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.

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