TITLE: Does iterating a certain function related to the sums of divisors eventually always result in a prime value?
QUESTION [34 upvotes]: Let define the following function for integers (from 2): $f(x)=\sigma(x)-1$, where $\sigma$ is the sum of the divisors of $x$.
For example $f(6)=6+3+2=11$, $f(5)=5$.
Note that $x$ is a fixed point for $f$ if and only if, $x$ is prime.
If we iterate starting at any integer $x$ we get a dynamical system.
Computations with Maple showed that for all integer $x$ in $[2,2000000]$ there exists an integer $n$ such that $(f \circ f \circ \dots \circ f)(x)$ is prime; that is for each integer $x$ the sequence generated is stationary.
The question is: can anyone help proving the conjecture ?
Thank you for any help on the subject.

REPLY [5 votes]: Every prime occurs at least once (obviously) as the end of a trajectory. However many occur only once and some occur quite often. Since $f$ is non-decreasing it is easy by simple computation to find all $x$ which end up at a particular prime $p$. 
Primes such that $p+1$ has many small prime factors are likely to occur often. For example, there are $125$ starting points which end at $5039$ but the nearby primes $5011,5023,5051, 5059$ and $5057$ can be reached only from themselves while $5021$ can be reached starting from itself and one other place, $2650=2\ 5^2\ 53$ with $\sigma(2650)=(2+1)(25+5+1)(53+1)=2\ 3^4\ 31=5022.$ It is not too hard to establish that $\sigma(x)=2651$ has no solutions.
On the other hand, $5040=2^33^25\ 7$ can be factored in may ways as the product of several   factors, which must have only small prime factors. Here are all the ways so that each factor is one less than a prime.
$[3, 4, 14, 30], [3, 6, 14, 20], [3, 4, 420], [3, 12, 140], [3, 20, 84], [4, 14, 90], $$[4, 30, 42], [4, 1260], [6, 14, 60], [6, 20, 42], [12, 14, 30], [6, 840],$$ [14, 18, 20], [12, 420], [14, 360], [20, 252], [30, 168], [60, 84]$
These, with $5039$ itself, give the $19$ square free integers with $\sigma(x)-1=5039.$ There are others which are not square free, such as $x=4y$ for odd $y$ with $\sigma(y)=720$ (also examples coming from $\sigma(3^3)=40$ and from  $\sigma(2^5)=63$.) There are as well starting points which land at $5039$ after several steps.
I'd expect $6719=(2^6\ 3\ 5 \ 7)-1$ to occur quite often, but I haven't checked.