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.