Thank you to
Chris for this solution. Chris has just left school and is about to
start his course at Oxford University this month. Balakhrishnan
also found the cycles but did not give proofs.
Recursive functions like this can be represented graphically by
considering the line $y=x$ and all the points of $f(x)$ as in the
graph below:

(1) Take your first point from $f(x)$.
(2) Draw a line horizontally to meet the line $y=x$.
(3) Draw a line vertically to meet a point on the line $f(x)$.
There will never be more than one point with the same $x$ value as
$f(x)$ because $f(x)$ has a unique value.
(4) Repeat steps two and three.
If we take a look at the graphs of $y=x$ and $y=f(x)$ we can
see that only a few points of $f(x)$ lie above the line $y=x$. For
these points alone $f(x)> x$ as for all other points step 2 will
draw a line to the left i.e decreasing in value.

We can see the same thing analytically. The terms increase if
and only if $2a + 3b > 10a + b$, that is iff $b > 4a$. As $b$
is a single digit number the only terms that are followed by larger
terms are 1, 2, 3, 4, 5, 6, 7, 8, 9, 15, 16, 17, 18, 19 and 29 as
seen on the graph.
Large terms reduce by one fifth in order of magnitude, and
once the sequence reaches 2 digit numbers the largest value is
given by $a=b=9$ which maps to 45. For numbers less than 45 $2a+3b
< 33$. As there are a finite number of values the sequence can
take, values are repeated and cycles occur.
There are three points that lie on the line $y=x$. These are
0, 14 and 28 for which values $x=f(x)$. Alternatively, we see that
for a fixed point $10a + b = 2a + 3b$ so $4a=b$. Hence the only
fixed points are 0, 14 and 28.
There are eight distinct loops (five 6cycles, one 2cycle and
three fixed points) which are:
[0, 0]
[2, 6, 18, 26, 22, 10, 2]
[4, 12, 8, 24, 16, 20, 4]
[5, 15,17, 23, 13, 11, 5]
[7, 21, 7]
[9, 27, 25, 19, 29, 31, 9]
[14, 14]
[28, 28]
Other terms which are not in the cycles but lead into the
cycles are:
1,3, $\to$ 9 etc.
30 $\to$ 6 etc.
32 $\to$ 12 etc.
33 $\to$ 15 etc.
We have proved that all higher numbers reduce to 33 or less so
sequence with all other starting points enter one of the listed
cycles or end up at a fixed point.
Why are all sequences all odd or all even?
even= 0 (mod 2)
odd=1 (mod 2)
When deciding if a number is odd or even we only need to look
at the last digit, in this question defined as '$b$'. The only
operation on $b$ is $b\to 3b$ and $3\times 0 \equiv 0{\rm mod}\ 2$
so it is even and $3\times 1 =3 \equiv1 {\rm mod}\ 2$ so it is odd.
Therefore $f(x) \equiv x {\rm mod}\ 2$.
The sequences are all odd or all even because $$2a + 3b \equiv
10a + b \quad {\rm mod}\ 2.$$