How many integers $\{xy \colon 1\le x\le y\le 2x\}$? Find the asymptotic upper density of the set $\{xy\, \colon\, 1\le x\le y\le 2x\}$.

In other words, let $S$ be the set of all integers which can be expressed as $xy$, for some positive integers $x,y$ such that $x\le y\le 2x$. Then evaluate
$$
\limsup_{x\to \infty}\frac{|S\cap [1,x]|}{x}
$$
 A: The asymptotic upper density is zero.
If $n = xy$ with $1 \leq x \leq y \leq 2x$ then $y^2 \leq 2n$, whence $n$
is contained in the $\sqrt{2n} \times \sqrt{2n}$ multiplication table.
Hence $|S \cap [1,X]| \leq M(\sqrt{2X})$, where for any $N$ 
we denote by $M(N)$ be the number of distinct entries in the $N \times N$ 
multiplication table (i.e. the size of $\{xy \mid 1 \leq x,y \leq N \}$.
It is known that $M(N) / N^2 \rightarrow 0$ as $N \rightarrow \infty$.
Therefore $|S\cap[1,X]| \,/\, X \rightarrow 0$ as claimed.
According to these lecture notes by Carl Pomerance,
the result $M(N) / N^2 \rightarrow 0$
is a theorem of Erdős from 1955 that follows easily from
a 1917 result by Hardy Ramanujan on the "normal order" of the 
$\Omega$ function.  Recall that $\Omega(k)$ is the number of prime factors
of $k$ with multiplicity, so $\Omega(\prod_i p_i^{a_i}) = \sum_i a_i$
(e.g. $\Omega(k)=3,2,2,1$ for $k=8,9,10,11$).  Thus
$\Omega(xy) = \Omega(x) + \Omega(y)$.  But Hardy and Ramanujan proved
that for each $\epsilon>0$ the set of integers $k$ for which
$$|\Omega(k) - \log\log k| > \epsilon \log\log k$$ has density zero.
Hence asymptotically 0% of the $N^2$ pairs $(x,y)$ with $1 \leq x,y \leq N$
yield $xy$ with $\Omega(xy) < (2-2\epsilon) \log \log N$. 
Hence asymptotically 0% of the $N^2$ pairs $(x,y)$ with $1 \leq x,y \leq N$
yield $xy$ with $\Omega(xy) < (2-2\epsilon) \log\log N$, 
while 0% of the numbers in $[1,N^2]$ have
an $\Omega$ value greater than $(1+\epsilon) \log\log N^2$.
But the loglog function grows so slowly that
$$\log\log N^2 = \log\log N + \log 2,$$
so for large $N$ (and $\epsilon < 1/3$) the $\Omega$ value cannot be
both $\geq (2-2\epsilon) \log\log N$ and $\leq (1+\epsilon) \log\log N^2$.
This proves the density-zero claim.
I found the link to the Pomerance notes in Joe Silverman's recent answer
to Mathoverflow Question
31663 from five years ago.  According to those notes, as of 2012
Kevin Ford has the sharpest results known about $M(N)$ for large $N$ 
(as Gerry Myerson hinted in his comment here, and also wrote in his
answer to that Mathoverflow question); but it is still an open
problem to obtain an asymptotic formula for $M(N)$.
A: I do not have a complete answer, but I found an upper bound and evidence that it is sharp.
Let $S(x):=\{xy : x \leq y \leq 2x \} $ and $M(z) := \vert S \cap [1,z] \vert$.
It seems that if $S(x) \cap S(y) \neq \emptyset$ then $x=y$. If this is true, my bound on $M(z)$ seems to be sharp, but if it is false, the approximation still gives an upper bound. As Peter Košinár pointed out, $72 \in S(6) \cap S(8)$. However, it seems that a similar technique might be able to count duplicates.
Suppose the above condition. Note: $\vert S(x) \vert = x+1$, $\min S(x) = x^2$, and $\max S(x) = 2x^2$.
Let $x:=w^2$ and $w \in \mathbb{Z}$. Then the $x$'s such that $S(x)$ intersects $[1,z]$ are exactly $x \in [1,w]$. 
$S(w)$ only has one element in the range, namely $z$.  
For the purposes of an upper bound, suppose that all of the elements in the rest of the sets intersect $[1,w]$.
Thus our bound $f(z) = 1 + (w-1) + (w-2) +\dots+ 2 = w(w-1)/2$, the $(w-1)$-th triangular number.
From this, we may derive $M(z) = \mathcal{O}(w(w-1)/2) = \mathcal{O}((z-\sqrt{z})/2)$ and thus $\lim_{x \to \infty} \frac{M(z)}{z} \leq \frac{1}{2}$
My main evidence is a proof that the above holds if $y:=x+1$. 
Suppose $S(x) \cap S(x+1) \neq \emptyset$. 
Then $\exists a,b \in \mathbb{Z}^+$ such that $x(x+a)=(x+1)(x+1+b)$, $0 \leq a \leq x$, and $0 \leq b \leq x+1$. 
Expanding this equation gives $ax = x(2+b)+b+1$ and rearranging gives $x(a-2-b)=b+1$. Because $a$ and $b$ are integers, we have $b+1 \bmod x \equiv 0$.
Because $b$ is less than or equal to $x+1$ and congruent to $0$ modulo $x$, $b$ must be equal to $x$. Thus we may substitute $x$ for $b$ and get $x(a-2-x)=1+x$, which, solved for $a$ is $a=(1+3x+x^2)/x$. Thus, because $a \in \mathbb(Z)$, $1+3x+x^2 \bmod x \equiv 0$ and $x$ must be equal to $1$.
Additionally, consider the following small values:
$$
S(1) = \{1,2\}\\
S(2) = \{4,6,8\}\\
S(3) = \{9,12,15,18\}\\
S(4) = \{16,20,24,28,32\}\\
S(5) = \{25,30,35,40,45,50\}\\
S(6) = \{36,42,48,54,60,66,72\}\\
S(7) = \{49,56,63,70,77,84,91,98\}
$$
(Note that $M(25)=13$ and $M(36)=18$.)
I tried generalizing the condition in a similar way to how I proved the above case, but it requires more tools than I have. Also, the assumption that $S(w-1)$ was entirely in the range should be improved for a complete proof.
This is a definitely an interesting problem and I hope this (still) helps. 
