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%I #22 Apr 18 2017 16:32:33
%S 1,2,3,1,4,1,5,2,5,1,2,6,1,3,7,1,3,7,2,4,8,2,1,3,9,2,1,4,9,3,1,4,10,3,
%T 1,5,11,3,2,4,11,3,1,2,5,12,3,1,2,5,13,3,1,2,6,13,4,1,3,5,14,4,1,3,6,
%U 15,3,2,3,6,15,4,2,1,2,7,16,4,2,1,3,6,17,4,2,1,3,7,17,5,2,1,3,7,18,4,3,1,3,8,19,4,3,1,4,7
%N Irregular triangle read by rows in which T(n,k) is the number of cells in the k-th horizontal bar of the n-th row of a diagram which is similar to the diagram of A237591, but here the even-indexed zig-zag paths are in the right hand part of the structure.
%e Triangle begins:
%e 1;
%e 2;
%e 3, 1;
%e 4, 1;
%e 5, 2;
%e 5, 1, 2;
%e 6, 1, 3;
%e 7, 1, 3:
%e 7, 2, 4;
%e 8, 2, 1, 3;
%e 9, 2, 1, 4;
%e ...
%e Illustration of initial terms:
%e Row _
%e 1 _|1|
%e 2 _|2 |_
%e 3 _|3 |1|
%e 4 _|4 |1|_
%e 5 _|5 _| 2|
%e 6 _|5 |1| 2|_
%e 7 _|6 |1| 3|
%e 8 _|7 _|1| 3|_
%e 9 _|7 |2 |_ 4|
%e 10 _|8 |2 |1| 3|_
%e 11 _|9 _|2 |1| 4|
%e 12 _|9 |3 |1| 4|_
%e 13 _|10 |3 |1|_ 5|
%e 14 _|11 _|3 _| 2| 4|_
%e 15 _|11 |3 |1| 2| 5|
%e 16 _|12 |3 |1| 2| 5|_
%e 17 _|13 _|3 |1| 2|_ 6|
%e 18 _|13 |4 |1| 3| 5|_
%e 19 _|14 |4 _|1| 3| 6|
%e 20 _|15 _|3 |2 |_ 3| 6|_
%e 21 |15 |4 |2 |1| 2| 7|
%e ...
%e For n = 6 the 6th row of the diagram has three horizontal bars (or parts) that contain 5, 1 and 2 cells respectively, so the 6th row of the triangle is [5, 1, 2].
%e Note that the number of horizontal line segments in the n-th row of the structure equals A001227(n), the number of odd divisors of n.
%Y Row sums give A001651.
%Y Row n has length A003056(n) hence column k starts in row A000217(k)
%Y Cf. A001227, A196020, A235791, A236104, A237048, A237591, A237593, A245092, A259176, A261699, A262626.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Dec 19 2016
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