Rational Counting Function

Haskell, 78 characters

q(r,f)=[(r-b,b)|b<-f[1..r-1],r`gcd`b==1]
d=reverse:id:d
f=((zip[2..]d>>=q)!!)

Sample run:

> map f [0..10]
[(1,1),(2,1),(1,2),(1,3),(3,1),(4,1),(3,2),(2,3),(1,4),(1,5),(5,1)]
> f 100
(17,1)
> f 1000
(3,55)

---

• Edit: (100 → 87) silly me, just testing the gcd is enough!
• Edit: (87 → 85) clever trick with cycle and functions to alternate row order
• Edit: (85 → 82) replace cycle with the hand-built infinite list d
• Edit: (82 → 78) applied gcd identity as suggested by Matías

J, 41 36 characters

Takes an integers and returns a vector comprising two integers. My first solution that is neither entirely tacit nor entirely explicit.

{3 :'~.;<`(<@|.)/.(,%+.)"0/~1+i.1+y'

Here is the solution with spaces inserted where appropriate:

{ 3 : '~. ; <`(<@|.)/. (, % +.)"0/~ 1 + i. 1 + y'

An explanation:

1. x (, % +.) y–a vector of length 2 representing the fraction with numerator x and denominator y reduced to the smallest denominator
2. 1 + i. 1 + y–a vector of integers from 1 to y + 1
3. (, % +.)"0/~ 1 + i. 1 + y–a matrix of all reduced fractions with unreduced denominator and numerator in the range from 1 to y + 1.
4. <`(<@|.)/. y–an array of the oblique diagonals of matrix y, each other diagonal flipped
5. ~. ; y–an array of diagonals collapsed into a vector of elements with duplicates removed
6. x { y–the item at position x in y
7. (u v) y–the same as y u v y. In this particular use case, u is { and v is 3 : '~. ; <`(<@|.)/. (, % +.)"0/~ 1 + i. 1 + y'

Python, 144 chars

def F(i):
 r,d,z=[1],1,[]
 while z[:i]==z:z+=[(x,y)for x,y in zip(r[::d],r[::-d])if all(x%j+y%j for j in r[1:])];d=-d;r+=[r[-1]+1]
 return z[i]