A curious sequence of rationals: finite or infinite? Consider the following function repeatedly applied to a rational
$r = a/b$ in lowest terms:
$f(a/b) = (a b) / (a + b - 1)$.
So, $f(2/3) = 6/4 = 3/2$. $f(3/2) = 6/4 = 3/2$.
I am wondering if it is possible to predict when the sequence is finite,
and when infinite.  For example, $f(k/11)$ seems infinite for
$k=2,\ldots,10$, but, e.g., $f(4/13)=f(13/4)$ is finite.
Several more examples are shown below.

$$ \frac{1}{k} \to \frac{k}{k} {=} 1 $$

$$ \frac{3}{7} \to \frac{21}{9} {=} \frac{7}{3} \to \frac{21}{9} {=} \frac{7}{3} $$
$$ \frac{5}{12} \to \frac{60}{16} {=} \frac{15}{4} \to \frac{60}{18} {=} \frac{10}{3}
\to \frac{30}{12} {=} \frac{5}{2} \to
\frac{10}{6} {=} \frac{5}{3} \to
\frac{15}{7} \to \frac{105}{21} {=} \frac{5}{1} \to \frac{5}{5} = 1 $$
$$ \frac{5}{23} \to \frac{115}{27} \to \frac{3105}{141} {=} \frac{1035}{47}
\to \frac{48645}{1081} = \frac{45}{1} \to 1 $$
$$ \frac{4}{11} \to \frac{44}{14} {=} \frac{22}{7} \to \frac{154}{28} {=} \frac{11}{2} \to
\frac{22}{12} {=} \frac{11}{6}
\to \frac{66}{16} {=} \frac{33}{8} \to \frac{264}{40} {=} \frac{33}{5}
\to \frac{165}{37} \to \cdots \to \infty ?$$

I am hoping this does not run into Collatz-like difficulties,
but it does seem straightforward to analyze...
If anyone recognizes it, or sees a way to tame it even partially, I would be interested to learn.  Thanks!

All fractions $a/b$ with $ a = b(b-1)+1 $
are fixed points  because then $(a,b)=1$. Moreover, all solutions to $f^{(2)}(r)=1$ I found so far  come from one of these families:
$$f\left(f\left(\frac{1}{n}\right)\right)=f\left(1\right)=1$$
$$f\left(f\left(\frac{(n-1)^2}{n^2}\right)\right)=f\left(\frac{n(n-1)}{2}\right)=1$$
$$f\left(f\left(\frac{2n-1}{2n}\right)\right)=f\left(n\right)=1$$
$$f\left(f\left(\frac{n+1}{n^2}\right)\right)=f\left(n\right)=1$$
$$f\left(f\left(\frac{3(2n+1)}{4(3n+1)}\right)\right)=f\left(4n+2\right)=1$$
If initially not much cancellation happens during the iteration and $a\approx b$, then$f^{(n)}(a/b)$ grows superexponential, probably close to $a^{\phi^{n-1}}$ which makes substantial cancellation more and more unlikely, especially that the numerator eventually hits by chance a perfect multiple of the denominator.
