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 2020-09-12, 19:39 #1 jrsousa2   Dec 2018 Miami 29 Posts The prime counting function conjecture Don't swallow me alive for this, but I was wondering if somebody can be kind enough to test this conjecture with a large enough say $x>50$ using a powerful computer: $\pi(x) \sim -16\sum _{h=1}^{\infty}\frac{x^{2h+1}}{2h+1}\sum _{i=1}^h \log\zeta(2i)\sum _{v=i}^{h}\frac{(-1)^{h-v}(4\pi )^{2h-2v}}{\zeta(2v-2i)(2h+2-2v)!} \text{, if }x\text{ is sufficiently large.}$ If you can let me know, and am willing to provide you with the typed out formula to be used in your math software. Notice this formula assumes 1 is not a prime (as it should.) The reasoning behind it can be found here. Last fiddled with by jrsousa2 on 2020-09-12 at 19:42