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: https://www.researchgate.net/publication/333393877_An_Exact_Formula_for_the_Prime_Counting_Function.