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 $o(n^2)$?

21$\begingroup$ Don't be so quick in closing this question. The estimate in Eric's answer is quite nontrivial. $\endgroup$ – Felipe Voloch Oct 5 '12 at 17:42

13$\begingroup$ Isn't being an elementary, simple, difficult question motivation enough? $\endgroup$ – Will Sawin Oct 6 '12 at 5:13

4$\begingroup$ Which of those three adjectives is immediately obvious? $\endgroup$ – François G. Dorais♦ Oct 6 '12 at 6:56

13$\begingroup$ 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' $\endgroup$ – user9072 Oct 6 '12 at 8:52

5$\begingroup$ Seems to be a duplicate of mathoverflow.net/questions/31663/… $\endgroup$ – Gerry Myerson Apr 28 '15 at 23:39
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).


17$\begingroup$ 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. $\endgroup$ – Kevin P. Costello Oct 6 '12 at 4:44

3$\begingroup$ @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? $\endgroup$ – Jernej Oct 6 '12 at 8:08

2$\begingroup$ @jernej: There's a very nice small book by Mark Kac about the number of prime factors of integers. Otherwise, try Hardy and Wright. $\endgroup$ – Anthony Quas Jun 25 '13 at 8:03

$\begingroup$ @FedorPetrov: Thank you for the correction  it is now fixed. $\endgroup$ – Eric Naslund Aug 14 '15 at 13:10
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 ErdősKac 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 $\{1,2,\dots,n^2\}$

2$\begingroup$ This is Kevin P. Costello's comment: mathoverflow.net/questions/108912/… $\endgroup$ – Eric Naslund Aug 14 '15 at 14:20


1$\begingroup$ But let me leave it here, as it is easier to see an answer than a comment and it has microadvantage by counting the number of prime divisors with multiplicity, which makes this function satisfying $f(xy)=f(x)+f(y)$ without errors. $\endgroup$ – Fedor Petrov Aug 14 '15 at 14:33
The answer is yes, for further infos see the references given at the OnLine Encyclopedia of Integer Sequences.