Conference rusure::math

    This pattern appears to work; it's two cycles across by one cycle down.
    * marks the single infinite strip referenced in .0. 
    
         3 4 5 6 7 1 2 3 4 5 6 7 1 2 
    *     1 2 3 4 5 6 7 1 2 3 4 5 6 7    
         5 6 7 1 2 3 4 5 6 7 1 2 3 4 
          3 4 5 6 7 1 2 3 4 5 6 7 1 2
         7 1 2 3 4 5 6 7 1 2 3 4 5 6 
          5 6 7 1 2 3 4 5 6 7 1 2 3 4
         2 3 4 5 6 7 1 2 3 4 5 6 7 1 
          7 1 2 3 4 5 6 7 1 2 3 4 5 6 
         4 5 6 7 1 2 3 4 5 6 7 1 2 3 
          2 3 4 5 6 7 1 2 3 4 5 6 7 1
         6 7 1 2 3 4 5 6 7 1 2 3 4 5 
          4 5 6 7 1 2 3 4 5 6 7 1 2 3 
    *    1 2 3 4 5 6 7 1 2 3 4 5 6 7 
          6 7 1 2 3 4 5 6 7 1 2 3 4 5
    					 The mirror image also works, but
    -----------------------------------  it's probably cheating to count
    					 it as a separate tiling.
          6 7 1 2 3 4 5 6 7 1 2 3 4 5
    *    1 2 3 4 5 6 7 1 2 3 4 5 6 7 
          4 5 6 7 1 2 3 4 5 6 7 1 2 3 
         6 7 1 2 3 4 5 6 7 1 2 3 4 5 
          2 3 4 5 6 7 1 2 3 4 5 6 7 1
    	   etc., 
    
    It appears to be very difficult to arrange anything in other than the
    order 1234567, if one line cycles thru that order.
    
      John

Assume you start by painting only 7 hexagons in a row like this:

    1 2 3 4 5 6 7

Prove that a "nice" painting must be extended so as to be periodic in that row,
so that you end up in the case given in .0.

Prove that, up to renumbering of colors, there are only two patterns that are
nice, which are mirror images (the ones given in .1).

My (alleged) proof works by proving these lemmas.  Can anyone shorten the proof?

1. Infinite repeated 7-strip in one line => one of the two given patterns.
2. 7-strip => infinite repeated 7-strip in one line.
3. 6-strip => 7-strip.
4. 5-strip => 6-strip.
5. 4-strip => 5-strip.
6. 3-strip => 4-strip.

Here, an n-strip is n consecutive hexagons in a straight line with different
colors.  Of course, any 3 hexagons in a line are of different colors, so these 6
lemmas together imply the theorem.

Interestingly, the two patterns are *not* equivalent under renumbering,
rotation, or translation, but only under mirror imaging.

I have found a much shorter proof, but I'm still hoping for an improvement from
the readers.  This lemma is the first part:

Lemma.  In a nice pattern, any 4 consecutive hexes in a straight line are
different colors.

Proof.  Assume otherwise; the color pattern, up to renumbering, of the 4 hexes
must be 1 2 3 1.  We can then renumber colors if necessary so the pattern looks
like this around the 2:

       4 5
      1 2 3 1
       6 7

The two remaining hexes next to the 3 must have 6 on top and 4 on bottom,
otherwise there are two 4s next to the 5:

       4 5 6
      1 2 3 1
       6 7 4

Now one of the hexes next to the 5 must be colored 1, but either position
conflicts with one of the 1s.

Define distance between hexes in the obvious way; a nice coloring is simply one
where d(x,y) = 1 or 2 implies x and y are different colors (and there are 7
colors total).  Take any hex, and call its color 1; consider its neighborhood:

       x y
    y o + o x
   x o o o o y
    o o 1 o o
   y o o o o x
    x o o o y
       y x

The + and the o's cannot be colored 1; since one of the neighbors of + must be
colored 1, it must be either the x or the y next to the +.  Similarly, in every
pair of adjacent xy hexes, one of them must be colored 1.  WLOG, we may assume
the x next to the + is colored 1 by taking a mirror image if needed.  But if the
x is colored 1, it excludes the y that is distance 2 in the xy group
counterclockwise from the +; so (continuing around) all the x's must be colored
1 and none of the y's.

It is then easy to see that the pattern for the color 1 is completely
determined, and must replicate the following pattern:

     o 1 o o
    o o o o 1
   1 o o o o o
  o o o 1 o o o
   o o o o o 1
    1 o o o o
     o o 1 o

It is also easy to see, by considering neighbors of the central hex, that every
other color must use the same pattern and not the mirror image one, since they
would conflict.  Extending the pattern to the right shows that the nearest hex
in a straight line with color 1 is at distance 7; so 7 consecutive hexes in a
line must be different colors (because all colors occur with the same pattern). 
Renumbering, we see that the pattern is unique up to reflection.