<s>[INST] Is there a closed form of this sequence? The sequence is  $\frac{1}{1}$ , $\frac{1}{2}$ , $\frac{2}{1}$ , $\frac{1}{3}$ , $\frac{2}{2}$ , $\frac{3}{1}$ , $\frac{1}{4}$ , $\frac{2}{3}$ , $\frac{3}{2}$ , $\frac{4}{1}$ , ...
                   and I need to find n th..
Here's my approach..
I bound them with who has same sum of denominator and numerator.
So $\frac{1}{1}$ is set_1 . 
$\frac{1}{2}$ , $\frac{2}{1}$ are set_2..  
$\frac{1}{3}$ , $\frac{2}{2}$ , $\frac{3}{1}$ are set_3 . 
$\frac{1}{4}$ , $\frac{2}{3}$ , $\frac{3}{2}$ , $\frac{4}{1}$ are set_4 and so on..
If I suppose that n is in set_K , I can conclude n th number is $$ \frac{1+( (n-1)-\sum_{i=1}^K (i-1) ) }{ K - ( (n-1)-\sum_{i=1}^K (i-1) ) } $$
( because set_K starts with $\frac{1}{K}$ and n is ( n - $\sum_{i=1}^K (i-1)$ ) th member of set_K.. ) 
As a result my conclusion is $$\frac{2n-(K-1)K}{K(K+1) - 2(n-1) }$$ but I don't know the relationship between n and K.. 
Is there anyone to help me? I really want to know.. :(
 [/INST] The $n$-th term can be expressed as ${\large{\frac{a}{b}}}$, where 
\begin{align*}
a&=n-\frac{k(k+1)}{2}\\[4pt]
b&=k+2-a\\[4pt]
k&=
\left\lfloor
{
\frac
{
\left\lfloor
\sqrt{8n-7}
\right\rfloor
-1
}
{2}
}
\right\rfloor\\[4pt]
\end{align*}
</s>