Generating a 2D coordinate from a 1D Index Imagine you have a 5x5 matrix (zero indexed), where only the upper right coordinates are valid. I am unsure if this is the terminology to use, so the following coordinates should be the valid ones:
(0;1)
(0;2)
(0;3)
(0;4)
(1;2)
(1;3)
(1;4)
(2;3)
(2;4)
(3;4)

The coordinates where x<=y are invalid. In this case the matrix has 10 valid elements, and for a N sized matrix the valid elements are
Elements = (N * (N - 1)) / 2

What I wanted to know is if there is a way of generating a 2D coordinate based on a zero indexed identifier, which corresponds to one of the valid coordinates.
For example, taking into account the coordinates I mentioned above, there are 10 identifiers (0 through 9), and each of them would map to a specific coordinate: index 0 would map to (0;1), and 9 would map to (3;4). I am looking for a formula, that given a specific index and the size of the matrix, would return the corresponding index.
Is this possible? so far, the only way I could think of this was through bruteforce, trying every number combination for X and Y that make this valid:
Index = 2x + 1 + y

where Y has to be between x+1 and the size of the matrix - 1
I was hoping that someone with math knowledge than me could know of a formula that would calculate this.
I hope this makes sense and is clear enough. If not, please let me know.
Thank you in advance for your help.
 A: Yes it is possible. The reason for this is that $\mathbb{N}^2$ is countable and so is a subset.
You could use a similar enumeration like used in those proofs, e.g.:
$$
I(x,y) = \pi(N-1 -x,y)
$$
where 
$$
\pi(x, y) := y + \sum_{i=0}^{x+y} i = y+\frac{1}{2} (x + y) (x + y + 1)
$$
is the Cantor pairing function, the enumeration used for the Cantor diagonal enumeration of $\mathbb{N}^2$.
This is a bijective function. The inverse of $\pi$ can be given like this:
$$
\pi^{-1} = (\pi_1, \pi_2)
$$
with
$$
\pi_2(z) = z - f(q(z))
\quad
\pi_1(z) = q(z) - \pi_2(z)
$$
and
$$
f(w) = \frac{w(w+1)}{2} \quad 
q(z) = \left\lfloor \frac{\sqrt{8z + 1} - 1}{2} \right\rfloor
$$
The main application is to understand that $\mathbb{N}$ has the same cardinality as $\mathbb{N}^k$.
Now the translation from $\pi$ to $I$:
$$
I^{-1} = (I_1, I_2)
$$
and
$$
x = I_1(z) = N - 1 - \pi_1(z)  \quad
y = I_2(z) = \pi_2(z)
$$
Example:
$N = 5$, index coordinates $x$, $y$ from $0$ to $4$ each, then $I$ would index like this:
$$
\left(
\begin{matrix}
* & 6 & 3 & 1 & 0 \\
* & * & 7 & 4 & 2 \\
* & * & * & 8 & 5 \\
* & * & * & * & 9 \\
* & * & * & * & * \\ 
\end{matrix}
\right)
$$
First going from $(x,y) = (3,1)$ to the enumeration index:
$I(3,1) = \pi(5 - 1 - 3, 1) = \pi(1,1) = 1 + \frac{1}{2}(1+1)(1+1+1) = 4$. 
And then back from the index $4$ to $x$ and $y$:
$$
y = I_2(4) = \pi_2(4) = 4 - f(q(4)) = 4 - f(2) = 4 - 3 = 1 \\
x = I_1(4) = 5 - 1 - \pi_1(4) = 4 - q(4) + \pi_2(4) = 4 - 2 + 1 = 3
$$
With a bit more tinkering regarding index transformation, one can end up with the enumeration you probably want, e.g.
$$
\left(
\begin{matrix}
* & 0 & 1 & 3 & 6 \\
* & * & 2 & 4 & 7 \\
* & * & * & 5 & 8 \\
* & * & * & * & 9 \\
* & * & * & * & * \\ 
\end{matrix}
\right)
$$
