Faye tackled this problem :

F1 : 0 1 , 1 1
F2 : 0 1 , 1 2 , 1 1
F3 : 0 1 , 1 3 , 1 2 , 2 3 , 1 1
F4 : 0 1 , 1 4 , 1 3 , 1 2 , 2 3 , 3 4 , 1 1
F5 : 0 1 , 1 5 , 1 4 , 1 3 , 2 5 , 1 2 , 3 5 , 2 3 , 3 4 , 4 5 , 1 1
F6 : 0 1 , 1 6 , 1 5 , 1 4 , 1 3 , 2 5 , 1 2 , 3 5 , 2 3 , 3 4 , 4 5 , 5 6 , 1 1
F7 : 0 1 , 1 7 , 1 6 , 1 5 , 1 4 , 2 7 , 1 3 , 2 5 , 3 7 , 1 2 , 4 7 , 3 5 , 2 3 , 5 7 , 3 4 , 4 5 , 5 6 , 6 7 , 1 1
F8 : 0 1 , 1 8 , 1 7 , 1 6 , 1 5 , 1 4 , 2 7 , 1 3 , 3 8 , 2 5 , 3 7 , 1 2 , 4 7 , 3 5 , 5 8 , 2 3 , 5 7 , 3 4 , 4 5 , 5 6 , 6 7 , 7 8 , 1 1
F9 : 0 1 , 1 9 , 1 8 , 1 7 , 1 6 , 1 5 , 2 9 , 1 4 , 2 7 , 1 3 , 3 8 , 2 5 , 3 7 , 4 9 , 1 2 , 5 9 , 4 7 , 3 5 , 5 8 , 2 3 , 5 7 , 3 4 , 7 9 , 4 5 , 5 6 , 6 7 , 7 8 , 8 9 , 1 1
F10 : 0 1 , 1 10 , 1 9 , 1 8 , 1 7 , 1 6 , 1 5 , 2 9 , 1 4 , 2 7 , 3 10 , 1 3 , 3 8 , 2 5 , 3 7 , 4 9 , 1 2 , 5 9 , 4 7 , 3 5 , 5 8 , 2 3 , 7 10 , 5 7 , 3 4 , 7 9 , 4 5 , 5 6 , 6 7 , 7 8 , 8 9 , 9 10 , 1 1 .
The sequence you get by putting in those mediants is the fifth Farey sequence.
When I tried it with the others, I discovered that if you're doing the nth Farey sequence and you put in the mediants that will give you n+1 on the denominator, you get the n+1th Farey sequence.