I was coding an algorithm that calculates $g(n)$, the number of decomposition of 2n into ordered sums of two odd primes (A002372), or the number of Goldbach partitions. I noticed i can express the algorithm as a sum involving the prime-counting function $\pi(n) $ and the nth prime $p(n)$
$$g(n) = \sum_{i=2}^{\pi(2n)} \pi(2n-p(i))-\pi(2n-1-p(i))$$
This sum gives the exact value of $g(n)$, so a lower bound to this sum would prove the Goldbach conjecture. Since the sum is entirely prime-related, shouldn't it be possible to get a lower bound using the prime number theorem?
Example:
$$g(n) = \sum_{i=2}^{\pi(2n)} \pi(2n-p(i))-\pi(2n-1-p(i))$$
Here, in order to get the lowest value for $g(n)$, we want $p(i)$ to be small, because the density of primes decreases as $n$ grows. We also want $\pi(x)$ to be small, so we use:
$$\pi(x) \gt \frac{x}{ln(x)}$$
$$p(i) \gt iln(i)$$
to get:
$$g(n) \gt \sum_{i=2}^{\frac{2n}{ln(2n)}} \frac{2n-iln(i)}{ln(2n-iln(i))}-\frac{2n-1-iln(i)}{ln(2n-1-iln(i))}$$
