How would this combinatorial process be written in conventional mathematical notation?

Question

Now, I am going to define an operation on a list of numbers, which I will treat as a set, for want of a better approach. This may not be ideal, or even conventional, so apologies for lack of clarity from the start - clarity of notation os my goal in this question.

Example:

Given the set $\{1,2,3\}$ (for example the indices of the prime factorisation for $2250=2^1 3^2 5^3$), the first operation on that set is division into a specific subset-type, as illustrated below:

\begin{array}{l} &\{\emptyset,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}\\ &\{1\},\{1,2\},\{1,3\},\{1,2,3\}\}\\ &\{2\},\{1,2\},\{2,3\},\{1,2,3\}\}\\ &\{3\},\{1,3\},\{2,3\},\{1,2,3\}\}\\ &\{1,2\},\{1,2,3\}\}\\ &\{1,3\},\{1,2,3\}\}\\ &\{2,3\},\{1,2,3\}\}\\ &\{1,2,3\}\}\\ \end{array}

Then the products of each subset are taken:

\begin{array}{l} &\{1, 1, 2, 3, 2, 3, 6, 6\}\\ &\{1, 2, 3, 6\}\\ &\{2, 2, 6, 6\}\\ &\{3, 3, 6, 6\}\\ &\{2,6\}\\ &\{3,6\}\\ &\{6,6\}\\ &\{6\}\\ \end{array}

Then each row is summed:

$$\{24, 12, 16, 18, 8, 9, 12, 6\}$$

Note:

Duplicates are not treated any differently.

eg $\{1,1,2\}$ results in $\{12, 6, 6, 8, 3, 4, 4, 2\}$.

There are further operations on these, but these are trivial.

Could someone help me in notating the above sequence of operations using succinct, clear, and conventional notation please?

Update

The thing is, the operation in its entirity is terribly straightforward to code:

op[n_] := If[n == 1, {{1, 0}, {1, 1}}, 
With[{a = Transpose@FactorInteger[n]}, 
With[{b = Times @@@ Subsets[a[[1]]]}, 
With[{c = x[#] & /@ Range@Length@a[[2]]}, 
Transpose@{b, (-1)^PrimeOmega@b Reverse[(-1)^PrimeNu@n* 
Total[Times @@@ #] & /@ -(aa[c, #] & /@ Subsets[c] 
/. Thread[c -> a[[2]] + 1])]}]]]]

,but not at all straightfarward (as far as I can see) to denote using conventional notation Anyway, the operation gives

$$ \{\{ 1 ,24 \}, \{2 ,-12 \}, \{ 3 ,-16 \}, \{ 5 ,-18 \}, \{ 6 ,8 \}, \{ 10 ,9 \}, \{ 15 ,12 \}, \{ 30 ,-6 \}\} $$

Then using these coefficients for some sum.

Is this a hurdle for many combinatorial operations, preferably avoiding set notation?

Answer 1

A multiset is just like a set in that order doesn't matter, except repeated elements are allowed. As far as I know there is no standard notation for them, but you could use square brackets (as the other answer indicates, it's not correct to use curly braces). For example $$[1,1,2,3,4,4]$$ is a multiset where $1$ occurs twice, $2$ occurs once, etc.

If you have a multiset $A$, a submultiset $B$ has all of its elements contained in $A$ with the same or smaller multiplicity. There's no harm in writing $B\subseteq A$ to mean that $B$ is a submultiset of $A$.

Consider the lattice of submultisets of $A$ (this is just the set of all submultisets of $A$; a lattice is a particular type of partially ordered set where every pair of elements has a least upper bound and greatest lower bound). There are plenty of ways you could write this, but one way to do it would be $\mathcal{S}(A)$ (after defining what it means). $\mathcal{S}(A)$ is a partially ordered set where the partial ordering is given by $\subseteq$.

If $B\in \mathcal{S}(A)$, then the (principal) filter or upper order ideal generated by $B$ is the set of all elements of $\mathcal{S}(A)$ (so submultisets of $A$) that contain $B$ as a submultiset. This is what you're doing when you split things up into "subset types"; you're taking the principal filter generated by each element of $\mathcal{S}(A)$ separately. For $B\in\mathcal{S}(A)$ we can call this filter $\langle B\rangle$.

Now we can define $$\pi(B):=\prod_{b\in B}{b}$$ for $B\in\mathcal{S}(A)$, which is the product of all of the elements in the multiset. Then the sum of all these products in a filter (the "sum of the row") is given by $$\sum_{B'\in\langle B\rangle}{\pi(B')}$$ Equivalently (may even be easier, don't have to mention filters at all) you could just write $$\sum_{B\subseteq B'\subseteq A}{\pi(B')}$$ Then you indicate somehow how you put these in the appropriate order.

Answer 2

You should not use set-notation for sequences because $\{1,1,2\}=\{1,2\}$ as this denotes the set of objects listed or defined within the brackets. You should use subscripts or function-notation. A sequence of $n$ objects, not necessarily distinct, is written $(x_j){0\leq j<n}.$ This can also be described as the list of the values of a function $f$ with $dom (f)=\{0,...,n-1\},$ where $f(j)=x_j.$ The empty sequence ($n=0$) is the same thing as the empty set.) Each of your "specific types" is a set $T_A=\{(f(x))_{x\in B} :A\subset B\subset dom (f)\}.$ The "products taken" is the sequence $P_A= (\prod_{x\in B}f(x))_{A\subset B\subset dom (f)}.$(The product of the members of the empty set is $1$.) It is assumed here that you have a way to list $\{B : A\subset B\subset dom (f)\}$ as a sequence.