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On complexity of the sieve, start crossing at the square of p - Part II

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Ok, this is a follow up question to this, about the Erathostenes Sieve.

If we look at this paper (page 3, footnotes), the author says:

If we start crossing off at $p^2$ rather than $p$, the number of composites we cross off is $\frac{n}{p}−p+1$, but it makes no significant difference to our sum, because it only subtracts an irrelevant $O(n/ \log n)$ factor

The total cost counting crossings is:

$$\sum_{p\leq n}\bigl(\frac{n}{p}-p+1\bigr)=n\sum_{p\leq n}\frac{1}{p}-\sum_{p\leq n}p+\sum_{p\leq n}1$$

The first sum would give us the known $O(n\log\log n)$. But I don't see the second sum to be an "irrelevant $O(n/\log n)$", so how is that?

According to this post in math overflow an upper bound for the sum of the primes up to $n$ is: $$\frac{n^2}{2\log n} + O\left(\frac{n^2}{\log^2 n}\right).$$

Asked By : Wyvern666
Answered By : Yuval Filmus

The sum only goes up to $\sqrt{n}$. You can see that in the big display at the bottom of page 3, where only the first $\pi(\sqrt{n})$ primes are considered. Substituting $\sqrt{n}$ for $n$ in your formula, we get $n/\log n + O(n/\log^2 n)$.

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