On the numbers divisible by all the Integers not exceeding their $r^{th}$ roots. Consider the set of all numbers which are divisible by all natural numbers not exceeding their square root, and denote this set by $S_2=\{1,2,3,4,6,8,12,24\}$ (Here the subscript indicates that we're taking the 2nd root of the numbers). Thus $|S_2|=8$.
Similarly, the set of all numbers which are divisible by all natural numbers not exceeding the cube root is $S_3 = \{1,2,3,4,5,6,7,8,10,12,14,16,18,20,22,24,26,30,36,42,48,54,60,72,84,96,108,120,‌​180,240,300,420\}$, with $|S_3|=32$.
Now define $S_r$ similarly as the set of all positive numbers divisible by all the naturals not exceeding their $r^{th}$ roots.  Then I have the folowing questions:
Q-1 What is the general formula for finding $|S_r|$ (ie. Cardinality of $S_r$)?
Q-2 Is there an expression for the greatest element of $S_r$?
Asymptotics will also be encouraged.
 A: I don't have asymptotics, but I coded this quick program
a(n)=my(s,old,cur,mult=1,k);while(mult<=2*(cur=(k+++1)^n-1)+9,mult=lcm(mult,k);s+=cur\mult-old\mult;old=cur);s

in PARI/GP and it finds $|S_1|=2, |S_2|=8, |S_3|=32, |S_4|=149, \ldots, |S_{1000}|\approx1.8077655\cdot10^{2571}$
which suggests that $|S_n|$ grows at roughly a factorial rate.
A: It's quite easy to calculate if you turn things around: Divisible by 1 = every number. Divisible by 2 = multiple of 2. Divisible by 1, 2, 3 = multiple of 6. Divisible by 1 to 4 = multiple of 12. Divisible by 1 to 5 or 1 to 6 = multiple of 60 etc. 
You don't really need PARI/GP, it's quite trivial: For n from 1 upwards you calculate L(n) = "LCM of the numbers 1 to n". L (1) is 1 obviously, and you multiple the previous value by p if n is a power of the prime p. The result is not far from $e^n$. 
Then you calculate how many of the numbers $n^r <= x < (n+1)^r$ are divisible by L (n), which is $((n+1)^r - 1) / L(n) - (n^r - 1) / L(n)$, both divisions rounded down. $n^r$ grows only polynomial, not exponential, so this will become zero quite soon. (For example n = 41 when r = 10). 
$n^r/e^n$ has its maximum roughly around n = r, where $n^r/e^n = r^r/e^r$ which is not far from Stirling's formula for r!. So very roughly the result is about r!. 
