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
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.
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Posted by Richard
on 2006-08-19 21:54:57 |
re(4): Just a guess... -- proof
