1, 3, 4, 1, 7, 3, 1, 6, 4, 3, 1, 1, 12, 7, 4, 3, 3, 1, 1, 8, 6, 7, 4, 4, 3, 3, 1, 1, 1, 1, 15, 12, 6, 7, 7, 4, 4, 3, 3, 3, 3, 1, 1, 1, 1, 13, 8, 12, 6, 6, 7, 7, 4, 4, 4, 4, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 18, 15, 8, 12, 12, 6, 6, 7, 7, 7, 7, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1
Triangle begins:
1;
3;
4, 1;
7, 3, 1;
6, 4, 3, 1, 1;
12, 7, 4, 3, 3, 1, 1;
8, 6, 7, 4, 4, 3, 3, 1, 1, 1, 1;
15, 12, 6, 7, 7, 4, 4, 3, 3, 3, 3, 1, 1, 1, 1;
13, 8, 12, 6, 6, 7, 7, 4, 4, 4, 4, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1;
...
From _Omar E. Pol_, Jan 13 2022: (Start)
Illustration of the first six rows of triangle showing the growth of the symmetric tower described in A221529:
Level k: 1 2 3 4 5 6 7
Stage
n _ _ _ _ _ _ _ _
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1 | |_| |
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2 | |_ _| |
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3 | |_|_ _ | |
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|_ _ _ _ _ _ _ _|_ _ _ _ _ _|_ _ _ _ _
| _ | _ | _ |
| | | | | |_ | |_| |
4 | | |_ | |_ _| | |
| |_ |_ _ | | |
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|_ _ _ _ _ _ _ _|_ _ _ _ _ _|_ _ _ _ _|_ _ _ _ _ _ _ _
| _ | _ | _ | _ | _ |
| | | | | | | | |_ | |_| | |_| |
| | | | |_|_ _ | |_ _| | | |
5 | |_|_ | |_ _| | | | |
| |_ _ _ | | | | |
| |_ _ _| | | | | |
|_ _ _ _ _ _ _ _|_ _ _ _ _ _|_ _ _ _ _|_ _ _ _|_ _ _ _|_ _ _ _ _ _
| _ | _ | _ | _ | _ | _ | _ |
| | | | | | | | | | | |_ | | |_ | |_| | |_| |
| | | | | |_ | |_|_ _ | |_ _| | |_ _| | | |
| | |_ _ | |_ |_ _ | |_ _| | | | | |
6 | |_ | | |_ _ _| | | | | | |
| |_ |_ _ _ | | | | | | |
| |_ _ _ _| | | | | | | |
|_ _ _ _ _ _ _ _|_ _ _ _ _ _|_ _ _ _ _|_ _ _ _|_ _ _ _|_ _ _|_ _ _|
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Every cell in the diagram of the symmetric representation of sigma represents a cubic cell or cube.
For n = 6 and k = 3 we add four cubes at 6th stage in the third level of the structure of the tower starting from the base so T(6,3) = 4.
For n = 9 another connection with the tower is as follows:
First we take the columns from the above triangle and build a new triangle in which all columns start at row 1 as shown below:
.
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3;
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4;
7, 7, 7, 7, 7, 7, 7;
6, 6, 6, 6, 6;
12, 12, 12;
8, 8;
15;
13;
.
Then we rotate the triangle by 90 degrees as shown below:
_
1; | |
1; | |
1; | |
1; | |
1; | |
1; | |
1; |_|_
1, 3; | |
1, 3; | |
1, 3; | |
1, 3; |_ _|_
1, 3, 4; | | |
1, 3, 4; | | |
1, 3, 4; | | |
1, 3, 4; |_ _|_|_
1, 3, 4, 7; | | |
1, 3, 4, 7; |_ _ _| |_
1, 3, 4, 7, 6; | | |
1, 3, 4, 7, 6; |_ _ _|_ _|_
1, 3, 4, 7, 6, 12; |_ _ _ _| | |_
1, 3, 4, 7, 6, 12, 8; |_ _ _ _|_|_ _|_ _
1, 3, 4, 7, 6, 12, 8, 15; 13; |_ _ _ _ _|_ _|_ _|
.
Lateral view
of the tower
. _ _ _ _ _ _ _ _ _
|_| | | | | | | |
|_ _|_| | | | | |
|_ _| _|_| | | |
|_ _ _| _|_| |
|_ _ _| _| _ _|
|_ _ _ _| |
|_ _ _ _| _ _|
| |
|_ _ _ _ _|
.
Top view
of the tower
.
The sum of the m-th row of the new triangle equals A024916(j) where j is the length of the m-th row, equaling the number of cubic cells in the m-th level of the tower. For example: the last row of triangle has 9 terms and the sum of the last row is 1 + 3 + 4 + 7 + 6 + 12 + 8 + 15 + 13 = A024916(9) = 69, equaling the number of cubes in the base of the tower. (End)
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