Finding the row and column number of the number $20096$

Question

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.

Answer 1

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 $(1,200)+(+4,-4)=\color{red}{(5,196)}$

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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 $$\frac12 \left[(m+n)^2-3m-n+2\right]$$