[seqfan] Re: Add to an integer its distinct factors and loop
franktaw at netscape.net
franktaw at netscape.net
Sat Jun 6 01:11:07 CEST 2009
I've been playing around with this a bit in PARI. It appears that
there are infinite paths from 60, including infinite paths of all even
numbers. Nothing smaller works. After 30 steps, there are 34610
possible numbers, and this seems to be growing at about 50% with each
step.
The first odd number that (apparently) has infinite paths leading from
it is 945 (coincidently, the smallest odd abundant number). After 32
steps, there are 8754 possible nodes, and this seems to be growing by
25%-75% with each step.
Franklin T. Adams-Watters
-----Original Message-----
From: franktaw at netscape.net
Sorry, that isn't quite correct.
If n is divisible by exactly 2^{2k+1}, then the next term can be
divisible by at most 2^k. However, if n is divisible by exactly 2^{2k}
for k > 0, the next term can be divisible by at least 2^{k+1}, and
possibly more. Still, this is a strong tendency for paths to wind up
with odd numbers, and once you get there there is no escape.
Franklin T. Adams-Watters
-----Original Message-----
From: franktaw at netscape.net
...
This more restrictive definition is of some interest, though. To
extend Hagen's remarks, note that if n is even, everything pointed to
by n is divisible by a smaller power of two than n; so any path will
fairly quickly come to only odd numbers. I think there still are
infinite paths, but it isn't as clear as for the original problem.
Franklin T. Adams-Watters
-----Original Message-----
From: Hagen von EItzen <math at von-eitzen.de>
Ah, now I see:
We are considering the directed graph with the natural numbers as
vertices
and an edge from n to m iff there are natural numbers a,b such that
1<a<b and n=a*b and m=n+a+b.
...
More information about the SeqFan
mailing list