Sunil sent us his work on this problem:

When I tried this out, no matter what numbers I started with, I always seemed to find that the 6th term was the same as the 1st term, and the 7th the same as the 2nd, so it kept repeating. I used a spreadsheet to help me do the calculations! Then I used algebra to try to explain why it worked.
The first term is a₁.
The second term is a₂.
The third term is

1+a2 a1

.
The fourth term is

1+a1+a2 a1 a2

.
The fifth term is

æ ç è

1+a1+a2+a1 a2 a1 a2

ö ÷ ø

×

æ ç è

a1 1+a2

ö ÷ ø

=

æ ç è

(1+a1)(1+a2) a1 a2

ö ÷ ø

×

æ ç è

a1 1+a2

ö ÷ ø

=

1+a1 a2

.
The sixth term is

æ ç è

1+a1+a2 a2

ö ÷ ø

×

æ ç è

a1 a2 1+a1+a2

ö ÷ ø

=a1

.
The seventh term is

(a1+1)×

æ ç è

a2 1+a1

ö ÷ ø

=a2

.

This explains why the pattern always works, no matter what you start with! (As long as the bottom is never 0, so we can't have a₁=0, or a₂=0, or a₁=-1, or a₂=-1, or a₁+a₂=-1.)