(Music) List of all possible "types of set" of 12 musical notes I have looked into trying to figure what are all the possible "types" of note set combinations there are and how I would go by listing them if possible. It turns out this is harder than I thought. The 12 notes are:
$$1, 2^{\flat}, 2, 3^{\flat}, 3, 4, 4^{\sharp}/5^{\flat}, 5, 6^{\flat}, 6, 7^{\flat}, 7$$
A type of set would be something like this:
$$1,3,5,7 .$$
The thing is, that set has inversion possibilities such as $3,1,5,7$, and I do not want to count any of them as part of the list. No repeating notes, either, e.g., $1,1,3,5$ or $1,3^{\flat},5,3^{\flat}$.
I've thought about in the inversion case of $3,1,5,7$ and can convert that same set into $1,6^{\flat},3^{\flat},5$, which is actually the same notes (which I'm trying to avoid as well). But what might be useful from this is using $1$ as a base reference and instead try using each interval in order (from lowest to highest) for every single set on the list for example. Here are some examples:
$$1,2^{\flat},2,3, \qquad 1,2^{\flat},3,4, \qquad 1,2,3,4, \qquad 1,4,5,6^{\flat} .$$
I could use $1,3^{\flat},5,6^{\flat}$, but then I would have to not list $1,3,5,7$ in the list, because they are the same 'set type'. So in the end it's kind of difficult to find which note sets are already there and which are not, which leaves me to believe that I am missing a certain condition. Or maybe rather theres actually an easier way to look at all this that I haven't thought of yet?
So the question is: 

How do I list all the different set types of notes without unintentionally repeating one?

(Edit: Added the last question to hopefully clear things up)
 A: This is really just an extended comment, since it gives a manually practical way to compute the number of "set types" using powerful combinatorial tools, rather than a way to produce a list of set types. See my later answer, which

*

*describes an algorithm for doing the former and

*applies that algorithm to give the answer for the corresponding $3$-note problem, leaving the more involved but procedurally similar $4$-note problem.

