I was wondering if somebody can be kind enough to test this conjecture with a large enough, say x>50, using a powerful computer and Mathematica:
\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.}
Note this formula assumes 1 is not a prime, as it should. Notice that x=50 will require increasing the precision and the number of terms, M.
$MaxExtraPrecision = 500; M = 230; x = 12;
N[{PrimePi[x], x1 = -16*Sum[(x^(2*h + 1)/(2*h + 1))*
Sum[Log[Zeta[2*i]]*Sum[((-1)^(h - v)*(4*Pi)^(2*h - 2*v))/
(Zeta[2*v - 2*i]*(2*h + 2 - 2*v)!), {v, i,
h}], {i, 1, h}],
{h, 1, M}]}, 10]
The reasoning behind it can be found here.