(In reply to
True but unprovable by Steve Herman)
another cycle is 3,6,3
s(3)=3+3=6
s(6)=6/3=3
s(3)=3+3=6
And this can be proven easily for all integers n.
first prove for n=1
s(1)=4
s(4)=2
s(2)=1
thus it cycles.
Now assume it cycles with either 4,2,1 or 3,6 for all integers less than n
if n is even the we have n=2k
s(n)=k<n and thus by assumption will eventually cycle
on the other hand if n is odd then n=2k+1
s(n)=2k+4
s(2k+4)=k+2
k+2<2k when k>2 and since 2 is part of one of the cycles then n will eventually cycle for all integer values.
Edited on March 2, 2009, 8:06 pm

Posted by Daniel
on 20090302 19:34:39 