I have the pairs $[i, j]$ with $i, j = 1,...,n$ and $j > i$. I know that there are $\frac{n(n-1)}{2}$ possible pairs. Is there a function $f_2(l) \rightarrow [i,j]$, $l=1,...,\frac{n(n-1)}{2}$ that returns all these pairs?
As above, but triples $[i, j, k]$ with $i,j,k=1,...,n$ and $k > j > i$. Number of triples is $\frac{n(n-1)(n-2)}{6}$. Is there a function $f_3(l) \rightarrow [i,j,k]$ that returns all triples?
I'm going to give you what you asked for and a little more that you didn't ask for (but may have been expecting). If you don't see what you want, you might consider editing the question to clarify your needs.
For $f_2(l),$ arrange all pairs $[i,j]$ such that $1 \leq i < j \leq n$ in sequence such that $[i_1, j_1]$ is before $[i_2, j_2]$ if and only if one of the following two statements is true: (a) $j_1 < j_2$ or (b) $j_1 = j_2$ and $i_1 < i_2.$ Then $f_2(l)$ is the $l$th pair in the sequence.
For $f_3(l),$ arrange all triples $[i,j,k]$ such that $1 \leq i < j <k \leq n$ in sequence such that $[i_1, j_1, k_1]$ is before $[i_2, j_2, k_2]$ if and only if one of the following three statements is true: (a) $k_1 < k_2,$ (b) $k_1 = k_2$ and $j_1 < j_2,$ (c) $k_1 = k_2,$ $j_1 = j_2,$ and $i_1 < i_2.$ Then $f_3(l)$ is the $l$th triple in the sequence.
There are other functions that can be obtained by changing the sequences of pairs or triples.
$f_2(l) = [i,j]$ where \begin{align} j &= \min\{t \in \mathbb N : \tfrac12 t(t-1) \geq l\}, \\ i &= l - \tfrac12 (j-1)(j-2). \end{align}
$f_3(l) = [i,j,k]$ where \begin{align} k &= \min\{t \in \mathbb N : \tfrac16 t(t-1)(t-2) \geq l\}, \\ [i,j] &= f_2\left(l - \tfrac16 (k-1)(k-2)(k-3)\right). \end{align}
It is easy enough to come up with a computational algorithm for the first version of the functions. A nested loop can construct the sequence of pairs or triples in the order specified by those definitions, without needing to sort them afterward. For example, to generate a list of triples in python,
triples = []
for k in range(3, n + 1):
for j in range(2, k):
for i in range(1, j):
triples.append([i, j, k])
But it may take a long time and a lot of memory to build the list. If you construct the list once and then call $f_2$ or $f_3$ many times, the amortized cost may be small enough; otherwise you probably want something more efficient.
For the second version, you need some way to compute $j$ for $f_2$ and $k$ for $f_3.$ An obvious approach is to start with the smallest possible value of $t$ and count up until you find the correct value. This is not as bad as it sounds, because $\tfrac12 t(t-1)$ grows like $t^2$ and $\tfrac16 t(t-1)(t-2)$ grows like $t^3$; computed this way, the computational complexity of $f_2$ is $O(\sqrt n)$ and that of $f_3$ is $O(\sqrt[3] n).$
Even better, there is a relatively straightforward closed-form formula for $j$ in $f_2,$ which you can deduce from the answer to Largest Triangular Number less than a Given Natural Number. There is even a closed-form formula for $k$ in $f_3,$ though I would not like to use it: $$ k = \frac{(\sqrt3 \sqrt{243 n^2 - 1} + 27 n)^{1/3}}{3^{2/3}} + \frac{1}{{3^{1/3} (\sqrt3 \sqrt{243 n^2 - 1} + 27 n)^{1/3}}} + 1 $$ A simpler but still efficient approach, I think, is to estimate a preliminary value of $k$, say $k_1 = 1 + \sqrt[3]{6l},$ and then increase or decrease the value until it is correct.
