Fibonacci DubTrip Reduxed
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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".

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.

Lines with 3 or more cells that can see each other can have the possible locations of the 3 marked.
Immediately, a 3 is placed in r6c7, and [25] and [18] go on the other 2 cells.
The 3 in box 8 forces the 3 in box 9 to r7c8, [25] and [18] are also on the line.
These two 3s force a 3 onto r1c4, [25] and [18] follow.
The two [18]s in col 8 place a [25] in r4c8, so a [18] is in r5c9.
Col 9's 3 can only go in r2c9 as it can't abut the [25] in box 6.
Box 9's other [25] can only go in r9c9.
r6c6 can only be [18], and the rest of the line labelled [25] and [18], apart from the 3 in the centre.
Box 5's 3 can only go in r4c4 as it mustn't abut the [25]
The other cell on its line is therefore the other [25].
The other [18] can only go in r6c4.
Do NOT be tempted to [46] mark the X-sum in the non-fibonacci cells, and [79] on the other two at this point. It's easier to just ignore them and do them in a sweep at the end.
Box 2's second [25] must be in r3c6.

I hope you've been colouring the [18] and [25] cell pairs such that you can later resolve them into those higher than 3 and those lower than 3. Do it now if not.

Box 1's other [25] must also be in row 2, but cannot abut the 3. This forces the 3 into r3c1.
Because it abuts the [18] in box 2, it must be of opposite colour (see!).
This also places the missing [18] in box 2 at r2c4.
The other [18] in box 1, avoiding its matching [25], can only go in r1c1.

The other [18] can be placed in row 3, and box 3 can be fib coloured.
Again, resist the temptation to mark up the [46] and [79] pairs that the X-sum dot gives you.
3 and a [25] can be placed in row 5. The other [25] must avoid 3s, so goes in r6c1.
The [18] of the same colour as the [25] on r5c1 cannot abut it, so must goes in r4c3.
The second [18] in box 4 must be in r4c2.

At this point, the DubTrip kropke dot south of r3c3 gives us all the informtion we need to separate the 1s from 8s, and the 2s from 5s.

Finally looking at box 7, The final 3 can be placed in r9c2.
The 2 and 5 in box 7 are placeable in row 7.
The 1 in row 7 can only go in col 6.
The 1 in box 7 can only go in r9c3.

The final 8s can be placed by noting that r8c7 can't be an 8 on an X-sum.
Therefore r8c1 is the final 8.

Finally, we can blast [46]s and [79]s into the boxes with X-sums, occasionally encountering a 4 or a [69] from a DubTrip kropke dot, and everything resolves.

Sorry for not putting a 3 in the corner. Them's the breaks.