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# Bash+Linux command line utils, 43 bytes

 jot -w%x $[16**9]|egrep -v "[0ef]|(.)1円{3}"

This uses a similar technique to my answer below, but just counts in base 16, and strips out all "names" containing `0`, `e` or `f` as well those with more than 3 same consecutive digits.

Convert to the monk's alphabet as follows:

 jot -w%x $[16**9]|egrep -v "[0ef]|(.)1円{3}" | tr 1-9a-d A-M

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# Bash+coreutils (dc and egrep), 46 bytes

**Edit - corrected version**

 dc<<<Edo9^[p1-d0\<m]dsmx|egrep -v "0|(.)1円{3}"

This'll take a while to run but I think its correct.

`dc` counts downwards from 14^9 to 1 and outputs in base 14. egrep filters out the numbers with more than 3 consecutive same digits. We also filter out any names with "0" digits, so we get the correct set of letters in the names.

The question specifies that any alphabet may be used, so I am using [1-9][A-D]. But for testing, this can be transformed to [A-M] using tr:

 dc<<<Edo9^[p1-d0\<m]dsmx|egrep -v "0|(.)1円{3}" | tr 1-9A-D A-M

This yields the sequence:

 MMMLMMMLM MMMLMMMLL MMMLMMMLK ... AC AB AA M L K ... C B A

*Note this `dc` command requires tail recursion to work. This works on dc version 1.3.95 (Ubuntu 12.04) but not 1.3 (OSX Mavericks).*