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Consider multiplication table for numbers $1,2,\cdots, n$. How many different numbers are there? That is how many different numbers of the form $ij$ with $1 \le i, j \le n$ are there?

I'm interested in a formulae or an algorithm to calculate this number in time less than $O(n^2)$.

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A nice paper on this with plenty of references: Multiples and Divisors, by Steven Finch, available online from CiteSeer. – SJR Jul 13 2010 at 6:04

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This is the multiplication table problem of Erdos. According to Kevin Ford, Integers with a divisor in $(y,2y]$, Anatomy of integers, 65-80, CRM Proc. Lecture Notes, 46, Amer Math Soc 2008, MR 2009i:11113, the number of positive integers $n\le x$, which can be written as $n=m_1m_2$, with each $m_i\le\sqrt x$, is bounded above and below by a constant times $x(\log x)^{-\delta}(\log\log x)^{-3/2}$, where $\delta=1-(1+\log\log2)/\log2$.

Erdos' work on this problem can be found (in Russian) in An asymptotic inequality in the theory of numbers, Vestnik Leningrad Univ. Mat. Mekh. i Astr. 13 (1960) 41-49.

Another reference is http://www.research.att.com/~njas/sequences/A027424 where a PARI program is given.

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Thank you for references! The PARI program implements straightforward $O(n^2)$ algorithm. Is there any algorithm which works faster than $O(n^2)$? – falagar Jul 13 2010 at 8:43
Ford's Annals of Mathematics preprint is on the arXiv: arxiv.org/abs/math/0401223 – Charles Jul 22 2010 at 4:35

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