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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.)
Solution re: Just a guess... -- proof | Comment 2 of 7 |
(In reply to Just a guess... by Dej Mar)

LCM(a(n),a(n+1)) = GCD(a(n),a(n+1)) * u1 * u2,

where u1 is a(n)/gcd and u2 is a(n+1)/gcd.

so
a(n+2) = GCD(a(n),a(n+1)) * u1 * u2 + a(n)

            = a(n) * u2 + a(n)

            = a(n) *(u2 + 1)

The only gcd this shares with a(n+1) is the same GCD(a(n),a(n+1)) as u2+1 can't chare common divisors with u2, and a(n+1) is in fact just GCD(a(n),a(n+1))  * u2. 

This confirms 19 as the answer.


  Posted by Charlie on 2006-08-19 16:21:41
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