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Irregular triangle read by rows: row n lists the indices of rows of the Christmas tree pattern (A367508) of order n, sorted by row length and, in case of ties, by row index.
3

%I #19 Jun 12 2024 03:37:50

%S 1,1,2,1,2,3,1,3,2,4,5,6,1,2,4,5,7,3,6,8,9,10,1,3,7,9,13,2,4,5,8,10,

%T 11,14,15,17,6,12,16,18,19,20,1,2,4,5,7,11,12,14,15,17,21,22,24,28,3,

%U 6,8,9,13,16,18,19,23,25,26,29,30,32,10,20,27,31,33,34,35

%N Irregular triangle read by rows: row n lists the indices of rows of the Christmas tree pattern (A367508) of order n, sorted by row length and, in case of ties, by row index.

%C Row n is a permutation of the integers in the interval [1, binomial(n,floor(n/2))].

%C See A367508 for the description of the Christmas tree patterns, references and links.

%H Paolo Xausa, <a href="/A368399/b368399.txt">Table of n, a(n) for n = 1..13494</a> (rows 1..15 of the triangle, flattened).

%e Triangle begins (vertical bars separate indices of rows having different lengths):

%e .

%e [1] 1;

%e [2] 1| 2;

%e [3] 1 2| 3;

%e [4] 1 3| 2 4 5| 6;

%e [5] 1 2 4 5 7| 3 6 8 9|10;

%e [6] 1 3 7 9 13| 2 4 5 8 10 11 14 15 17| 6 12 16 18 19|20;

%e ...

%e For example, the order 4 of the Christmas tree pattern is the following:

%e .

%e 1010 Row 1 length = 1

%e 1000 1001 1011 Row 2 length = 3

%e 1100 Row 3 length = 1

%e 0100 0101 1101 Row 4 length = 3

%e 0010 0110 1110 Row 5 length = 3

%e 0000 0001 0011 0111 1111 Row 6 length = 5

%e .

%e and ordering the rows by length (and then by row index) gives 1, 3, 2, 4, 5, 6.

%t With[{nmax=8},Map[Flatten[Values[PositionIndex[#]]]&,SubstitutionSystem[{1->{2},t_/;t>1:>{t-1,t+1}},{2},nmax-1]]]

%Y Cf. A001405, A363718 (row lengths), A367508, A368400.

%K nonn,tabf

%O 1,3

%A _Paolo Xausa_, Dec 23 2023