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1's complement of A103582.
5

%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.