Ordering of a curious set Suppose n is a perfect square. Let us look at the 
set of all numbers which is
the product of two numbers, not necessarily distinct, both of which are at
least n. 
Is it possible to find an expression for  the n-th smallest number in this set in terms of n
 A: Having written some python to pump these out (see code here: https://repl.it/CnHE/5), I get the same sequence mentioned in the comments:
\begin{matrix} \\ 
n & n^{th} \text{ term} & n^{th} \text{ term} \\ \hline
1 & 1 & (1+0)^2 \\
4 & 25 &  (4+1)^2 \\
9 & 121 & (9+2)^2 \\
16 & 361 & (16+3)^2 \\
25 & 841 & (25+4)^2 \\
36 & 1681 & (36+5)^2 \\
49 & 3025 & (49+6)^2 \\
64 & 5041 & (64+7)^2 \\
81 & 7921 & (81+8)^2 \\
100 & 11881 & (100+9)^2 \\
\end{matrix}
We can observe that the $n^{th}$ term is $$(n+\sqrt{n}-1)^2$$  Of course this is just an observation, not a proof...
A: Let's arrange the terms in the following order:
$n^2$
$n(n+1)$
$n(n+2), (n+1)^2$
$n(n+3), (n+1)(n+2)$
$n(n+4), (n+1)(n+3), (n+2)^2$
$n(n+5), (n+1)(n+4), (n+2)(n+3)$
....
Some simple remarks you can prove easily yourself:


*

*The terms are in increasing order from left to right.

*Suppose $n=k^2$. When $i<k$ we have $(n+i)^2 < n(n+2i+1)$, the terms are also in increasing order from up to down.

*The number of terms in each row are 1,1;2,2;3,3;4,4;... (notice that the general term in a row is $(n+i)(n+j)$, where $i+j$ is fixed.)

*The term (n+i)(n+j) lies in $(i+j+1)^{th}$-row

*The term $(n+k-1)^2$ lies at the end of $(2k-1)^{th}$-row, therefore, its position is:
$1+1+2+2+3+3+...+(k-1)+(k-1)+k =n^2$ 
So we have confirmed the observation of @Carser.
