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 a4096 and a4097.
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(n-2) and a(n-1), which applied enough times gives "what is the GCD of a(1) and a(2)"
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Posted by Gamer
on 2006-08-19 21:55:03 |
Reduction
