Finding Value of Pascal's Triangle Given Single Index I know that you can find the value of any number in Pascal's Triangle using the row and column with binomial coefficients, but cannot find a method to find the value given an index.
For instance, a Pascal's Triangle with 5 rows looks like:
\begin{gather}
1\\
1 \quad 1\\
1 \quad 2 \quad 1\\
1 \quad 3 \quad 3 \quad 1\\
1 \quad 4 \quad 6 \quad 4 \quad 1
\end{gather}
If you were to index this triangle, it would look like this:
\begin{gather}
00\\
01 \quad 02\\
03 \quad 04 \quad 05\\
06 \quad 07 \quad 08 \quad 09\\
10 \quad 11 \quad 12 \quad 13 \quad 14
\end{gather}
Without using previous values or referencing the row and column, can you find the value at any given index?
 A: If you do not want to use the row number explicitly from the position index  $i$  then you can use the inefficient $$\frac{\Big\lfloor \dfrac{\sqrt{8i+1}-1}{2}\Big\rfloor!}{\left(i-\dfrac{\big\lfloor \frac{\sqrt{8i+1}-1}{2}\big\rfloor\big\lfloor \frac{\sqrt{8i+1}+1}{2}\big\rfloor}{2}\right)!\left(\Big\lfloor \dfrac{\sqrt{8i+1}-1}{2}\Big\rfloor+\dfrac{\big\lfloor \frac{\sqrt{8i+1}-1}{2}\big\rfloor\big\lfloor \frac{\sqrt{8i+1}+1}{2}\big\rfloor}{2}-i\right)!}$$
That is just a substituted version of a more efficient method of finding the row $r=\Big\lfloor \dfrac{\sqrt{8i+1}-1}{2}\Big\rfloor$ using the floor function and and the value of the index of the initial position in the row $s=\dfrac{r(r+1)}{2}$, so the value corresponding to index position $i$ is $\displaystyle {r\choose i-s}=\dfrac{r!}{(i-s)!(r+s-i)!}$
A: It might not be immediately obvious to you, but the answer I wrote out here shows how to compute the row and column index from the "linear index" and vice versa; once you know these, you've got your answer. 
