How do we list all points in the Middle $3/7$-th Cantor Set?

Question

Suppose we were to list all points in the $3/7$th Middle Cantor set where we first remove $3/7$th of $[0,1]$:

$$[0,1/7]\cup[2/7,3/7]\cup[4/7,5/7]\cup[6/7,1]$$

then $3/7$th of remaining intervals, $3/7$th that still remain, $3/7$th of those, etc.

If I'm correct, we can list all points in the set as:

$$\left\{\left(\sum_{q=1}^{w}\frac{j}{7^q}\right)+\frac{h}{7^{w+1}}:j\in\left\{0,2,4,6\right\},h\in\left\{0,1,2,3,4,5,6,7\right\},w\in\mathbb{N}\right\}$$

I tried recreating this in Mathematica:

  Table[Sum[Subscript[j, 3][[q, w]]/7^q, {q, 1, w}] + 
  Subscript[h, 3]/7^(w + 1), {w, 1, 2}, {Subscript[j, 
  3], {0, 2, 4, 6}}]

But I realized they treated $j$ (or in this case j3) as a fixed number with q. As q changes j3 remains the same which is not what I intended. This gives an unusual graph.

  Subscript[B, 3] = 
 Select[Sort[
   DeleteDuplicates[
    Flatten[Table[
      Sum[Subscript[j, 3]/7^q, {q, 1, w}] + 
       Subscript[h, 3]/7^(w + 1), {w, 1, 2}, {Subscript[j, 
       3], {0, 2, 4, 6}}, {Subscript[h, 
       3], {0, 1, 2, 3, 4, 5, 6, 7}}]]]], Between[#, {0, 1}] &]
Length[Subscript[B, 3]]
ListPlot[Table[{Subscript[B, 3][[x]], 0}, {x, 1, 
   Length[Subscript[B, 3]]}]]

Which I doubt is correct. How do we correct my method?

Answer 1

By the links provided by @Flinty, it seems the answer to my question should be

cantor = {a_, 
    b_} :> {{a, a + (b - a)*1/7}, {a + (b - a)*2/7, 
     a + (b - a)*3/7}, {a + (b - a)*4/7, 
     a + (b - a)*5/7}, {a + (b - a)*6/7, b}};
CantorRegion[n_Integer?NonNegative] := 
 Module[{ints}, 
  ints = Flatten[
    Nest[Flatten[Map[Function[s, s /. cantor], #], 1] &, {{0, 1}}, n]]]
ListPlot[Table[{CantorRegion[3][[n]], 0}, {n, 1, 
   Length[CantorRegion[3]]}]]

Giving the desired result

I'm not completely sure why this is the case. The code seems very advanced. If anyone can explain this mathematically I will be grateful. Thanks!