Let $p$ prime and define function $f$ as
$$f(n)=\sum_{p-1\mid2n}1$$
Claim
Let $\pi(n)$ shows prime counting function
Can it be shown that
$$\pi(n)\sim f(n)$$
And that there exist real number $a$ such that
$$\mid\pi(n)- f(n)\mid<a$$
I am a beginner in analytic number theory and I'm stuck here with curiosity. Thanks.
On
The function $f(n)$ behaves more like $\omega(n)$ than $\pi(n)$. For example, $\pi(n)\sim n/\log n$ and in contrast, the number of $1\leq n \leq x $ with $f(n)> (\log \log n )^2$ is $$ \ll \frac{1}{(\log \log x)^2} \sum_{n\leq x } f( n) .$$ This can be proved to be $O(x/\log\log x)$ as follows: the sum is $$ [x]+\sum_{2< p \leq 1+2x} \#\{n\leq x : n\equiv 1/2 \mod {p} \}\ll x+\sum_{p\leq 3x}x/p\ll x/\log \log x.$$
Let $n\in\mathbb{N}^*$, then $$ f(2^n)=\sum_{p-1|2^{n+1}}1 $$ But if $p$ is a prime number such that $p-1$ divides $2^{n+1}$, there exists $k\leqslant n+1$ such that $p=2^k+1$. Since $p$ is a prime number, there exists $\ell$ such that $k=2^{\ell}$ and since $k\leqslant n+1$, we have $\ell\leqslant\frac{\log(n+1)}{\log 2}$. Thus $f(2^n)\leqslant 1+\left\lfloor\frac{\log(n+1)}{\log 2}\right\rfloor$ whereas $\pi(2^n)\underset{n\rightarrow +\infty}{\sim}\frac{2^n}{n\log 2}$, thus we can't have $f(n)\underset{n\rightarrow +\infty}{\sim}\pi(n)$, and $\lim\limits_{n\rightarrow +\infty}|f(2^n)-\pi(2^n)|=+\infty$ so there doesn't exist a constant $a$ such that $|f(n)-\pi(n)|\leqslant a$ for all $n$.