# [seqfan] Re: Add to an integer its distinct factors and loop

franktaw at netscape.net franktaw at netscape.net
Fri Jun 5 22:24:35 CEST 2009

```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.

-----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.

-----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.
...

```