First, note that if the pair is (a>b), a = qb + r gives the next pair (q,r). By definition of the remainder theorem, r<b and q<a. Thus, one of r or q will be lower than the lowest of a and b. Then the ideal is to make sure the lowest number in the pair decreases by no more than 1.
The smallest possible pair is (1,0). The smallest possible pair which produces this one is either a = 0b + 1 or a = 1b + 0. (1,1) happens to satisfy both of these.
The next pair solves a = b + 1. b must be at least 2, so the next pair is (3,2).
The fourth pair solves a = 3b + 2. b must be 3, so this pair is (11,3).
The fifth pair solves a = 11b + 3, where b = 4, and this pair is (47,4).
The sixth pair solves a = 47b + 4, where b = 5, and this pair is (239,5).
The chain, backwards, is (1,0), (1,1), (3,2), (11,3), (47,4), (239,5).
Notice that incrementing the larger number of each pair creates the sequence 2,2,4,12,48,240...
Halving each term in this new sequence creates 1,1,2,6,24,120..., the familiar sequence of factorials.
So if we consider (1,0) the 0th pair in the ideal chain, the general formula for the nth pair in this chain is (2n!-1, n).