Indexing a set of objects with a symmetry or antisymmetry property Suppose I have a collection of objects $\{X(i, j)\}_{i,j=1}^{n}$ (which could be anything - numbers, functions, matrices, $\ldots$) indexed by $1 \leq i,j \leq n$, such that the following holds:
$$ X(i,j) = X(j,i).$$
Up to the above constraint, all of the objects are different. If I want to index every object, such that no equal object is indexed more than once, I can impose the condition that $i \leq j$ in my indexing.
My question is, how do I generalize this to sets of objects with more than two indices. For example, if I have $\{X(i,j,k)\}$ such that
$$ X(i,j,k) = X(k,j,i),$$
or $\{X(i,j,k,l)\}$ such that
$$X(i,j,k,l) = X(l,k,j,i),$$
what condition on the indices can I impose so that I index each non-equal object exactly once?
 A: If you have $k$ indices $(i_1,\dots,i_k)$, then for an object $X(i_1,\dots,i_k)$, either the values of the indices are palindorimic (ie $i_j=i_{n+1-j}$ for all $1\le j\le k$) in which case there is no choice to be made; or there is some minimal $1\le d\le k/2$ such that $i_d\neq i_{n+1-d}$ but for any $1\le j<d$, $i_j=i_{n+1-j}$. Choose $X(i_1,\dots,i_k)$ if $i_d< i_{n+1-d}$, and choose  $X(i_k,\dots,i_1)$ if $i_d> i_{n+1-d}$.
A: You want the upper triangular elements of a symmetric matrix.

*

*Symmetric
For $i \leq j$ you have $\frac{n (n+1)}{2}$ unique elements that are indexed as $$\mathrm{index}(i,j) = n (i-1) - \frac{i (i-1)}{2}+j $$
Here is an example for $n=4$, with $i=1 \ldots n$ the row index and $j=1 \ldots n$ the column index
$$ \begin{array}{|c|c|c|c|}\hline  1 & 2 & 3 & 4 \\ \hline & 5 & 6 & 7 \\ \hline& & 8 & 9 \\ \hline& & & 10 \\ \hline \end{array} $$


*Anti-symmetric
For $i<j$ you have $ \frac{n (n-1)}{2}$ unique elements that are indexed as $$\mathrm{index}(i,j) = n (i-1) - \frac{i ( i+1)}{2} +j  $$
Here is an example for $n=4$, with $i=1 \ldots n$ the row index and $j = 1 \ldots n$ the column index
$$ \begin{array}{|c|c|c|c|}\hline    & 1 & 2 & 3 \\ \hline &  & 4 & 5 \\ \hline& &  & 6 \\ \hline& & &  \\ \hline \end{array} $$
