# Upper bound of prime counter function

I am trying to prove a relatively simple bound for the problem below. I want someone to check if my solution is good :)

For a $$P=\{ p_1,p_2,...,p_k \}$$ a set of prime factors, we define as $$N_P= \{ n\in \mathbb{N}: p|n \Rightarrow p\in P \}$$. We also define as $$N_P(x)= \# \{n\in N_P : n \leq x \}$$.

I want to prove that exist a constant $$C$$ and a $$x_0$$, which depend on $$P$$, such that: $$N_P(x) \leq C (\log x)^k \; \text{for x \geq x_0}$$

My attempt:

My first idea was to use somehow use the Prime Number Theorem (PMT): $$\pi(x) \sim x/\log x$$, for a sufficiently large $$x$$. But since the $$\log$$ is on the denominator, I think this is hopeless.

So I made this argument: Let $$n\in N_P$$, then $$n$$ has prime factorization that contains only primes from the set $$P$$.

Thus $$n=p_1^{n_1}p_2^{n_2}...p_k^{n_k}$$. For $$n \leq x$$, we must have that: $$p_i^{n_i} \leq x \Rightarrow n_i \leq \frac{\log x}{\log p_i}$$ for $$\forall i$$.

Thus for $$x\geq p_k$$ $$N_p \leq \frac{\log x}{\log p_1} \frac{\log x}{\log p_2} ... \frac{\log x}{\log p_k} = \frac{1}{\log p_1 \log p_2 ... \log p_k} (\log x)^k = C (\log x)^k$$ for $$x\geq x_0=p_k$$.

What do you think? Thanks in advance for your time!

• This... looks reasonable to me. I'm not 100% certain; others may have quibbles. Commented Jan 20, 2023 at 0:32
• "a set of prime factors"? factors of what? "a set of prime numbers" Commented Jan 20, 2023 at 0:32
• Why do you think you have to consider $x \ge p_k$? Commented Jan 20, 2023 at 0:37
• Write $n=\prod_j p_j^{e_j}$, if $n\le x$ then $e_j \le \log_{p_j}(x)$ so there are at most $f(x)=\prod_j (1+\log_{p_j}(x))$ choices for the $e_j$ ie. at most $f(x)$ integers $\le x$ whose all prime factors are in the $p_j$. Commented Jan 20, 2023 at 0:38
• A detail about MathJax: Don't write log\; x ($log\; x$) write \log x ($\log x$). Commented Jan 20, 2023 at 0:47

After you get $$n_i \le \frac{\log x}{\log p_i}$$ you need to consider that $$n_i$$ can take $$\left\lfloor \frac{\log x}{\log p_i}\right\rfloor \color{red}{+1}$$ values, because $$n_i$$ can be $$0$$.
• Hardy discussed the case $k=2$ when discussing a result of Ramanujan. I think the general result was done by Landau. Commented Jan 20, 2023 at 1:23