Problem:
The sequence {an​} is defined by
a0​=1,a1​=1, and an​=an−1​+an−2​an−12​​ for n≥2.
The sequence {bn​} is defined by
b0​=1,b1​=3, and bn​=bn−1​+bn−2​bn−12​​ for n≥2
Find a32​b32​​.
Solution:
Observe that if xn​=xn−1​+xn−2​xn−12​​ and yn​=xn−1​xn​​, then yn​=1+yn−1​, so
x0​xn​​=xn−1​xn​​⋅xn−2​xn−1​​⋯x0​x1​​=y1​y2​⋯yn​=y1​(y1​+1)⋯(y1​+n−1)
In particular, for the first sequence, y1​=a0​a1​​=1, and so an​=n!. Similarly, for the second sequence, y1​=b0​b1​​=3, and so bn​=21​(n+2)!. The required ratio is then 32!34!/2​=17⋅33=561​.
The problems on this page are the property of the MAA's American Mathematics Competitions