
Hi!
  Here is what I already know from sample output recently send by 3200 user.
See questions bellow:

Lexmark 3200
------------
Very strange. All Escapes are 8 bytes long:

1b 2a 80 00 00 00 00 00 
1b 33 00 00 00 00 00 33
1b 30 80 0c 02 00 00 be 
1b 21 00 00 00 00 00 21
1b 23 80 01 fc 00 00 a0 
1b 40 11 00 e0 01 e8 1a
1b 42 00 00 00 01 a0 e3 
1b 43 00 05 d6 1b a0 d9

Data packets follow:
83 ?? ?? FF pixels
   ^^ ^^ this defines number of pixel bytes: 1 = no pixel data
                                             0 = 1 byte of pixel data follow
c3 ?? ?? FF pixels
   ^^ ^^ as above
  pixel format unknown

Example:
83 ff ff ff 
   ^^ ^^ no pixel data (white on paper)
83 f7 ff ff 10 
   ^^ ^^ 1111 0111 1111 1111 - 1 zero bit = 1 data byte follows (0x10)
83 f7 7f ff 38 14 41
   ^^ ^^ 1111 0111 0111 1111 - 2 zero bits = 2 data byte follow (0x14 0x41)

These were seen on epilogue:
1b 22 80 3a f4 00 00 d0
1b 31 10 01 e8 00 00 2a 
1b 33 00 00 00 00 00 33

===============================================================================
What I need to know:
-  escape for paper shift (look for escapes between two printing passes)
-  escape for paper eject (probably one of the 3 above - verify)
-  how are coded data for 83 ?? ?? ff & C3 ?? ?? ff packets
-  how is coded horizontal offset start and end, where is coded number
   of 83 ?? ?? ff & C3 ?? ?? FF packets (probably in some packet before these
   bytes)

Decoding paper shift:
    Paint something on differnet vertical position of paper and 
    look for appropriate escape.
Decoding paper eject:
    Print more pages and see for escapes...
Data coding:
    Modify data bits (not 4 byte packets) and see difference
Horizontal offsets:
    Try to print the same painting at the different horizontal offset and
    examine escapes.

Sincerely
  ...........


