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I would like to know whether, given a positive integer $m$, the quantity $\displaystyle{\sum_{p_{i}\leqslant\sqrt{m}}\lfloor\frac{m}{p_{i}}\rfloor-\sum_{p_{i}<p_{j}\leqslant\sqrt{m}}\lfloor\frac{m}{p_{i}p_{j}}\rfloor+\sum_{p_{i}<p_{j}<p_{k}\leqslant\sqrt{m}}\lfloor\frac{m}{p_{i}p_{j}p_{k}}\rfloor}-\dots$ giving the number of positive integers below $m$ divisible by the primes below $\sqrt{m}$ through the inclusion-exclusion principle is always the nearest integer to the same expression without the $\lfloor\ \ \rfloor$ symbol (which equals $\displaystyle{m\left(1-\prod_{p\leqslant\sqrt{m}}\left(1-\dfrac{1}{p}\right)\right)}$ if I'm not mistaken).
Thanks in advance.

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  • $\begingroup$ I'm afraid without step function it does not work, since you could add/subtract relatively big rationals and the error you commit is usually bigger than 1, so taking the nearest integer doesn't work. $\endgroup$ – PITTALUGA Oct 7 '13 at 16:18
  • $\begingroup$ I see. Do we know an upper bound depending on $m$ for the absolute value of the difference of these two expressions? $\endgroup$ – Sylvain Julien Oct 7 '13 at 16:27

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