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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(4): Just a guess... -- proof | Comment 5 of 7 |
(In reply to re(3): Just a guess... -- proof by Charlie)

OK then, I see where you are coming from.  If we suppose that P is a prime that divides both a(n+2) and a(n+1), but does not divide a(n), then it must divide [lcm(a(n+1),a(n))/a(n)]+1 and hence cannot divide R=lcm(a(n+1),a(n))/a(n). But P does divide R since it divides a(n+1) and not a(n), and this contradiction then proves that P does divide a(n) as well as a(n+1) and a(n+2).

The easily seen fact that a prime P that is contained in A but not in B must be contained in lcm(A,B)/B is the crux, then.

  Posted by Richard on 2006-08-19 21:54:57

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