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Let $A_n=\{a\cdot b : a,b \in \mathbb{N}, a,b\leq n\}$. Are there any estimates for $|A_n|$? Will it be $o(n^2)$?

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Don't be so quick in closing this question. The estimate in Eric's answer is quite non-trivial. – Felipe Voloch Oct 5 '12 at 17:42
Isn't being an elementary, simple, difficult question motivation enough? – Will Sawin Oct 6 '12 at 5:13
Which of those three adjectives is immediately obvious? – François G. Dorais Oct 6 '12 at 6:56
I upvoted Voloch's comment yet not the question. I agree this is not an optimal example how to write an MO question, but it is a quite precise mathematical question; what saves it for me is the second question, asking specifically for o(n^2), giving a quite clear idea what type of estimates the OP is after. Also it is tagged very well. And searching for it in the literature without a starting point could be tricky. That it is not easy can be tested by trying to solve it. And (legitimately) the only motivation might well be 'it seems like a natural problem and I could not do it' – quid Oct 6 '12 at 8:52
Seems to be a duplicate of… – Gerry Myerson Apr 28 at 23:39

3 Answers 3

up vote 62 down vote accepted

This question is known as the multiplication table problem, and was originally posed by Erdős in 1955. Erdős proved that $|A_n|=o(n^2)$, and this was sharpened by Tenenbaum in 1984. In 2008, Ford gave the exact magnitude and proved that $$\left|\lbrace a\cdot b:\ a,b\leq N\rbrace\right|\asymp \frac{N^2}{(\log N)^c(\log\log N)^{3/2}},$$ where $$c=1-\frac{(1+\log \log 2)}{\log 2}.$$

In 2010 Koukoulopoulos gave multidimensional generalizations of Ford's result, proving that $$\left|\lbrace a_1\cdots a_{k+1}\ :\ a_i\leq N \text{ for all } \ i\rbrace\right|\asymp \frac{N^{k+1}}{(\log N)^{c_k}(\log\log N)^{3/2}},$$ where $$c_{k}=\int_{1}^{\frac{k}{\log(k+1)}}\log x\text{d}x=\frac{\log(k+1)+k\log\left(k\right)-k\log\log(k+1)-k}{\log(k+1)}.$$

Some references:

  • Ford 2008: The distribution of integers with a divisor in a given interval. arXiv link

  • Koukoulopoulos 2010: Localized Factorization of Integers. arXiv link

  • Koukoulopoulos 2012: On the number of integers in a generalized multiplication table. arXiv link

Remark: The dates used above refer to the publication dates (not necessarily the date posted to the arXiv).

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Presumably $n=N$. – Felipe Voloch Oct 5 '12 at 17:40
@Felipe: Fixed, thank you. – Eric Naslund Oct 6 '12 at 1:12
If all you're looking to show is $o(n^2)$, then I believe Erdos' original argument can get you there somewhat faster. As a very rough sketch, the idea is: 1. By a result of Hardy and Ramnujan, almost all integers between $1$ and $n$ have roughly $\ln \ln (n^2)=\ln \ln n+O(1)$ prime factors. 2. Almost all pairs $(x,y)$ have approximately $2\ln \ln n$ prime factors dividing $xy$ (since $x$ and $y$ usually won't share many factors). 3. Since most products lie in only a small subset of $\{1,…,n^2\}$ (the numbers having an unusually large number of factors), most of the rest must remain uncovered. – Kevin P. Costello Oct 6 '12 at 4:44
@Kevin P. Costello Do you happen to know a book or survey paper covering these types of results as the one of Hardy and Ramanujan about almost all integers and their prime factors? – Jernej Oct 6 '12 at 8:08
@jernej: There's a very nice small book by Mark Kac about the number of prime factors of integers. Otherwise, try Hardy and Wright. – Anthony Quas Jun 25 '13 at 8:03

Let me give here an answer with a quick argument why it is $o(n^2)$. I do not know whether it is the same as Erdős original proof. UPD: it really is, and is mentioned above in a comment by Kevin P. Costello.

Most numbers from 1 to $n$ have $\log \log n (1+o(1))$ prime divisors (counted with multiplicity) by Erdős-Kac theorem. Then most their products have $2\log \log n (1+o(1))$ prime factors, while most numbers from 1 to $n^2$ have again $\log \log n (1+o(1))$ prime factors. It proves that products of two numbers from 1 to $n$ are rare in $\{1,2,\dots,n^2\}$

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This is Kevin P. Costello's comment:… – Eric Naslund Aug 14 at 14:20
ops, indeed, I have to be more careful – Fedor Petrov Aug 14 at 14:27
But let me leave it here, as it is easier to see an answer than a comment and it has micro-advantage by counting the number of prime divisors with multiplicity, which makes this function satisfying $f(xy)=f(x)+f(y)$ without errors. – Fedor Petrov Aug 14 at 14:33

The answer is yes, for further infos see the references given at the On-Line Encyclopedia of Integer Sequences.

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