I know there are several bounds for $\pi(x)$, which is the prime counting function. But for my thesis I want to calculate a very rough estimation. I'd appreciate any feedback on my calculation. So I have the proth number $n = h\cdot2^{k}+1$ where $h<2^k$. I want to know how many primes (again: Very roughly) I had to check if I did trial division. So I need a lower bound for the amount of primes up to $\sqrt{n}$. I use the formula
$$ \pi(x) \geq \frac{x}{\log(x)-1}. $$
I have
$$ \frac{\sqrt{h\cdot 2^k+1}}{\log(\sqrt{h\cdot 2^k+1)}-1} \geq \frac{\sqrt{h\cdot 2^k}}{\frac{1}{2}\log(h\cdot 2^k+1)} = \frac{2\sqrt{h}\cdot 2^\frac{k}{2}}{\log(h\cdot 2^k+1)} = \frac{\sqrt{h}\cdot 2^{\frac{k}{2}+1}}{\log(h\cdot 2^k+1)} > \frac{\sqrt{h}\cdot 2^{\frac{k}{2}+1}}{\log(h\cdot 2^{k+1})} $$ Now I use $h<2^k$: $$ \frac{\sqrt{h}\cdot 2^{\frac{k}{2}+1}}{\log(h\cdot 2^{k+1})} > \frac{\sqrt{h}\cdot 2^{\frac{k}{2}+1}}{\log(2^{2k+1})} > \frac{\sqrt{h}\cdot 2^{\frac{k}{2}+1}}{\log(2^{2k+2})} = \frac{\sqrt{h}\cdot 2^{\frac{k}{2}+1}}{(k+1)\log(4)} > \frac{\sqrt{h}\cdot 2^{\frac{k}{2}+1}}{(k+1)\cdot 2} = \frac{\sqrt{h}\cdot 2^{\frac{k}{2}}}{(k+1)} $$
So for $h = 2^{5001} + 5473$ and $k=10.000$ I receive $$ \pi(x) > 2^{7486} \geq 10^{2253}. $$