Let's look at the sequence with terms
a
_{1}=19,
a
_{2}=95, and a
_{n+2}=LCM(a
_{n+1},a
_{n})+a
_{n}
LCM stands for Least Common Multiple, and n is a positive integer.
Find the Greatest Common Divisor (GCD) of terms a_{4096} and a_{4097}.
Let X(n+1) be a(n+1) divided by the GCD of a(n) and a(n+1). Note that X(n+1) times a(n) equals the LCM of a(n) and a(n+1), as the product of two numbers equals their LCM times their GCF.
Thus, a(n+2)=a(n) + a(n+1)*a(n)/GCD(a(n),a(n+1)) or (a(n)*(1+a(n+1)/GCD())=a(n)*(1+X(n)) and note that X(n)+1 is relatively prime to a(n+1).
Thus, the problem can be simply reduced to "What is the GCD of a(n2) and a(n1), which applied enough times gives "what is the GCD of a(1) and a(2)"

Posted by Gamer
on 20060819 21:55:03 