%I #15 Sep 08 2013 19:54:49
%S 0,1,0,0,0,0,1,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,1,1,0,0,0,
%T 0,0,0,1,1,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,
%U 0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0
%N 1's complement of A103582.
%C Comment from Jared Benjamin Ricks (jaredricks(AT)yahoo.com), Jan 31 2009: (Start)
%C This sequence be also be obtained in the following way. Write numbers in binary from left to right and read the resulting array by antidiagonals upwards:
%C 0 : (0, 0, 0, 0, 0, 0, 0, ...)
%C 1 : (1, 0, 0, 0, 0, 0, 0, ...)
%C 2 : (0, 1, 0, 0, 0, 0, 0, ...)
%C 3 : (1, 1, 0, 0, 0, 0, 0, ...)
%C 4 : (0, 0, 1, 0, 0, 0, 0, ...)
%C 5 : (1, 0, 1, 0, 0, 0, 0, ...)
%C 6 : (0, 1, 1, 0, 0, 0, 0, ...)
%C 7 : (1, 1, 1, 0, 0, 0, 0, ...)
%C ... (End)
%H David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [<a href="http://neilsloane.com/doc/slopey.pdf">pdf</a>, <a href="http://neilsloane.com/doc/slopey.ps">ps</a>].
%e Triangle begins:
%e 0
%e 1 0
%e 0 0 0
%e 1 1 0 0
%e 0 1 0 0 0
%e 1 0 0 0 0 0
%e 0 0 1 0 0 0 0
%e 1 1 1 0 0 0 0 0
%e 0 1 1 0 0 0 0 0 0
%e 1 0 1 0 0 0 0 0 0 0
%e 0 0 0 0 0 0 0 0 0 0 0
%e 1 1 0 1 0 0 0 0 0 0 0 0
%Y Cf. A103582, A103581, A103589. Considered as a triangle, obtained by reversing the rows of the triangle in A103589.
%K nonn,easy,tabl
%O 0,1
%A _Philippe Deléham_, Mar 24 2005
%E More terms from _Robert G. Wilson v_ and _Benoit Cloitre_, Mar 26 2005
%E Corrected by _N. J. A. Sloane_, Apr 19, 2005
%E Rechecked by _David Applegate_, Apr 19 2005.