A002372)
The following sum gives the exact value for g(n):
$$g(n) = \sum_{i=2}^{\pi(2n)} \pi(2n-p(i))-\pi(2n-1-p(i))$$ 
where $p(n)$ is the nth prime and $\pi(n)$ is the prime-counting function.
Now if you are looking for something that will work without knowing the primes up to $n$, we can get a pretty good estimate from the sum above. Let's consider the sum of all $g(i)$ from $i=3\to n$:
$$ \sum_{i=3}^{n} g(i)$$
From the first sum, we can get:
$$\sum_{i=3}^{n} g(i) = 1+\sum_{i=3}^{\pi(2n)} \pi(2n-p(i))$$
If we replace $\pi(n)$ with $\frac{n}{ln(n)}$ and $p(n)$ with $nln(n)$, we know from the prime number theorem that the bigger the interval $n-3$ gets, the more precise our estimation gets:
$$\sum_{i=3}^{n} g(i) \approx  1+\sum_{i=3}^{\frac{2n}{ln(2n)}} \frac{2n-iln(i)}{ln(2n-iln(i))}$$
Since:
$$ g(n) = \sum_{i=3}^{n} g(i) - \sum_{i=3}^{n-1} g(i)$$
$$ g(n) \approx  (1+\sum_{i=3}^{\frac{2n}{ln(2n)}} \frac{2n-iln(i)}{ln(2n-iln(i))})-(1+\sum_{i=3}^{\frac{2(n-1)}{ln(2(n-1))}} \frac{2(n-1)-iln(i)}{ln(2(n-1)-iln(i))})$$
$$ g(n) \approx  \sum_{i=3}^{\frac{2n}{ln(2n)}} \frac{2n-iln(i)}{ln(2n-iln(i))}-\sum_{i=3}^{\frac{2n-2}{ln(2n-2)}} \frac{2n-2-iln(i)}{ln(2n-2-iln(i))}$$

The results is quite surprising, yet not very practical. The value of the estimation seems to behave like the real g(n), only in a more regular way:


