What is the cardinality of the set of values corresponding to the first $n$ rationals generated in Cantor's enumeration scheme for proving their countability?


Edit:
following the [suggestion](https://mathoverflow.net/questions/204536/redundancy-of-the-cantor-enumeration-of-the-rationals#comment508412_204536) of Todd, here I integrate the clarification from the comments here:
By "Cantor's enumeration scheme" I mean the one, that Cantor originally used (I am however not aware, which that was, so that would also be part of the question).  

In case of doubt, sort the fractions according to numerator+denominator or, in case of equal sum, accoding to numerator/denominator in ascending order in both cases.


As the rationals, whose value is encountered for the first time, are those for which numerator and denominator are relatively prime and because the probability that two natural numbers are relatively prime, is $\frac{6}{\pi^2}$, it would not be suprising if the number of different values encountered in the first $n$ fractions of the Cantor sequence would approach $n\frac{6}{\pi^2}$, so the interesting question is, whether the exact value of the factor for a given $n$ can be calculated or at least the asymptotic growth.



Redundancy of the Cantor enumeration of the rationals
I'm not sure, but it seems to me that you are asking about an asymptotic formula for the cardinality of the set $\left\{\frac{n}{m} : n + m \leq x\right\}$, as $x \to +\infty$, where $n$ and $m$ are positive integers. If so, the answer is the following


$$\#\left\{\frac{n}{m} : n + m \leq x\right\} = \sum\_{k \leq x} \sum\_{\substack{n + m = k \\ \gcd(n,m) = 1}} 1 = \sum\_{k \leq x} \sum\_{\substack{n \leq k - 1 \\ \gcd(n,k) = 1}} 1$$
$$= \sum\_{k \leq x} \varphi(k) - 1 = \frac{3}{\pi^2} x^2 + O(x\log x),$$
as $x \to +\infty$, see *F. Mertens, Uber einige asymptotische Gesetze der Zahlentheorie. Crelle’s Journal 77 (1874), 289–338*.


