The $n^\text{th}$ term in the sequence $\frac{1}{1},\frac{1}{2},\frac{2}{1},\frac{1}{3},\frac{2}{2},\frac{3}{1},\frac{1}{4},\frac{2}{3},\dots$ $\begin{array} {|r|r|}\hline R\text{ \ }C & 1 & 2 & 3 & \dots \\ \hline 1 & \frac{1}{1} & \frac{1}{2} & \frac{1}{3} &  \\ \hline 2 & \frac{2}{1} & \frac{2}{2} & \frac{2}{3} &  \\ \hline 3 & \frac{3}{1} & \frac{3}{2} & \frac{3}{3} &  \\ \hline \vdots &  &  &  & \ddots \\ \hline  \end{array}$
Considering the sequence of fractions that is made by moving diagonally, with respect to the entries above, in the following manner:
$$\frac{1}{1},\frac{1}{2},\frac{2}{1},\frac{1}{3},\frac{2}{2},\frac{3}{1},\dots$$
where the entries are obtained by dividing the value of the row $R$ by the value of the column $C$.
So the next few terms are $\frac{1}{4},\frac{2}{3},\frac{3}{2},\frac{4}{1},\frac{1}{5},\dots$
Now I have two question:


What is the $n^\text{th}$ term in the sequence (including all
  fractions)?



What is the $n^\text{th}$ term in the sequence (excluding repeated
  values)?

For example, $\frac{4}{2}$ has to be excluded since $\frac{2}{1}$ is already considered. Similarly, $\frac{2}{2},\frac{3}{3},\frac{4}{4},\dots$ are to be excluded since $\frac{1}{1}$ is already there.

What I observed is; for any natural number $m$, the $(2m^2+2m+1)^\text{th}$
term is to be excluded since that term is equal to $1$ which is already considered.
Also, it is easy to find an expression in $m$ to exclude other fractions
I have no idea to solve these two questions.

Any help would be appreciated. THANKS!
 A: § 1 Frame of reference
It is useful to adopt a two index identification of the fractions.
Decompose the fractions into groups according to a scheme of numerators and denominators as shown for the first 4 groups below
$$(1)(1 2)(1 2 3)(1 2 3 4)...$$
$$(1)(2 1)(3 2 1)(4 3 2 1)...$$
The position $p_s(k)$ of the start of the k-th group in a linear sequence of fractions is given by 
$$p_s=1+\frac{1}{2} k(k-1), k=1, 2, …\tag{1}$$
which is the sequence $1, 2,4,7,11,16,...$.
We can write the fractions in group $k$ as
$$f(k,m) = \frac{m}{k-m+1}, m=1..k\tag{2}$$
The position of fraction $f(k,m)$ in the sequence of all fractions is given by 
$$p(k,m) = p_s+m-1=m+\frac{1}{2} k(k-1) \tag{3}$$
This can be inverted by
$$k(p)=\lfloor \left( \frac{1}{2}+\sqrt{\frac{1}{4}+2p-2} \right) \rfloor \tag{4a}$$
$$m(p) =p-\frac{k(p)}{2}(k(p)-1)\tag{4b}$$
Here $\lfloor . \rfloor$ is the floor function.
§ 2 Drop repeated fractions
If we wish to avoid double counting we have to check if the fraction is reducible. In other words we have to drop fractions with $GCD(m, k-m+1)\gt 1$.
I don't know how to express this condition in an explicit formula. Maybe others can.
Here is a related result:
Let $r(n)$ be the number of different rationals in our set of groups of fractions up to the element with the fraction $\frac{n}{1}$ 
Then $r(n)$ can be found by direct calculation to start like this
$$r(n) = {1, 3, 5, 9, 11, 17, 21, 27, 31, 41, 45, 57, 63, 71, 79, 95, 101}$$
This sequence is easily found in OEIS: http://oeis.org/A015614. And the formula is
$$r(n) =  -1 + \sum_{i=1..n} \phi(i)\tag{5}$$
where $\phi(i)$ is Euler's totient function (https://en.wikipedia.org/wiki/Euler%27s_totient_function).
The asymptotic behaviour is
$$r(n\to \infty) \simeq \frac{3}{\pi^2}n^2+O(n \log(n))$$
Since the number of primarily different fractions is $1+2+3+...+n = \frac{1}{2}n(n+1)$ the percentage of different fractions is given asymptotically by $\frac{6}{\pi^2} \simeq 0.607927$.