We can just as well think of the set choices of $4$ notes out of the $12$ as the set $S$ of $12$-bead strings containing $4$ black beads ($\blacksquare$) and $8$ white beads ($\square$), where black beads mark the chosen notes. So, for example, the given set $\{1, 3, 5, 7\}$ of notes (the $0$th, $4$rd, $6$th, and $11$th notes) corresponds to the string $$\blacksquare\square\square\square\blacksquare\square\square\blacksquare\square\square\square\blacksquare .$$ (There are ${12 \choose 4} = \frac{12!}{4! 8!} = 495$ such sets.)
Now, we want to view the notes as arranged in a circle and count two sets of notes as equivalent if one can be rotated to produce the other. Equivalently, we declare two bead strings in $S$ to be equivalent the same if one can be cyclically permuted into the other, so that we count as equivalent $$\blacksquare\square\square\square\blacksquare\square\square\blacksquare\square\square\square\blacksquare, \quad \square\square\square\blacksquare\square\square\blacksquare\square\square\square\blacksquare\blacksquare, \quad\square\square\blacksquare\square\square\blacksquare\square\square\square\blacksquare\blacksquare\square, \ldots$$ (imagine tying the two ends of the string together, so that we get a necklace of beads).
Put another way, the cyclic group $C_{12}$ acts on the set $S$ of strings of beads by the above rotation, and we have declared two strings to be equivalent iff they are in the same orbit of that action. So, we are equivalently asking how many orbits this action has, so we can apply the general version of the Pólya Enumeration Formula:
We declare black beads $\blacksquare$ to have weight $1$ and white beads $\square$ to have weight $0$, giving the generating function $f(t) = 1 + t$ encoding the colors and weights; so, we are interested in counting the necklaces of total weight $4$. The cycle index for $C_{12}$ encodes the the cycles of each size of the permutations of the elements and is
$$Z(a_1, a_2, \ldots, a_{12}) = \frac{1}{12} \sum_{d \mid 12} \phi(d) a_d^{12 / d} = \frac{1}{12}(a_1^{12} + a_2^6  + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12}) .$$ Here, $\phi$ denotes Euler's totient function.
Now, the Enumeration Formula gives that the generating function $F(t) = \sum_w n_w t^w$ for the numbers $n_w$ of necklaces of weight $w$ (that is, with $w$ black beads) is
\begin{align}
F(t) = Z(f(t), f(t^2), \ldots, f(t^{12}))
&= \frac{1}{12} \sum_{d \mid 12} \phi(d) (1 + t^d)^{12 / d} \\
&= \frac{1}{12}[(1 + t)^{12} + (1 + t^2)^6 + 2 (1 + t^3)^4 \\ &\qquad\qquad + 2 (1 + t^4)^3 + 2 (1 + t^6)^2 + 4 (1 + t^{12})] \\
&= 1 + t + 6 t^2 + 19 t^3 + \color{#df0000}{43} t^4 + 66 t^5 + 80 t^6 + \cdots
\end{align}
Thus, there are $\color{#df0000}{n_4 = 43}$ necklaces (sets up to equivalence) of size $4$, which agrees with the result in weee's answer. As a byproduct, this computation also gives us the counts of sets of notes of each size (again, up to equivalence). By reversing the colors of all of the beads in a necklace, we can see that there are just as many sets of size $k$ up to equivalence as there are of size $12 - k$.
If we cared only about computing $n_4$ we could save a few moments by noticing that the only $d$ that can contribute to the $t^4$ term in $$F(t) = \frac{1}{12} \sum_{d \mid 12} \phi(d) (1 + t^d)^{12 / d}$$ are those with $d \mid 4$, and the $t^4$ term of $(1 + t^d)^{12 / d}$ is ${12 / d}\choose{4 / d}$, so
\begin{align}
n_4
&= [t^4] F(t)\\
&= [t^4] \frac{1}{12} \sum_{d \mid 12} \phi(d) (1 + t^d)^{12 / d} \\
&= \frac{1}{12} \sum_{d \mid 4} \phi(d) {{12 / d}\choose{4 / d}} \\
&= \frac{1}{12} \left[ \phi(1) \cdot {12 \choose 4} + \phi(2) \cdot {6 \choose 2} + \phi(4) \cdot {3 \choose 1}\right] \\
&= \frac{1}{12} \left[1 \cdot 495 + 1 \cdot 15 + 2 \cdot 3\right] \\
&= \color{#df0000}{43} .
\end{align}
A: Here's a manual algorithm for producing an explicit list; I'll apply it to triples of notes, and let you modify it for the case of quadruples of notes. The triple case makes for a more compact explanation (there are 19 "set types" in this case but 43 in the quadruple case) but still captures all of the phenomena that occur in the quadruple case.
As in my other answer, we denote sets of notes by a sequence of black and white boxes. So, for example, we denote the set $\{1, 3, 5\}$ (the $0$th, $4$th, and $6$th notes) by $$\blacksquare \square \square \square \blacksquare \square \square \blacksquare \square \square \square \square .$$
For any set of notes we can always rotate to assume that the first box is black, $\blacksquare$. Now, we can encode any such set of notes by writing down the numbers $b_1, b_2, b_3$ of steps between successive black boxes numbers (including between the last and first); these numbers should add up to $12$, the total number of boxes.
For example, in the example sequence, there are $b_1 = 4$ steps between the first and second block, $b_2 = 3$ between the second and third, and similarly $b_3 = 12 - b_1 - b_2 = 5$, giving the decomposition (ordered partition) $4 + 3 + 5$ of $12$. We can recover the sequence from the partition $(b_1, b_2, b_3)$ by forming the sequence whose black boxes occur at positions $0, b_1, b_1 + b_2$.
Now, two decompositions give rise to the same sequence if there is a cyclic permutation taking one to the other---this just amounts to rotate the sequence so that we start a different black box---and we can choose a unique representative among these by taking the first partition lexicographically. In our running example, the four decompositions of $8$ we can get from cyclic permutation from $4 + 3 + 5$ are that permutation, $3 + 5 + 4$, and $5 + 4 + 3$. The second is first lexicographically, and corresponds to the triple $(0, 3, 8)$, or the sequence $$\underset{(1, 3^{\flat}, 6^{\flat})}{\blacksquare \square \square \blacksquare \square \square \square \square \blacksquare \square \square \square} .$$
So, we can encode uniquely all of the equivalent triples by writing decompositions that are lexicographically earliest among their cyclic permutations. Note that if a decomposition is lexicographically earliest, the smallest number in the decomposition must appear first, and so the first number at most $\frac{1}{3} \cdot 12 = 4$, significantly reducing the number of decompositions we need to check.
Proceeding naively, the 10 decompositions starting with $1$ are $$1 + 1 + 10, \quad 1 + 2 + 9, \quad \ldots, \quad 1 + 9 + 2, \quad 1 + 10 + 1 ,$$ and all of these are in lexicographical order except the last one, respectively giving $9$ sequences:
$$\underset{(1, 2^{\flat}, 2)}{\blacksquare \blacksquare \blacksquare \square \square \square \square \square \square \square \square \square}, \ldots, \underset{{(1, 2^{\flat}, 7^{\flat})}}{\blacksquare \blacksquare \square \square \square \square \square \square \square \square \blacksquare \square} .$$
The decompositions of $12$ into numbers at least $2$ are $$2 + 2 + 8, \quad 2 + 3 + 7, \quad \ldots, \quad 2 + 8 + 2 ,$$ and all but the last are in lexicographical order, giving $6$ more sequences, respectively,
$$\underset{(1, 2, 3)}{\blacksquare \square \blacksquare \square \blacksquare \square \square \square \square \square \square \square}, \ldots, \underset{(1, 2, 6)}{\blacksquare \square \blacksquare \square \square \square \square \square \square \square \blacksquare \square \square} .$$
The decompositions of $12$ into numbers at least $3$ are $$3 + 3 + 6, \quad 3 + 4 + 5, \quad 3 + 5 + 4, \quad 3 + 6 + 3 ,$$ and again all but the last are in order, giving $3$ more sequences:
$$\underset{(1, 3^{\flat}, 4^{\sharp}/5^{\flat})}{\blacksquare \square \square \blacksquare \square \square \blacksquare \square \square \square \square \square}, \ldots, \underset{(1, 3^{\flat}, 6^{\flat})}{\blacksquare \square \square \blacksquare \square \square \square \square \blacksquare \square \square \square}.$$
Finally, there is only a single decomposition of $12$ into numbers at least $4$, namely $$4 + 4 + 4 ,$$ giving the sequence
$$\underset{(1, 3, 6^{\flat})}{\blacksquare \square \square \square \blacksquare \square \square \square \blacksquare \square \square \square} .$$
In total this gives $9 + 6 + 3 + 1 = 19$ sequences, which agrees with the count in the other answers.
