A279388 -id:A279388 - cp's OEIS frontend

A279387 Irregular triangle read by rows: suppose the symmetric representation of sigma(n) consists of m = A250068(n) layers of width 1, arranged in increasing order; then T(n,k) (n >= 1, 1 <= k <= m) is the number of subparts in the k-th layer.

1, 1, 2, 1, 2, 1, 1, 2, 1, 3, 2, 2, 1, 1, 2, 2, 3, 1, 1, 2, 1, 2, 2, 1, 1, 4, 2, 2, 1, 1, 3, 2, 4, 1, 1, 2, 1, 3, 2, 1, 4, 2, 3, 1, 1, 2, 2, 2, 4, 1, 1, 2, 1, 3, 2, 2, 3, 3, 2, 2, 1, 1, 3, 3, 4, 2, 2, 1, 3, 4, 1, 1, 4, 2, 2, 1, 1, 2, 2, 2, 5, 1, 1, 4, 1, 3, 2, 2, 4, 3, 1, 2, 1, 1, 1, 2, 2

Offset: 1

Omar E. Pol, Dec 12 2016

The "subparts" of the symmetric representation of sigma(n) are defined to be the regions that arise after the dissection of the symmetric representation of sigma(n) into successive layers of width 1.
The number of layers of width 1 in the symmetric representation of sigma(n) is given in A250068.
The number of subparts in the first layer of the symmetric representation of sigma(n) is equal to A237271(n).
We can find the symmetric representation of sigma(n) as the terraces at the n-th level (starting from the top) of the stepped pyramid described in A245092.
(All above comments are essentially the same as the comments dated Nov 05 2016 at the old version of A275601, which was the same as A001227).
The sum of row n equals the number of subparts in the symmetric representation of sigma(n).
Conjecture:
The number of subparts in the symmetric representation of sigma(n) equals A001227(n), the number of odd divisors of n.
From Hartmut F. W. Hoft, Dec 16 2016: (Start)
Proof:
Each row of the irregular triangle of A262045 can be interpreted as a step function of step sizes 1, 0, and -1. The numbers in row n are the widths of the segments in the parts of the symmetric representation of sigma(n). Each new subpart in a segment (in the left half) of row n starts at the same odd index that represents an odd divisor d of n in the irregular triangle of A237048. Either a subpart ends at an even index e, representing a second odd divisor, which satisfies d * e = oddpart(n), and thus the entire subpart is duplicated in the symmetric portion of the representation, or a subpart runs through the center and continues contiguously into the right half of the symmetric portion of the representation. In other words, the number of subparts in row n equals the number of odd divisors of n, i.e., the conjecture is true. (End)

