Problem:
Consider the sequence (ak)k≥1 of positive rational numbers defined by a1=20212020 and for k≥1, if ak=nm for relatively prime positive integers m and n, then
ak+1=n+19m+18.
Determine the sum of all positive integers j such that the rational number aj can be written in the form t+1t for some positive integer t.
Solution:
Note that all the terms in the sequence (ak)k≥1 are strictly between 1918 and 1 . Call an integer j simple if the rational number aj can be written in the form t+1t for some integer t>18. Suppose j is a simple positive integer and term j of the sequence is aj=t+1t. Let t=p1p2⋯pℓ+18 with p1≤p2≤⋯≤pℓ being the primes in the prime factorization of t−18. Note that for any positive integer k, the greatest common divisor of t+18k and t+1+19k is
gcd(t+18k,t+1+19k)=gcd(t+18k,k+1)=gcd(t−18,k+1)
Thus this greatest common divisor is first greater than 1 when k=p1−1, in which case the greatest common divisor is equal to p1. At that point,
aj+p1−1=t+1+19(p1−1)t+18(p1−1)=p1p2⋯pℓ+19p1p1p2⋯pℓ+18p1=p2p3⋯pℓ+19p2p3⋯pℓ+18
so j+(p1−1) is the next integer greater than j that is simple. By the same reasoning, the numbers
j+(p1−1)+(p2−1),…,j+(p1−1)+(p2−1)+(p3−1)+⋯+(pℓ−1)
are all the simple numbers exceeding j+(p1−1).
The first simple number is j=1 for which t=2020=2002+18=2⋅7⋅11⋅13+18. Therefore the sequence of simple numbers is 1,2,8,18, and 30 . The requested sum is
1+2+8+18+30=59