TITLE: A curious sequence of rationals: finite or infinite?
QUESTION [19 upvotes]: 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!

REPLY [6 votes]: Note:  I've appended some additional material that extends the analysis of, in effect, the pre-images of the pre-images of the fixed point $n/m=1$ of the function $f$, to look at general pre-images and pre-images of other fixed points (i.e., $n/m = (m^2-m+1)/m$ for $m>1$.  The new stuff follows a boldface "Added later."  End of note
The families Karl Fabian found can be subsumed under a single rule:  

$f^{(2)}(a/b)=1$ if and only if $${a\over b}={n+d\over n+d'}$$ where
  $n>0$,
  $dd'=n(n-1)$, and $(n+d,n+d')=1$.

It's not hard to show that if $a/b$ has the given form, then $f(a/b)=n$, and it's easy to show that $f(n)=f(n/1)=1$.  (The assumption that $n+d$ and $n+d'$ are relatively prime is crucial.)  The tricky part is showing that nothing else gets to $1$ in two steps.
It's clear that $f(a/b)=1$ if and only if $a+b-1=ab$, which can be rewritten as $(a-1)(b-1)=0$, so the only numbers that get to $1$ in one step are integers $n$ and their reciprocals $1/n$.  It's also clear that $f(a/b)\ge1$ for all $a/b$.  Therefore, the numbers that get to $1$ in two steps are those that get to some integer $n$ in one step.  Let's see how that can happen.
If $ab/(a+b-1)=n$, then we must have $ab=nk$ and $a+b-1=k$ for some integer $k$.  Writing $b=nk/a$, we wind up with $a^2-(k+1)a+nk=0$, so
$$a={k+1\pm\sqrt{(k+1)^2-4nk}\over2}.$$
For $a$ to be an integer, we must have a square inside the square root:
$$(k+1)^2-4nk = m^2,$$
which can be rewritten as
$$(k+1-2n)^2-m^2 = 4n(n-1),$$
or
$$(k+1-2n+m)(k+1-2n-m)=4n(n-1).$$
The two terms on the left hand side have the same parity (they differ by $2m$), hence they must both be even, i.e., $k+1-2n+m=2d$ and $k+1-2n-m=2d'$ where $dd'=n(n-1)$.  From this we get $k+1-2n=d+d'$ and $m=d-d'$, so
$$a={2c+d+d'\pm(d-d')\over2}.$$
Choosing the positive sign gives $a=n+d$.  (Choosing the negative sign gives $b=n+d'$.  If you like, you can assume $a\ge b$ and $d\ge d'$.)
To do just one example, let $n=16$.  The choices for $dd'$ are $240\cdot1$, $80\cdot3$, $48\cdot5$, and $16\cdot15$, leading to the four possibilites for $a/b$ (with $a>b$) for which $f(a/b)=16$:
$${a\over b} = {256\over 17}, {96\over19}, {64\over21}, {32\over31}.$$
Note how a factorization like $dd'=6\cdot40$ fails:
$$f((16+6)/(16+40)) = f(22/56)=f(11/28)= {11\cdot28\over38}={154\over19}.$$
Added later:  If I've done everything correctly, a similar analysis gives the following nice result about pre-images:

Let $n/m$ be a fraction with $n\ge m$ and
  $(n,m)=1$.  Then $f(a/b)=n/m$ if and
  only if
$${a\over b} = {n+d\over n+d'}$$
where $dd'=n(n-m)$ and $(n+d,n+d')=m$.

Let's see how this applies to the other fixed points for $f$, namely when $n=m^2-m+1$, for the first few values of $m$.
Skipping the case $m=1$ (which you can check gives the result noted earlier), let $n/m = 3/2$, so that we need $dd'=3$ and $(3+d,3+d')=2$.  The only factors are $3$ and $1$, which indeed satisfy $(6,4)=2$, but this only gives $a/b = 6/4=3/2$.  In other words, the fixed point $3/2$ has no pre-image other than itself.
For $n/m=7/3$, we have $n(n-m)=28$, for which the possible factorizations are $28\cdot1$, $14\cdot2$, and $7\cdot4$.  But the only one of these for which $(7+d,7+d')=3$ is $14\cdot2$, which gives $a/b = 21/9 = 7/3$, so $7/3$ also has no pre-image other than itself.
You can check that the same thing happens for $n/m = 13/4$:  The only factorization $dd'$ of $13(13-4)$ for which $(13+d,13+d')=4$ is $39\cdot3$, giving $a/b=52/16 = 13/4$.
But $n/m = 21/5$, finally, is interesting:  There are two factorizations that work, namely $84\cdot4$ and $24\cdot14$.  The first, as before, gives the fixed point again, $$a/b = (21+84)/(21+4) = 105/25 = 21/5.$$  But the other one gives $$a/b = (21+24)/(21+14) = 45/35 = 9/7.$$  Note, however, that $9/7$ has no pre-image:  The factorizations of $9(9-7)=18$ are $18\cdot1$, $9\cdot2$, and $6\cdot3$, none of which produce anything divisible by $7$, much less a pair $(9+d,9+d')$ with $7$ as a common divisor.
In summary (for now), the only fraction with infinitely many pre-images is $n/m=1$ (since $n(n-m)=0$ has infinitely many divisor pairs!); the size of the pre-image set for all other fractions is bounded by the number of divisor pairs of $n(n-m)$.  For some of the fixed points of $f$, the pre-image set is just the fixed point itself, while for others (e.g. $n/m = 21/5$), the pre-image set contains additional points.  There may be some simple criterion that identifies the fixed points that have no additional pre-images, but I don't offhand see one, possibly because I haven't thought hard enough about it (but maybe because I'm just blind to the obvious).