If we have a positive integer $n$ and a function $f(n)$, such that $f(n)=3n$ , we can see that the prime factorisation of $n$ simply changes to includes another 3 within its prime factorisation.
E.g $n = 12 = 2 \cdot 2 \cdot 3$ becomes $f(n)=3(12)=2 \cdot 2 \cdot 3 \cdot 3$
If we have a positive integer $n$ and a function $f(n)$, such that $f(n) = n/2$ and we stop when the end result is not a whole number, we can see that the prime factorisation of $n$ simply changes to remove another 2 within its prime factorisation.
E.g $n = 12 = 2 \cdot 2 \cdot 3$ becomes $f(n)=12/2= 2 \cdot 3$
Any function that multiplies an integer $n$ by an integer $a$ $[f(n)=a \cdot n]$ simply includes more primes to the prime factorisation of $n$ (aka, the primes of a).
Any function that divides an integer $n$ by an integer $a$ $[f(n)=a/n]$ simply removes more primes from the prime factorisation of $n$ (aka, the primes of a), given that the end result remains an integer.
This is simple. On the other hand, when there is a positive integer $n$ and a function $f(n)$, such that $f(n) = n + 1$, why does the prime factorisation of the input and output differ so much?
E.g $n = 12 = 2 \cdot 2 \cdot 3$ becomes $f(n) = 12 + 1 = 13$
Primes are supposed to be the atoms of number theory and of the natural numbers. Yet, when such a simple operation such as adding 1 is applied to any positive integer, the Prime Factorisation suddenly... changes? How can the atoms of a number just change like that? Is there any correlation between the input prime factorisation and the output prime factorisation?