Is there a closed form of this sequence?

Question

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.. :(

Answer 1

You can break the sequence in groups of $N$ terms with $N=1,2,3,\dots$ of the form $$\frac{k}{N+1-k}\quad \text {for $k=1,2,\dots ,N$}.$$ So the first $N-1$ groups have $\sum_{i=1}^{N-1}i=\binom{N}{2}$ terms. Now, given $n$, there is a unique $N$ such that $\binom{N}{2}<n\leq \binom{N+1}{2}$, which means that the $n$-term is in the $N$-th group and it is equal to $\frac{k}{N+1-k}$ with $k=n-\binom{N}{2}$. For an explicit formula solve $\binom{N}{2}<n\leq \binom{N+1}{2}$ with respect to $N$: $$N=\left\lfloor\frac{1+\sqrt{8n-7}}{2}\right\rfloor$$

Answer 2

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*}