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A347529
Irregular triangle read by rows: T(n,k) is the sum of the 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).
10
1, 3, 4, 7, 6, 11, 1, 8, 0, 15, 0, 10, 3, 18, 0, 12, 0, 23, 5, 14, 0, 24, 0, 16, 7, 1, 31, 0, 0, 18, 0, 0, 35, 4, 0, 20, 0, 0, 39, 0, 3, 22, 10, 0, 36, 0, 0, 24, 0, 0, 47, 13, 0, 26, 0, 5, 42, 0, 0, 28, 12, 0, 55, 0, 0, 1, 30, 0, 0, 0, 59, 6, 7, 0, 32, 0, 0, 0, 63, 0, 0, 0, 34, 14, 0, 0
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
---------------------------
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:
_
_| |_
_| _ |_
_| | | |_
_| _| |_ |_
_| | _ | |_
_| _| | | |_ |_
_| | | | | |_
_| _| _| |_ |_ |_
_| | | _ | | |_
_| _| | | | | |_ |_
_| | _| | | |_ | |_
_| _| | | | | |_ |_
_| | | _| |_ | | |_
_| _| _| | _ | |_ |_ |_
|_ _ _ _ _ _ _ _|_ _ _|_ _|_|_|_|_ _|_ _ _|_ _ _ _ _ _ _ _|
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:
_
_| |_
_| _ |_
_| | | |_
_| _| |_ |_
_| | _ | |_
_| _| | | |_ |_
_| | | | | |_
_| _| _| |_ |_ |_
_| | | | | |_
_| _| | | |_ |_
_| | _| |_ | |_
_| _| | | |_ |_
_| | | | | |_
_| _| _| _ |_ |_ |_
|_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _|
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).
Sequence in context: A332993 A126253 A057032 * A279388 A292288 A347273
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Sep 05 2021
STATUS
approved