I am trying to count all possible relations on a set $S$ whose cardinality is $m$. The relations can vary from $1$ to $n$-ary relations where $n > m$. For example if $S = \{a, b, c\}$ and if $n = 4$, then both $(a)$ and $(a,a,a,a)$ are to be counted as well as $(a,b,c)$ and $(c, b, a)$ etc.
I know the number of ordered pairs is the cardinality of the cartesian product: $$ |S \times S| = m^2$$
The number of combinations of binary relations on a set is the size of the powerset of the cartesian product: $$|\mathcal P(S \times S)| = 2^{|S|^2} = 2^{m^2}$$
And I've read on Quora that the number of $n$-ary relations are the number of subsets of the $n$-fold cartesian product:
$$|\mathcal P(\underbrace{S\times S \times\cdots \times S}_{n\text{ times}} )| = 2^{|S|^n} = 2^{m^n}$$
What I am asking is a formula to count the number $1$-ary, $2$-ary... etc up to $n$-ary relations definable on a set size $m$.
My intuition is that the solution is something like the powerset of the union of the $n$-fold cartesian products from $1$ to $n$:
$$|\mathcal P(\{S\} \cup \{S \times S\} \cup \{S \times S \times S\} \cup \cdots \cup \{\underbrace{S\times S \times\cdots \times S}_{n\text{ times}}\})|$$.
If this is right can I count size of the set of all relations $R$ as something like: $$ \left| \bigcup_{?}^{?}\prod_{i=1}^n S_i \right| \stackrel{?}=\sum_{i=1}^n m^i$$
and the total number of subsets of R as $2^R$?
Is this right? I'm unsure how to convert this into a formula in terms $m$ and $n$. I am also unsure if the method I am using is the correct one. Or if this method misses any relations or double counts them? I apologize if I am using notation incorrectly. I am teaching myself set theory and started to wonder what the answer to this question is.
As Greg Martin suggested in his comment, if $k \neq k'$, the sets of $k$-ary and $k'$-ary relations are disjoint. It follows that the number of $1$ to $n$-ary relations on a set of size $m$ is $$ S(m,n) = \sum_{i=1}^n 2^{m^i} $$ There is no close form for this sum. However, it is easy to write the result in binary (which would make any computer happy...). For instance, $S(2,4) = 2^2 + 2^4 + 2^8 + 2^{16}$ can be writen in binary as \begin{array}{ccccccccccccccccccccccc} \color{red}{\scriptsize{16}}& \scriptsize{15}& \scriptsize{14}& \scriptsize{13}& \scriptsize{12}& \scriptsize{11}& \scriptsize{10}& \scriptsize{9}& \color{red}{\scriptsize{8}}& \scriptsize{7}& \scriptsize{6}& \scriptsize{5}& \color{red}{\scriptsize{4}}& \scriptsize{3}& \color{red}{\scriptsize{2}}& \scriptsize{1}& \scriptsize{0}\\ 1&0&0&0&0&0&0&0&1&0&0&0&1&0&1&0&0 \end{array}