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I want to ask some question related with the sieve of Eratosthenes.

The sieve of Eratosthenes: write it as $E_1(x) (=\pi(x)-\pi(\sqrt x)+1)$.

Then we have an obvious result $$E_1(x)/x\ln^{-1}x = 1,$$ as $x\rightarrow \infty$ by PNT.

The question comes, we can think weight $a$ (positive integer) to each summation of the series, and write it as $E_a(x)$. (It is not a "sieve" when $a>1$.) In detail, $$E_a(x):=x- a \sum \lfloor \frac{x}{p_i} \rfloor + a^2 \sum \lfloor \frac{x}{p_i p_j} \rfloor - \cdots ,$$ for same index of the sieve of Eratosthenes.

Then the question is that : Are there some constants $c_a$ such that satisfies $$E_a(x)/x\ln^{-a}x = c_a ?$$ And, are there any papers or discussions on this function?

I'd been searched on it, but nothing found.

Thanks for reading.

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  • $\begingroup$ The problem is solved for $a=2$, immediately after I asked, by using several lemmae and the estimate of some summatory functions which are similar to the Mertens function. Additionally, as I guess, for any fixed $a\geq 2$, $E_a(x)=O(x\exp(-c\sqrt{\log x}))$ for some positive constant $c$. $\endgroup$
    – B . O
    Commented Mar 22, 2016 at 15:35

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Just as the original sieve of Eratosthenes formula is a rewriting of $$ E_1(x) = \sum_{n\le x} \prod_{\substack{p\le\sqrt x \\ p\mid n}} (1-1) = \sum_{\substack{n\le x \\ p\mid n\Rightarrow p>\sqrt x}} 1, $$ this modification is the same as $$ E_a(x) = \sum_{n\le x} \prod_{\substack{p\le\sqrt x \\ p\mid n}} (1-a) = \sum_{n\le x} (1-a)^{\omega_{\sqrt x}(n)}, $$ where $\omega_{\sqrt x}(n)$ is the number of distinct primes up to $\sqrt x$ that divide $n$. This sum should be quite similar to $\sum_{n\le x} (1-a)^{\omega(n)}$, which is classical even for $a$ a complex number (see Montgomery/Vaughan or Tenenbaum's books).

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  • $\begingroup$ Thanks for your answer, but I've obtained that representation and that's the reason of why I called it 'weight'. I think that the $k$ of $\omega_k$ changes the estimation of the arithmetic summation. $\endgroup$
    – B . O
    Commented Apr 15, 2014 at 19:44
  • $\begingroup$ It changes it, yes, but probably in a way that can be handled. After all, $\omega_{\sqrt x}(n)$ is either $\omega(n)$ or $\omega(n)-1$ in a predictable way. $\endgroup$
    – Greg Martin
    Commented Apr 15, 2014 at 20:17
  • $\begingroup$ Probably, the value changed very rapidly by $\omega(n)$ and $\omega(n)−1$ as $a$ grows up. And it seems to need some sharp techniques for deciding between them for given certain $k$, because the estimation be smaller while the weight(changeable values) bigger. I will have a try with your advise, Thanks! $\endgroup$
    – B . O
    Commented Apr 15, 2014 at 21:11

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