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I’m interested in estimating the sum $$ \sum_p \pi\left(\frac{x}{p^3}\right) $$ where the sum is over all primes $p$. (It’s a finite sum of course, you can cut off at $\sqrt[3]{x/2}$.) The goal is a decent estimate, ideally the leading asymptotic term. What is an appropriate method for solving problems like this? I’m less concerned with this particular example than the general technique.

It seems like it should be asymptotic to $P(3)\cdot x/\log x$ where P(3) = 0.1747... is the prime zeta function at 3.

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Charles
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  • Is that a Legendre symbol in parens? – Randall Dec 30 '17 at 18:19
  • @Randall No, just ordinary division. (Legendre uses primes down below, and for Jacobi p and p^3 would yield the same result.) – Charles Dec 30 '17 at 18:22
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    For primes more than $M$, use $\pi(x) < x$ so the tail is at most $x \sum_{p>M} \frac{1}{p^3} \le x \sum_{n>M} \frac{1}{n^3} \le \frac{x}{M^2}$, so you can ignore primes more than say $\log x$, which should make the leading constant indeed $P(3)$. – Rodrigo Dec 30 '17 at 19:43
  • @Charles: I am updating the answer - that old note is a little overly complicated with little gain – Eric Naslund Jan 05 '18 at 20:20
  • @Charles: Done - that should be clearer now – Eric Naslund Jan 05 '18 at 20:34

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Edit 1: Looking back at that note, which I wrote $6$ years ago, it's unclear and overly technical. Since $\pi(x)=\frac{x}{\log x}+O\left(\frac{x}{(\log x)^2}\right),$ we have that $$\sum_{p\leq (\log x)^2}\pi\left(\frac{x}{p^3}\right)\sim \frac{x}{\log x}\sum_{p\leq (\log x)^2}\frac{1}{p^3}+O\left(\frac{x\log \log (x)}{(\log x)^2}\right).$$ Now, $$\sum_{p\leq (\log x)^2}\frac{1}{p^3}=P(3)+O\left(\frac{1}{(\log x)^4}\right),$$ so all that remains is to bound the sum $$\sum_{(\log x)^2\leq p\leq x^{1/3}}\pi\left(\frac{x}{p^3}\right)$$ from above, which can be done in a straightforward way with only the bound $\pi(x)=O(x/\log x)$, and we conclude that $$\sum_{qp^3\leq x} 1 \sim P(3)\frac{x}{\log x}.$$

Edit 2: Here's a different convoluted way to do it: Sequence of numbers with prime factorization $pq^2$

This can be proven using partial summation. See this note I wrote for the full details. The sum $$\sum_{p} \pi\left(\frac{x}{p^3}\right)=\sum_{p\leq x} \sum_{q\leq \frac{x}{p^3}}1=\sum_{qp^3\leq x} 1$$ counts the number of integers $n\leq x$ such that $n=qp^3$ where $q$ and $p$ are both prime (and not-necessarily distinct). Let's call this $\sigma_{(1,3)}(x)$

You are correct that $$\sigma_{(1,3)}:=\sum_{p} \pi\left(\frac{x}{p^3}\right)\sim P(3)\frac{x}{\log x}.$$ More generally, consider the number of integers of the form $n=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ where $1=\alpha_1\leq \alpha_2\leq \dots\leq \alpha_k$ and $1=\alpha_1=\cdots=\alpha_r<\alpha_{r+1}$. Call this $\sigma_{(\alpha_1,\dots,\alpha_r)}$. Then

Theorem: We have that $$\sigma_{(\alpha_1,\dots,\alpha_r)}\sim \frac{x(\log \log x)^{r-1}}{\log x}\prod_{i=r+1}^k P\left(\alpha_i\right).$$

This all appears in the note I linked to above.

For example, this means that the number of integers of the form $n=pqr^2s^3$ less than $x$, where $p,q,r,s$ are prime, is asymptotic to $$\frac{x\log \log x}{\log x} P(2)P(3).$$

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