%I M0171 N0066
%S 2,1,4,7,24,62,216,710,2570,9215,34146,126853,477182,1802673,6853152,
%T 26153758,100215818,385226201,1485248464,5741275753,22246121356,
%U 86383454582,336094015456
%N Chessboard polyominoes with n squares.
%C Chessboardcolored polyominoes, considering to be distinct two shapes that cannot be mapped onto each other by any form of symmetry. For example, there are two distinct monominoes, one black, one white. There is only one domino, with one black square, and one white.  _John Mason_, Nov 25 2013
%D W. F. Lunnon, personal communication.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Joseph Myers, <a href="http://list.seqfan.eu/pipermail/seqfan/2010November/013893.html">Chessboard polyominoes</a>
%F For odd n, a(n) = 2*A000105(n).  _John Mason_, Nov 27 2013
%F For even n, a(n) = 2*A000105(n)  (M(n) + R90(n) + R180(n))
%F Where:
%F M(n) is the number of free polyominoes of size n that have reflectional symmetry on a horizontal or vertical axis that coincides with the edges of some of the squares. (Note for example that the 3 X 3 square nonomino is not included, as the axes of symmetry do not coincide with the edges of any squares.) M(n) = 0 for all n not multiples of 2.
%F R90(n) is the number of free polyominoes of size n that have 90degree rotational symmetry about a point that coincides with the corner of a square. Note that for polyominoes which have a hole in the center, the center of rotation will be the corner of a square within the hole, rather than being the corner of a square of the polyomino itself. R90(n) = 0 for all n not multiples of 4. Exclude from R90(n) any polyominoes already in M(n).
%F R180 (n) is the number of free polyominoes of size n that have 180degree rotational symmetry about a point that coincides with the midside of a square. Note that for polyominoes which have a hole in the center, the center of rotation will be the midside of a square within the hole, rather than being within a square of the polyomino itself. R180 (n) = 0 for all n not multiples of 2. Exclude from R180 (n) any polyominoes already in M(n).
%F  _John Mason_, Dec 05 2013
%Y Cf. A001071, A000105, A121198.
%K hard,nonn
%O 1,1
%A _N. J. A. Sloane_
%E a(14)a(17) from _Joseph Myers_, Oct 01 2011
%E a(18)a(23) from _John Mason_, Dec 05 2013
