$f(x) + f(x/2) + f(x/3) + ... = x$ and conjecture A : $\pi(x/a) > \frac{\pi(x)}{a}$

Question

Consider $f(x)$ :

$f(x) + f(x/2) + f(x/3) + f(x/4) + ... = x$ and $\lim_{n \to \infty} \frac{f(n)}{\pi(n)} = 1$?

and the edit that contains conjecture A :

$$\pi(x/a) > \frac{\pi(x)}{a}$$

Is there a name for conjecture A ?

Is there a proof or good argument ?

Any references or so ?

Answer 1

Here is a "solid ground" for a proof ...

From the Dusart's result, for $\forall x \ge 60184$ we have $$\frac{x}{\log{x}}<\frac{x}{\log{x}-1} < \pi(x) <\frac{x}{\log{x} - 1.1}$$

Now, $\forall a \ge e^{1.1}$ and $a<x$ we have $$\pi(x) <\frac{x}{\log{x} - 1.1}<\frac{x}{\log{x} - \log{a}}=a\cdot\frac{\frac{x}{a}}{\log{\frac{x}{a}}}$$ or $$\frac{\pi(x)}{a}<\frac{\frac{x}{a}}{\log{\frac{x}{a}}}<\pi\left(\frac{x}{a}\right)$$

A computer program could check the remaining $\forall x < 60184$.

Answer 2

Not an answer but maybe useful :

if

$$\pi(x/a) > \frac{\pi(x)}{a}$$

and

$$\pi(x/b) > \frac{\pi(x)}{b}$$

then

$$\pi(\frac{x}{ab}) > \frac{\pi(x)}{ab}$$

This might simplify the proof.