%I
%S 1,3,7,11,15,23,27,30,31,47,62,63,75,79,91,94,95,111,126,127,143,159,
%T 175,181,182,183,188,189,190,191,207,219,220,221,222,223,239,252,253,
%U 254,255,347,350,351,367,378,379,382,383,406,407,413,415,431,443,446
%N Smallest integers whose binary digits "1" define a polyomino if sorted into a quadrant shape.
%C Each integer k can be mapped onto a blackandwhite checkerboard pattern if we read the digits of its binary representation into the diagonals of a quadrant form, the leastsignificant digit b(0) into the corner, taking 1, then 2, then 3 etc. digits to fill consecutive diagonals of the quadrant:
%C b(0) b(2) b(5) b(9) ..
%C b(1) b(4) b(8) ..
%C b(3) b(7) ..
%C b(6) ..
%C b(10)
%C This will leave one last diagonal partially filled unless the number of binary digits in k is a triangular number.
%C Replace the "1" bits by black unit squares and the "0" or unset bits by white squares.
%C If the black squares define a singly connected polyomino considering position and updownleftright connectivity, and if the same polyomino cannot be created by a number smaller than k, we add k to the sequence. Polyominoes are considered the same if they can be matched by translations, rotations or flips.
%e (i) The triangular representations of 3= 11, 5 = 101, 10=1010 and are
%e 1
%e 1
%e and
%e 11
%e 0
%e and
%e 00
%e 1
%e 1
%e The 1's define the same 2omino in all 3 cases, so only the smallest representative, the 3, enters the sequence.
%e (ii) For k =181 = 10110101 the quadrant is filled with
%e 111
%e 01
%e 01
%e 0
%e No smaller number leads with this method to this Tshaped 5omino, so 181 enters the sequence.
%e (iii) the representations of 6=110, 9= 1001 and 29 =11101 are
%e 01
%e 1
%e and
%e 10
%e 0
%e 1
%e and
%e 11
%e 01
%e 1
%e In all of these 3 cases, the 1's are not singly connected and do not represent polyominoes, so neither 6 nor 9 nor 29 can make it into the sequence.
%K nonn,base
%O 1,2
%A _Leonhard Kreissig_, Sep 06 2009
%E Explanation expanded  _R. J. Mathar_, Sep 22 2009
