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Irregular triangle read by rows: T(n, k) = number of vertices with rank k in cocoon concertina n-cube.
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%I #18 Apr 03 2018 10:43:47

%S 1,1,1,1,3,3,1,1,6,13,6,13,6,1

%N Irregular triangle read by rows: T(n, k) = number of vertices with rank k in cocoon concertina n-cube.

%C Although the cocoon concertina n-cube has no ranks for n>2, its inner vertices can be forced on the rank layers of the convex solid.

%C Sum of row n is the number of vertices of a cocoon concertina n-cube, i.e., A000696(n).

%C The rows are palindromic. Their lengths are the central polygonal numbers A000124 = 1, 2, 4, 7, 11, 16, 22, ... That means after row 0 rows of even and odd length follow each other in pairs.

%C A300699 is a triangle of the same shape that shows the number of ranks in convex concertina hypercubes.

%H Tilman Piesk, ranks <a href="https://commons.wikimedia.org/wiki/File:Cocoon_concertina_cube;_ranks_1_and_5.png">1 / 5</a>, <a href="https://commons.wikimedia.org/wiki/File:Cocoon_concertina_cube;_ranks_2_and_4.png">2 / 4</a> and <a href="https://commons.wikimedia.org/wiki/File:Cocoon_concertina_cube;_rank_3.png">3</a> for n=3

%H Tilman Piesk, <a href="https://github.com/watchduck/concertina_hypercubes/blob/master/cocoon.py">Python code used to generate the sequence</a> (currently unfinished, does not find all ranks for n>3)

%e First rows of the triangle:

%e k 0 1 2 3 4 5 6

%e n

%e 0 1

%e 1 1 1

%e 2 1 3 3 1

%e 3 1 6 13 6 13 6 1

%Y Cf. A000696, A300699, A000124.

%K nonn,tabf,more

%O 0,5

%A _Tilman Piesk_, Mar 13 2018