# 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.

Without using previous values or referencing the row and column, can you find the value at any given index?

• you might be able to use the diagonals but that still may relate to the rows and columns. – user451844 Jul 13 '17 at 22:37
• I think you'd need to recover the row and column first, which is easily done of course. – Joffan Jul 13 '17 at 22:40

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 $\displaystyle {r\choose i-s}=\dfrac{r!}{(i-s)!(r+s-i)!}$