Let $x\geq 100$ then we have as conjecture :
$$\frac{\pi{(x)}}{\pi{(\pi{(x)}})}<\ln(x)$$
I have tested at $x=100$ to $x=5000000000$ without any counter-example.
The first fact :
It seems that the function :
$$f(x)=\frac{\pi{(x)}}{\pi{(\pi{(x)}})}-\ln(x)$$
Can be associated to a decreasing function $h(x)$ for $x\geq 100$.I have no informations on $h(x)$.
To show it I have tried to use the PNT wich states :
$$\lim_{x\to \infty}\frac{\pi(x)}{\frac{x}{\ln(x)}}=1$$
But this is an asymptotic information and I don't see any good outcome .
I have also a stronger statement :
Let $x\geq 10^{10}$ then we have :
$$\frac{\pi{(x)}}{\pi{(\pi{(x)}})}<\ln(x)-\ln(\ln(x))-1$$
But I have serious doubt on this second conjecture.
Edit :
Now I have an equality wich is also a conjecture :
Let $x\geq 10^{10}$ then we have :
$$\lfloor\frac{\pi{(x)}}{\pi{(\pi{(x)}})}\rfloor\pm 0.5=\lfloor \ln(x)-\ln(\ln(x))-1\rfloor\pm 0.5$$
Warning with the sign $\pm$ we can have 4 possibilities not simultaneously of course!
Question:
Can you prove or disprove it ? Is it well-know ?
Thanks in advance for your effort !
To prove the main inequality, which can be rewritten as $\pi(\pi(x))\ln x > \pi(x)$, we can use explicit bounds on $\pi(x)$. For example, we know that $\pi(u) > u/\ln u$ when $u\ge17$; therefore when $x\ge59$ (the $17$th prime), $$ \pi(\pi(x))\ln x > \frac{\pi(x)}{\ln\pi(x)}\ln x > \pi(x). $$ More precise inequalities (including these) will in principle determine whether the second conjecture is true.