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Jan 2, 2018 at 21:16 comment added Gerhard Paseman Indeed, with all the odd prime tapes, every square gets a finite number of colors, and only two squares get no colors. When I add the tape for two, exactly one of these uncolored squares gets a color. Thus this solution is different from my other solution where all squares get at least one and at most finitely many different colors. (It does resemble the explicit example posted by Achim Krause.) However, I find this exposition even more appealing. Gerhard "Math Is In The Telling" Paseman, 2018.01.02.
Jan 2, 2018 at 20:57 history answered Gerhard Paseman CC BY-SA 3.0