Disclaimer: I am not familiar with the technical definitions of ring theory or other possibly relevant concepts, so feel free to share the good definitions.
Let $n$ and $k$ be positive integers with $k\lt n$.
Then there exists an integer $r\ge 1$ so that $nk^{r-1}$ can be expressed as a product of $r$ factors such that for each factor $a$, $k\lt a\le n$ ($a=n$ if $r=1$) and for each integer $s\ge 2$, $ak^{s-1}$ cannot be expressed as a product of $s$ factors such that for each factor $b$, $k\lt b\lt a$.
For example, although $9(2^{2-1})=3\cdot 6$ and $2\lt 3,6 \le 9$, this factorization does not satisfy the second statement since $6(2^{2-1})=3\cdot 4$ where $2\lt 3,4 \lt 6$. A qualifying factorization would be $9(2^{3-1})=3\cdot 3\cdot 4$.
What we are actually looking at is the domains of the rational numbers that start at $1+\frac{1}{k}$ and differ by $1/k$ and we're interested in each of these numbers' irreducible factorizations into numbers within their domains. So in the above example we can say that the prime factorization of $\frac{9}{2}$ is $\frac{3}{2}\cdot\frac{3}{2}\cdot\frac{4}{2}$.
For $k=1$ we have the usual prime factorizations of the natural numbers $$2, 3, (2\cdot 2), 5, (2\cdot 3), 7, \dots$$
For $k=2$ we have $$\frac{3}{2}, \frac{4}{2}, \frac{5}{2}, \left(\frac{3}{2}\cdot\frac{4}{2}\right), \frac{7}{2}, \left(\frac{4}{2}\cdot\frac{4}{2}\right), \left(\frac{3}{2}\cdot\frac{3}{2}\cdot\frac{4}{2}\right), \left(\frac{4}{2}\cdot\frac{5}{2}\right), \frac{11}{2}, \left(\frac{3}{2}\cdot\frac{4}{2}\cdot\frac{4}{2}\right),\dots$$
For $k=3$ we have $$\frac{4}{3}, \frac{5}{3}, \frac{6}{3}, \frac{7}{3}, \left(\frac{4}{3}\cdot\frac{6}{3}\right), \frac{9}{3}, \left(\frac{5}{3}\cdot\frac{6}{3}\right), \frac{11}{3}, \left(\frac{4}{3}\cdot\frac{9}{3}=\frac{6}{3}\cdot\frac{6}{3}\right),\cdots$$
In the very last example, we see that the representation of $\frac{12}{3}$ is not unique so not all of or none of these rings for $k\ge 2$ are unique factorization domains. I suppose we should ask, is any of them a UFD?
But the second question I want to ask is, of those that are not UFD, do any numbers in those domains have multiple factorizations that differ in number of prime factors? So far I have not found any.
And also it is clear that if the numerator of a $1/k$-fraction is a prime from the natural numbers ($2$,$3$,$5$,...), then that fraction will be a prime in its domain. Then what about the sequences of $1/k$-fraction primes whose numerators are not natural primes? Are those sequences infinite?