OFFSET
1,2
COMMENTS
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.
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.
If n is prime then the only nonzero term in row n is T(n,1) = 1 + n.
If n is a power of 2 then the only nonzero term in row n is T(n,1) = 2*n - 1.
If n is an even perfect number then there are two nonzero terms in row n, they are T(n,1) = 2*n - 1 and the last term in the row is 1.
If n is a hexagonal number then the last term in row n is 1.
Row n contains a subpart 1 if and only if n is a hexagonal number.
First differs from A279388 at a(10), or row 9 of triangle.
EXAMPLE
Triangle begins:
---------------------------
n / k 1 2 3 4
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1 | 1;
2 | 3;
3 | 4;
4 | 7;
5 | 6;
6 | 11, 1;
7 | 8, 0;
8 | 15, 0;
9 | 10, 3;
10 | 18, 0;
11 | 12, 0;
12 | 23, 5;
13 | 14, 0;
14 | 24, 0;
15 | 16, 7, 1;
16 | 31, 0, 0;
17 | 18, 0, 0;
18 | 35, 4, 0;
19 | 20, 0, 0;
20 | 39, 0, 3;
21 | 22, 10, 0;
22 | 36, 0, 0;
23 | 24, 0, 0;
24 | 47, 13, 0;
25 | 26, 0, 5;
26 | 42, 0, 0;
27 | 28, 12, 0;
28 | 55, 0, 0, 1;
...
For n = 15 the calculation of the 15th row of the 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
.
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
.
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:
_
| |
_ | | _
_| | _| |_ | |_
_| | | | | |_
_| | | | | |_
_| | _| |_ | |_
_| | | | | |_
_| | | | | |_
_| | _| _ |_ | |_
|_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _|
1 3 5
.
The diagram is called "ziggurat of 15".
The number of steps in the staircase labeled 1 is 8. There is a pair of these staircases, so T(15,1) = 2*8 = 16, since the symmetric representation of sigma(15) is also the base of the three dimensional version of the ziggurat .
The number of steps in the double-staircase labeled 3 is equal to 7, so T(15,2) = 7.
The number of steps in the double-staircase labeled 5 is equal to 1, so T(15,3) = 1.
Therefore the 15th row of triangle is [16, 7, 1].
The top view of the 3D-Ziggurat of 15 and the symmetric representation of sigma(15) with subparts look like this:
_ _
|_| | |
|_| | |
|_| | |
|_| | |
|_| | |
|_| | |
|_| | |
_ _ _|_| _ _ _|_|
_ _|_| 36 _ _| | 8
|_|_|_| | _ _|
_|_|_| _| |_|
|_|_| 1 |_ _| 1
| 34 | 7
_ _ _ _ _ _ _ _| _ _ _ _ _ _ _ _|
|_|_|_|_|_|_|_|_| |_ _ _ _ _ _ _ _|
36 8
.
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.
.
CROSSREFS
Another (and more regular) version of A279388.
Row sums give A000203.
Row n has length A351846(n).
Cf. A347263 (analog for the ziggurat diagram).
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Sep 05 2021
STATUS
approved