Counting

For three consecutive tiles with numbers $a_i, a_{i+1}, a_{i+2}$, we require $a_i + a_{i+1} + a_{i+2} = 3$.

Since each number is in {-1, 0, 1}, the maximum sum is $1 + 1 + 1 = 3$, so $a_i = a_{i+1} = a_{i+2} = 1$.

So, all 7 tiles must have number 1 to satisfy for all consecutive triples.

Each of the 7 tiles, all numbered 1, can be red or blue, giving $2^7 = 128$ color assignments.

So there are 128 possible arrangements.

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