Minimal Fibonacci DubTrip 
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Firstly, see the hints in http://fatphil.org/Sudoku/fibdubtrip19_solve.txt, as the same techniques are used here. Colouring [12358] different from [4679] I shall call "fib colouring".
To keep the grid clean, there's absolutely no need to add markup to the [4679] cells apart from the fib-colouring - their resolution is best done in one sweep at the end.

Firstly, fib colour the cells on lines. The empty box centres can also be coloured, and the non-empty ones can be marked [18] or [235] depending on the number of lines leading out of them.
Connected cells alternate between [138] and [25] as long as 1-1 can be excluded.
Lines with 3 or more cells that can see each other can have the possible locations of the 3 marked.

The line segments lining up in column 8 force [138] into the line in box 9, which must be paired with a [25].
The two [25]s in row 8 force [138] into r8c4, also placing a [25] and second [138] onto its line.
This places the other [25] in box 8, which places a [18] in box 4.

Row 2's two [18]s force [25] above [18] in box 2.

Box 3's [25] forces its 3 onto row 3, which excludes them in box 1, whose 3 can be placed in r2c2.
Box 2's 3 can only go in r1c6, completing its fib-colouring.
Column 6's other [18] cannot go in row 5, as that would demand a line-segment into column 5.
Box 5's line is therefore 2-3-5, placing the 3.
Box 4's [138] is therefore 3.
Column 7's 3 can only go in r3c7.
Column 8's 3 can only go in r7c8.

It's worth colouring the [18]s and [25]s so that the cells above 3 and the cells below 3 can be distinguished. All cells marked [18] and [25] can be unambiguously coloured.

From this we see that r6c9 cannot be [18], so is 3.
Row 2's other [25] can only go in r2c9.
Box 1's other [18] can only go in r1c7, completing the fib-colouring.
Row 3's other [18] can only go in r3c3.

Because box 4's other [25] can only be in column 1, box 7's other [25] can only be in r7c3.
The lack of link from that cell places box 8's 3 in r9c4.
Likewise, box 3's other [18] is forced into r5c3, and box 6's other [18] is forced into r6c7.

We could now theoretically handle the ambiguous double/triple kropke dot, and that would split the 1s from the 8s, and the 2s from the 5s. However, we're in the groove placing unresolved ones, let's just continue as we are.

Box 6's other [25] can only go in r4c9.
Therefore box 9's other [25] can only go in r9c7.
Likewise, the unplaced [25] in box 4 is forced into r5c1.
The [18]s of the same colour in boxes 3 and 6 force box 9's [18] of that colour into column 9.
Box 9's other [18] must also go in column 9.

Box 7's two [18]s and 3 are all in column 1, completing the box's fib-colouring.
3 goes into r8c1, and the lack of links into box 4 restrict r7c1 to be the [18] of the opposite colour to r5c2.
This resolves the position of the other [18] in row 1 to r1c2, completing that box's fib-colouring.
It also resolves the position of the [18] of the same colour in box 9.
The [138] cross in rows 8 & 9 force r7c9 to be the other [18], completing that box's fib-colouring.

That's all the [12358] cells placed, and you're only seconds away from completion:

The ambiguous double-triple kropke dot in row 5 can't be a 1-2 or a 1-3, so must be an 8-4.
All the [18]s of that colour are 8, all of the other colour are 1, all the [25]s of that colour are 5, and the remaining ones are 2s.

The x-sums between [4679] cells can only be [46].
Propagate 4s, 6s, and [79]s and round the grid until you reach the ambiguous double-triple kropke dot with a 3 on it, which splits the 7s from the 9s.

Bosh!