Problem:
The sequence {an} is defined by
a0=1,a1=1, and an=an−1+an−2an−12 for n≥2.
The sequence {bn} is defined by
b0=1,b1=3, and bn=bn−1+bn−2bn−12 for n≥2
Find a32b32.
Solution:
Observe that if xn=xn−1+xn−2xn−12 and yn=xn−1xn, then yn=1+yn−1, so
x0xn=xn−1xn⋅xn−2xn−1⋯x0x1=y1y2⋯yn=y1(y1+1)⋯(y1+n−1)
In particular, for the first sequence, y1=a0a1=1, and so an=n!. Similarly, for the second sequence, y1=b0b1=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