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%I #60 Dec 11 2020 13:27:51
%S 1,2,2,1,2,0,2,2,2,0,1,2,2,0,2,0,0,2,2,2,2,0,0,1,2,2,0,0,2,0,2,0,2,2,
%T 0,0,2,0,0,2,2,2,2,0,1,2,0,0,0,0,2,2,0,0,0,2,0,2,2,0,2,2,0,0,0,2,0,0,
%U 0,2,2,2,2,0,0,1,2,0,0,2,0,0,2,2,0,0,0,0,2,0,2,0,0,0,2,2,0,0,2,0,2
%N Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the terms of A040000: 1, 2, 2, 2, ... interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.
%C T(n,k) is also the number of horizontal line segments in the n-th level of the k-th largest double-staircase of the diagram defined in A335616 (see example).
%C The partial sums of column k give the k-th column of A338721.
%e Triangle begins (rows 1..28):
%e 1;
%e 2;
%e 2, 1;
%e 2, 0;
%e 2, 2;
%e 2, 0, 1;
%e 2, 2, 0;
%e 2, 0, 0;
%e 2, 2, 2;
%e 2, 0, 0, 1;
%e 2, 2, 0, 0;
%e 2, 0, 2, 0;
%e 2, 2, 0, 0;
%e 2, 0, 0, 2;
%e 2, 2, 2, 0, 1;
%e 2, 0, 0, 0, 0;
%e 2, 2, 0, 0, 0;
%e 2, 0, 2, 2, 0;
%e 2, 2, 0, 0, 0;
%e 2, 0, 0, 0, 2;
%e 2, 2, 2, 0, 0, 1;
%e 2, 0, 0, 2, 0, 0;
%e 2, 2, 0, 0, 0, 0;
%e 2, 0, 2, 0, 0, 0;
%e 2, 2, 0, 0, 2, 0;
%e 2, 0, 0, 2, 0, 0;
%e 2, 2, 2, 0, 0, 2;
%e 2, 0, 0, 0, 0, 0, 1;
%e ...
%e For an illustration of the rows of triangle consider the infinite "double-staircases" diagram defined in A335616.
%e The first 15 levels of the structure looks like this:
%e .
%e Level "Double-staircases" diagram
%e n _
%e 1 _|1|_
%e 2 _|1 _ 1|_
%e 3 _|1 |1| 1|_
%e 4 _|1 _| |_ 1|_
%e 5 _|1 |1 _ 1| 1|_
%e 6 _|1 _| |1| |_ 1|_
%e 7 _|1 |1 | | 1| 1|_
%e 8 _|1 _| _| |_ |_ 1|_
%e 9 _|1 |1 |1 _ 1| 1| 1|_
%e 10 _|1 _| | |1| | |_ 1|_
%e 11 _|1 |1 _| | | |_ 1| 1|_
%e 12 _|1 _| |1 | | 1| |_ 1|_
%e 13 _|1 |1 | _| |_ | 1| 1|_
%e 14 _|1 _| _| |1 _ 1| |_ |_ 1|_
%e 15 |1 |1 |1 | |1| | 1| 1| 1|
%e .
%e For n = 15, in the 15th level of the diagram we have that the first largest double-staircase has two horizontal steps, the second double-staircase has two steps, the third double-staircase has two steps, there are no steps in the fourth double-stairce and the fifth double-staircase has only one step, so the 15th row of triangle is [2, 2, 2, 0, 1].
%Y Column 1 is A040000.
%Y Row sums give A335616.
%Y Row n has length A003056(n).
%Y Column k starts in row A000217(k).
%Y The number of positive terms in row n is A001227(n).
%Y Cf. A196020, A236104, A237048, A237270, A237591, A237593, A249351, A280850, A296508, A299484, A338721.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Dec 01 2020
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