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LCM Sequence (Posted on 2006-08-19) Difficulty: 3 of 5
Let's look at the sequence with terms a1=19, a2=95, and an+2=LCM(an+1,an)+an

LCM stands for Least Common Multiple, and n is a positive integer.

Find the Greatest Common Divisor (GCD) of terms a4096 and a4097.

No Solution Yet Submitted by atheron    
Rating: 4.0000 (1 votes)

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Solution Reduction | Comment 6 of 7 |

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)"


  Posted by Gamer on 2006-08-19 21:55:03
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