%I #115 Apr 18 2017 16:32:15
%S 1,1,1,0,0,1,1,0,1,1,0,1,1,0,0,0,0,0,0,1,1,0,0,1,0,0,1,0,0,1,1,0,0,0,
%T 0,1,1,0,0,0,0,1,1,0,0,0,1,0,0,0,0,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,
%U 0,0,0,1,1,0,0,0,0,1,0,1,0,0,1,0,1,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1
%N Triangle read by rows which is constructed with the diagram of the isosceles triangle of A279693 and filling the empty cells with zeros.
%C For the construction of this triangle we start with the diagram of A237048. Then with the diagram of the isosceles triangle of A279693 as shown below:
%C Row _ _
%C 1 _|1|1|_
%C 2 _|1 _|_ 1|_
%C 3 _|1 |1|1| 1|_
%C 4 _|1 _|0|0|_ 1|_
%C 5 _|1 |1 _|_ 1| 1|_
%C 6 _|1 _|0|1|1|0|_ 1|_
%C 7 _|1 |1 |0|0| 1| 1|_
%C 8 _|1 _|0 _|0|0|_ 0|_ 1|_
%C 9 _|1 |1 |1 _|_ 1| 1| 1|_
%C 10 _|1 _|0 |0|1|1|0| 0|_ 1|_
%C 11 _|1 |1 _|0|0|0|0|_ 1| 1|_
%C 12 _|1 _|0 |1 |0|0| 1| 0|_ 1|_
%C 13 _|1 |1 |0 _|0|0|_ 0| 1| 1|_
%C 14 _|1 _|0 _|0|1 _|_ 1|0|_ 0|_ 1|_
%C 15 _|1 |1 |1 |0|1|1|0| 1| 1| 1|_
%C 16 |1 |0 |0 |0|0|0|0| 0| 0| 1|
%C ...
%C And then filling with zeros the empty cells of the structure, as shown below:
%C Illustration of initial terms as an isosceles triangle:
%C Row _ _
%C 1 _|1 1|_
%C 2 _|1 0 0 1|_
%C 3 _|1 0 1 1 0 1|_
%C 4 _|1 0 0 0 0 0 0 1|_
%C 5 _|1 0 0 1 0 0 1 0 0 1|_
%C 6 _|1 0 0 0 0 1 1 0 0 0 0 1|_
%C 7 _|1 0 0 0 1 0 0 0 0 1 0 0 0 1|_
%C 8 _|1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1|_
%C 9 _|1 0 0 0 0 1 0 1 0 0 1 0 1 0 0 0 0 1|_
%C 10 _|1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1|_
%C 11 _|1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1|_
%C 12 _|1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1|_
%C 13 _|1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1|_
%C 14 _|1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1|_
%C 15 _|1 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 0 0 1 0 0 1 0 0 0 0 0 0 0 1|_
%C 16 |1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1|
%C ...
%C Note that the mentioned triangles are related to isosceles triangle A237593 and to the front view of the pyramid described in A245092.
%C The position of the 1's in the n-th row of the diagram is related to the subparts of the symmetric representation of sigma(n). For more information see A279387, A281010 and A281011.
%C For a right triangle which is the left hand part of this triangle see A279733.
%e Triangle begins:
%e 1, 1;
%e 1, 0, 0, 1;
%e 1, 0, 1, 1, 0, 1;
%e 1, 0, 0, 0, 0, 0, 0, 1;
%e 1, 0, 0, 1, 0, 0, 1, 0, 0, 1;
%e 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1;
%e 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1;
%e 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1;
%e ...
%Y Absolute values of A281011.
%Y Row n has length 2n.
%Y Row sums give A054844.
%Y One half of row sums gives A001227.
%Y Cf. A196020, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A239657, A245092, A249351, A261699, A262611, A262626, A279387, A279693, A279733, A281010.
%K nonn,tabf
%O 1
%A _Omar E. Pol_, Dec 17 2016