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?


$$g(n) = \sum_{i=2}^{\pi(2n)} \pi(2n-p(i))-\pi(2n-1-p(i))$$

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: $$g(n) \gt \sum_{i=2}^{\frac{2n}{ln(2n)}} \frac{2n-iln(i)}{ln(2n-iln(i))}-\frac{2n-1-iln(i)}{ln(2n-1-iln(i))}$$

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  • $\begingroup$ The prime number theorem is an asymptotic. $\pi(x) \sim \frac{x}{\log x}$ means $$\lim_{x \to \infty} \frac{\pi(x)}{x / \log x} =1,$$ so you can't use this for every term in the sum. It's only for large $x$. (And directly applying the asymptotic in the large terms makes them 0!) $\endgroup$ – Dzoooks May 26 '18 at 22:35
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    $\begingroup$ Heuristics are no proof. Of course, almost all (all ?) mathematicians are convinced that the Goldbach-conjecture is true because of overwhelming statstical evidence. But we cannot conclude because of that that Goldbach's conjecture must be true. We cannot even rule out a counterexample near, lets say, $10^{30}$. The prime number theorem is known already very long. It is unlikely that such an approach (if successful) would have been overseen. $\endgroup$ – Peter May 27 '18 at 8:51
  • $\begingroup$ I agree. I find this sum very interesting though. $$g(n) = \sum_{i=2}^{\pi(2n)} \pi(2n-p(i))-\pi(2n-1-p(i))$$Mostly because i haven't found anything similar on the web yet. This conjecture is hundreds of years old, and this is g(n), expressed as a simple sum involving the prime-counting function and the nth prime. It is unlikely that this sum was never mentioned in any work. $\endgroup$ – François Huppé May 27 '18 at 10:30

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