Björn,

This is an interesting and important topic. The short answer is "no" we don't have code for this situation out of the box.

However, the flexibility of Maxima means you could create a function which supports this situation. E.g. a function

`recurring(n,m)`

which holds the non-recurring bit n and then the recurring bit m. More recent versions of STACK allow you to then deal with this with tex.
Assuming you are happy with the recurring bit as an integer you would do

`texput(recurring, lambda([ex], sconcat(tex1(first(ex)),".\\bar{",tex1(second(ex)),"}") ));`

Then, in the PRT you can define the function "recurring" as a rational expression, i.e.

```
recurring(n,m) := block([k],
k:1+floor(float(log(m)/log(10))),
n+m/10^k);
```

You probably want a more flexible situation in which you can do more, so your display function becomes

`texput(recurring, lambda([ex],if integerp(first(ex)) then sconcat(tex1(first(ex)),".\\bar{",tex1(second(ex)),"}") else sconcat(tex1(first(ex)),"\\bar{",tex1(second(ex)),"}") ));`

This is, of course, all untested. I hope this gives you a pointer as to how you could define a Maxima function to represent a recurring decimal, and (a) display it, and (b) evaluate the function later in terms of the rational number it represents.
The input side of things is more difficult. We have ambitious plans to enable teachers to have much more flexible input options, to either interpret students' input or combine two boxes (start and recurring bit) into a single value to show the student "your answer is ...." For now input is somewhat fixed.
Anyway, I hope this gives you some idea of what would need to be done.