%I
%S 1,0,1,1,1,1,0,1,1,1,1,0,0,1,1,0,0,1,0,1,1,1,0,0,0,0,1,1,0,1,1,
%T 1,0,0,1,1,1,1,1,0,0,0,0,1,1,0,1,0,1,1,0,0,0,1,1,1,0,0,0,0,0,0,0,
%U 0,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,1,1,0,1,0,0,0,1
%N Triangle T(n,k) read by rows. A051731(n,k)A051731(n,k+1).
%C Matrix inverse of A134541.
%C The subsequence of A180430 beginning from the second column, divided by 2, equals this sequence. [From _Mats Granvik_, Sep 04 2010]
%e Triangle begins:
%e 1
%e 0...1
%e 1..1...1
%e 0...1..1...1
%e 1...0...0..1...1
%e 0...0...1...0..1...1
%e 1...0...0...0...0..1...1
%e 0...1..1...1...0...0..1...1
%e 1..1...1...0...0...0...0..1...1
%e 0...1...0..1...1...0...0...0..1...1
%e 1...0...0...0...0...0...0...0...0..1...1
%e 0...0...0...1..1...1...0...0...0...0..1...1
%Y Cf. A176702*A000012=A051731. Row sums equals A000012.
%K sign,tabl
%O 1,1
%A Mats Granvik, _Gary W. Adamson_, Apr 24 2010
