1/1, 1/1, 1/2, 1/3, 1/4, 1/9...
Sequence of denominators bn: 1,1,2,3,4,9,...
5(b3)-6(b1) = 5(2)-6(1) = 10-6 = 4 = b5
5(b4)-6(b2) = 5(3)-6(1) = 15-6 = 9 = b6
So a recursion equation for the sequence of denominators is:
b1=1,b2=1,b3=2,bn=5bn-2-6bn-4
Therefore a recursion equation for the given sequence an is
a1=1,a2=1,a3=1/2,an=5/an-2-6/an-4
Edwin