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.
(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.
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Posted by Charlie
on 2006-08-19 16:21:41 |
re: Just a guess... -- proof
