This computes 3x where x is a string of 1's.

We use example 9.10 from Linz, adding a few instructions to get xxx instead
of xx

These are from example 9.10...

(01x0r   Replace all 1's with x's
(0  1l   State 1:  replace an x with a 1 and go to state 2
(1111l
(1x12r
(2112r   State 2:  append a 1 to the end of the string
!(2 11l
(1  3r   State 3:  final state... all the x's were replaced.

Add the following to replace the quintuple which is commented out...

(2 18r  State 2:  append a 1, then
(8 11l  State 8:  append a second 1, then continue as in the replaced quint.