TITLE: Finding the row and column number of the number $20096$
QUESTION [5 upvotes]: Consider the numbers arranged in the following way $$\begin{array}{ccccccc} 1 & 3 & 6 & 10 & 15 & 21 & \cdots \\ 2 & 5 & 9 & 14 & 20 & \cdots & \cdots \\ 4 & 8 & 13 & 19 & \cdots & \cdots & \cdots \\ 7 & 12 & 18 & \cdots & \cdots & \cdots & \cdots \\ 11 & 17 & \cdots & \cdots & \cdots & \cdots & \cdots \\ 16 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array}$$The question is to find the row number and column number in which the number $20096$ occurs.
I tried to find a general expression for the number in $k^{th}$ column and $n^{th}$ row.$$\frac{k(k-1)}{2}-(n-1)=20096$$and $$\frac{n(n-1)}{2}+k=20096$$But I am getting fractional value of row which is incorrect.Any ideas?Thanks.

REPLY [4 votes]: Note that rotating the grid $45^\circ$ clockwise shows a triangle of natural numbers arranged sequentially. 
$$\begin{array}
&&&&&1\\
&&&2&&3\\
&&4&&5&&6&\\
&7&&8&&9&&10\\
11&&12&&\cdots&&&& \ddots \end{array}$$
In its rectangular form (as shown in the question), the element in the $n$-th column of the first row, i.e. in position $(1,n)$, corresponds to $\displaystyle\binom {n+1}2$. The number which is one less than this is found in the next row and the previous column, i.e. $(+1, -1)$. 
Hence $\displaystyle\binom {201}2=20100$ is located in position $(1,200)$. 
As the number $20096$ is $4$ less than $20100$, its position is $(1,200)+(+4,-4)=\color{red}{(5,196)}$

NB - the element in row $m$ and column $n$, i.e. $(m,n)$ has value
$$\binom {m+n-1}2+n$$
This can also be written in another form (attributable to a friend of mine), as follows: 
$$\binom {m+n}2-m+1$$
Both of these are equal to $$\frac12 \left[(m+n)^2-3m-n+2\right]$$