Problem:
The sequence (an​) satisfies a1​=1 and 5(an+1​−an​)−1=n+32​1​ for n≥1. Let k be the least integer greater than 1 for which ak​ is an integer. Find k.
Solution:
Rearranging the given equation and taking the logarithm base 5 of both sides yields
an+1​−an​=log5​(3n+5)−log5​(3n+2)
Successively substituting n=1,2,3,… and adding the resulting equations produces an+1​−1=log5​(3n+5)−1. Thus the closed form for the sequence is an​=log5​(3n+2), which is an integer only when 3n+2 is a positive integer power of 5 . The least positive integer power of 5 greater than 1 of the form 3k+2 is 53=125=3⋅41+2, so k=41​.
The problems on this page are the property of the MAA's American Mathematics Competitions