A351819 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).
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
Offset: 1
Examples
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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|_|_| 1 |_ _| 1
| 34 | 7
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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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