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A351819
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Irregular triangle read by rows: T(n,k) is the number of subparts of the symmetric representation of sigma(n) that arise from the (2*k-1)-th double-staircase of the double-staircases diagram of n described in A335616, n >= 1, k >= 1, and the first element of column k is in row A000384(k).
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7
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1, 1, 2, 1, 2, 1, 1, 2, 0, 1, 0, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 2, 1, 1, 1, 0, 0, 2, 0, 0, 1, 2, 0, 2, 0, 0, 1, 0, 1, 2, 2, 0, 2, 0, 0, 2, 0, 0, 1, 1, 0, 2, 0, 1, 2, 0, 0, 2, 2, 0, 1, 0, 0, 1, 2, 0, 0, 0, 1, 2, 1, 0, 2, 0, 0, 0, 1, 0, 0, 0, 2, 2, 0, 0, 2, 0, 0, 0, 2, 0, 1, 1, 1, 2, 0, 0, 2, 0, 0, 0
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OFFSET
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1,3
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COMMENTS
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Conjecture 1: the number of nonzero terms in row n equals A082647(n).
Conjecture 2: column k lists positive integers interleaved with 2*k+2 zeros.
T(n,k) is also the number of staircases (or subparts) of the ziggurat diagram of n (described in A347186) that arise from the (2*k-1)-th double-staircase of the double-staircases diagram of n (described in A335616).
The k-th column of the triangle is related to the (2*k+1)-gonal numbers. For further information about this connection see A347186 and A347263.
Terms can be 0, 1 or 2.
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LINKS
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EXAMPLE
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Triangle begins:
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n / k 1 2 3 4
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1 | 1;
2 | 1;
3 | 2;
4 | 1;
5 | 2;
6 | 1, 1;
7 | 2, 0;
8 | 1, 0;
9 | 2, 1;
10 | 2, 0;
11 | 2, 0;
12 | 1, 1;
13 | 2, 0;
14 | 2, 0;
15 | 2, 1, 1;
16 | 1, 0, 0;
17 | 2, 0, 0;
18 | 1, 2, 0;
19 | 2, 0, 0;
20 | 1, 0, 1;
21 | 2, 2, 0;
22 | 2, 0, 0;
23 | 2, 0, 0;
24 | 1, 1, 0;
25 | 2, 0, 1;
26 | 2, 0, 0;
27 | 2, 2, 0;
28 | 1, 0, 0, 1;
...
For n = 15 the calculation of the 15th row of triangle (in accordance with the geometric algorithm described in A347186) is as follows:
Stage 1 (Construction):
We draw the diagram called "double-staircases" with 15 levels described in A335616.
Then we label the five double-staircases (j = 1..5) as shown below:
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1 2 3 4 5
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Stage 2 (Debugging):
We remove the fourth double-staircase as it does not have at least one step at level 1 of the diagram starting from the base, as shown below:
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1 2 3 5
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Stage 3 (Annihilation):
We delete the second double-staircase and the steps of the first double-staircase that are just above the second double-staircase.
The new diagram has two double-staircases and two simple-staircases as shown below:
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1 3 5
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The diagram is called "ziggurat of 15".
The staircase labeled 1 arises from the double-staircase labeled 1 in the double-staircases diagram of 15. There is a pair of these staircases, so T(15,1) = 2, since the symmetric representation of sigma(15) is also the base of the three dimensional version of the ziggurat .
The double-staircase labeled 3 is the same in both diagrams, so T(15,2) = 1.
The double-staircase labeled 5 is the same in both diagrams, so T(15,3) = 1.
Therefore the 15th row of the triangle is [2, 1, 1].
The top view of the 3D-Ziggurat of 15 and the symmetric representation of sigma(15) with subparts look like this:
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_ _|_| 36 _ _| | 8
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36 8
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Top view of the 3D-Ziggurat. The symmetric representation of
The ziggurat is formed by 3 of sigma(15) is formed by 3 parts.
polycubes with 107 cubes It has 4 subparts with 24 cells in
in total. It has 4 staircases total. It is the base of the ziggurat.
with 24 steps in total.
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CROSSREFS
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Cf. A000384, A082647, A196020, A235791, A236104, A237048, A237591, A237593, A262626, A335616, A346875, A347186, A347263, A347529.
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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