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How many possible shapes can one make by rearranging $n$ square shaped blocks, with and without allowing rotational symmetry? For example, for $n = 4$, there are seven possible shapes after discounting rotational symmetry, as in Tetris.

How does this number change if one of the blocks is coloured – distinguishable from the rest? For instance, for $n = 2$ there are only two shapes one can make (with rotational symmetry admissible). However, if one block is coloured, one can make four shapes, with the second block left, right, above and below the coloured block.

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Have you seen ? – Matthew Conroy Sep 10 '12 at 17:07
You've seen this? – J. M. Sep 10 '12 at 17:08
Nope, I hadn't. This is perfect, thanks! I'd upvote the comments, but I'm new on StackMath. – hauntsaninja Sep 10 '12 at 18:10
I don't understand why you consider the four colored polyominoes the same: they can all be transformed into one another by rotations. With $n=3$, there should be 4 polyominoes with one colored square: the three squares can be in a straight line, either either an end square or a middle square colored, or they can be in an L-shape, with either and end square or the elbow colored. But with $n=2$ there is only one way to do it. – MJD Sep 10 '12 at 19:06
up vote 2 down vote accepted

There has been much work done on polyominoes. The number counting rotations and reflections as the same is given in OEIS A000105. Counting reflections as distinct but rotations as the same gives A00988

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You might also want to mention that the problem is still open, despite the large amount of work that has been done. – MJD Sep 10 '12 at 19:18

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