As we know from Riemann (by means of Derbyshire): $$ \pi(x) = J(x) - \frac{1}{2}J\left(\sqrt{x}\right) - \frac{1}{3} J\left(\sqrt[3]{x}\right) - \frac{1}{5} J\left(\sqrt[5]{x}\right) + \frac{1}{6} J\left(\sqrt[6]{x}\right) - \frac{1}{7}J\left(\sqrt[7]{x}\right) + \frac{1}{10}J\left(\sqrt[10]{x}\right) \ldots $$ And we recall that $J(x) = 0, x<2$ so we have absolute convergence (for $x>1$).
The question is about the number of terms, as a function $f$ of $x$. In prose: $f(x)$ is the smallest $i$ that makes $\sqrt[\large i]{x} < 2$. Please excuse me for not knowing how to state that without using prose.
The figure shows the number of terms, with one logarithmic scale and $f(10^{99})=329$. On the other hand, $\ln 10^{99}$ is $230$.
The question is: Is $O(f(x)) = \log x$ or above?
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