			Triangle begins (first 18 rows):
1;
1;
2;
1;
2;
1, 1;
2;
1;
3;
2;
2;
1, 1;
2;
2;
3, 1;
1;
2;
1, 2;
...
For n = 12, the 11th row of triangle A237593 is [6, 3, 1, 1, 1, 1, 3, 6] and the 12th row of the same triangle is [7, 2, 2, 1, 1, 2, 2, 7], so the diagram of the symmetric representation of sigma(12) = 28 is constructed as shown below in Figure 1:
.                          _                                    _
.                         | |                                  | |
.                         | |                                  | |
.                         | |                                  | |
.                         | |                                  | |
.                         | |                                  | |
.                    _ _ _| |                             _ _ _| |
.                  _|    _ _|                           _|  _ _ _|
.                _|     |                             _|  _| |
.               |      _|                            |  _|  _|
.               |  _ _|                              | |_ _|
.    _ _ _ _ _ _| |    28                 _ _ _ _ _ _| |    5
.   |_ _ _ _ _ _ _|                      |_ _ _ _ _ _ _|
.                                                       23
.
.   Figure 1. The symmetric            Figure 2. After the dissection
.   representation of sigma(12)        of the symmetric representation
.   has only one part which            of sigma(12) into layers of
.   contains 28 cells, so              width 1 we can see two "subparts"
.   A237271(12) = 1.                   that contain 23 and 5 cells
.                                      respectively, so the 12th row of
.                                      this triangle is [1, 1], and the
.                                      row sum is A001227(12) = 2,
.                                      equaling the number of odd divisors
.                                      of 12.
.
For n = 15, the 14th row of triangle A237593 is [8, 3, 1, 2, 2, 1, 3, 8] and the 15th row of the same triangle is [8, 3, 2, 1, 1, 1, 1, 2, 3, 8], so the diagram of the symmetric representation of sigma(15) = 24 is constructed as shown below in Figure 3:
.                                _                                  _
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                          _ _ _|_|                           _ _ _|_|
.                      _ _| |      8                      _ _| |      8
.                     |    _|                            |  _ _|
.                    _|  _|                             _| |_|
.                   |_ _|  8                           |_ _|  1
.                   |                                  |    7
.    _ _ _ _ _ _ _ _|                   _ _ _ _ _ _ _ _|
.   |_ _ _ _ _ _ _ _|                  |_ _ _ _ _ _ _ _|
.                    8                                  8
.
.   Figure 3. The symmetric            Figure 4. After the dissection
.   representation of sigma(15)        of the symmetric representation
.   has three parts of size 8          of sigma(15) into layers of
.   because every part contains        width 1 we can see four "subparts".
.   8 cells, so A237271(15) = 3.       The first layer has three subparts:
.                                      [8, 7, 8]. The second layer has
.                                      only one subpart of size 1, so
.                                      the 15th row of this triangle is
.                                      [3, 1], and the row sum is
.                                      A001227(15) = 4, equaling the
.                                      number of odd divisors of 15.
.
For n = 360, the 359th row of triangle A237593 is [180, 61, 30, 19, 12, 9, 7, 6, 4, 4, 3, 3, 2, 3, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1] and the 360th row of the same triangle is [181, 60, 31, 18, 13, 9, 7, 5, 5, 4, 3, 2, 2, 2, 2, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1], so have that the symmetric representation of sigma(360) = 1170 has only one part, five layers, and six subparts: [(719), (237), (139), (71), (2, 2)], so the 360th row of this triangle is [1, 1, 1, 1, 2], and the row sum is A001227(360) = 6, equaling the number of odd divisors of 360 (the diagram is too large to include).
From _Hartmut F. W. Hoft_, Dec 16 2016: (Start)
45 has 6 subparts of which 2 have symmetric duplicates and 2 span the center. Row length is 18 and "|" indicates the center marker for a row.
1 2 3 4 5 6 7 8 9|9 8 7 6 5 4 3 2 1  : position indices
1 0 1 1 2 1 1 1 2|2 1 1 1 2 1 1 0 1  : row 45 of A262045
1   1 1 1 1 1 1 1|1 1 1 1 1 1 1   1  : layer 1
        1       1|1       1          : layer 2
1 1 1 0 1 1 0 0 1|                   : row 45 of A237048 (odd divisors)
+ - + . + - . . +|                   : change in level ("." no change)
90 has 6 subparts and 3 layers (row length is 24).
1 2 3 4 5 6 7 8..10..12|.14..16..18..20..22..24 : position indices
1 1 2 1 2 2 2 2 3 3 3 2|2 3 3 3 2 2 2 2 1 2 1 1 : row 90 of A262045
1 1 1 1 1 1 1 1 1 1 1 1|1 1 1 1 1 1 1 1 1 1 1 1 : layer 1
    1   1 1 1 1 1 1 1 1|1 1 1 1 1 1 1 1   1     : layer 2
                1 1 1  |  1 1 1                 : layer 3
1 0 1 1 1 0 0 0 1 0 0 1|                        : row 90 of A237048
+ . + - + . . . + . . -|                        : change in level ("." no change)
The process of successive levels provides two "default" dissections of the symmetric representation into subparts from the boundary at n towards the boundary at n-1 or in the reverse direction. (End)
From _Omar E. Pol_, Nov 24 2020: (Start)
For n = 18 we have that the 17th row of triangle A237593 is [9, 4, 2, 1, 1, 1, 1, 2, 4, 9] and the 18th row of the same triangle is [10, 3, 2, 2, 1, 1, 2, 2, 3, 10], so the diagram of the symmetric representation of sigma(18) = 39 is constructed as shown below in Figure 5:
.                                     _                                      _
.                                    | |                                    | |
.                                    | |                                    | |
._                                   | |                                    | |
.                                    | |                                    | |
.                                    | |                                    | |
.                                    | |                                    | |
.                                    | |                                    | |
.                                    | |                                    | |
.                             _ _ _ _| |                             _ _ _ _| |
.                            |    _ _ _|                            |  _ _ _ _|
.                           _|   |                                 _| | |
.                         _|  _ _|                               _|  _|_|
.                     _ _|  _|                               _ _|  _|    2
.                    |     |  39                            |  _ _|
.                    |  _ _|                                | |_ _|
.                    | |                                    | |    2
.   _ _ _ _ _ _ _ _ _| |                   _ _ _ _ _ _ _ _ _| |
.  |_ _ _ _ _ _ _ _ _ _|                  |_ _ _ _ _ _ _ _ _ _|
.                                                              35
.
.   Figure 5. The symmetric               Figure 6. After the dissection
.   representation of sigma(18)           of the symmetric representation
.   has one part of size 39, so           of sigma(18) into layers of
.   A237271(18) = 1.                      width 1 we can see three "subparts".
.                                         The first layer has one subpart of
.                                         size 35. The second layer has
.                                         two subparts of size 2, so
.                                         the 18th row of this triangle is
.                                         [1, 2], and the row sum is
.                                         A001227(18) = 3.
(End)

The sum of row n equals A001227(n).
Hence, if n is odd, the sum of row n equals A000005(n).
Row n has length A250068(n).
Column 1 gives A237271.
For more information about "subparts" see A279388 and A279391.
Cf. A000203, A196020, A235791, A236104, A237048, A237270, A237591, A237593, A239657, A243982, A244050, A245092, A249223, A249351, A250070, A262045, A262611, A261699, A262626, A279693, A280850, A280851, A296508.
Cf. A235791, A237048, A237270, A237591, A237593, A247687, A249223, A250070, A264102, A280851, A341969.

(* function a341969[ ] is defined in A341969 *)
a279387[n_] := Module[{widthL=a341969[n], partL, cL, top, ft, sL}, partL=Select[SplitBy[widthL, #==0&], #!={0}&]; cL=Table[0, Max[widthL]]; While[partL!={}, top=Last[partL]; ft=First[top]; sL=Select[SplitBy[top, #==ft&], #!={ft}&];
cL[[ft]]++; partL=Join[Most[partL], sL]]; cL]
Flatten[a279387[74]] (* the first 74 rows of the table; Hartmut F. W. Hoft, Feb 24 2021 *)

Definition edited by Omar E. Pol and N. J. A. Sloane, Nov 25 2020

A296508 Irregular triangle read by rows: T(n,k) is the size of the subpart that is adjacent to the k-th peak of the largest Dyck path of the symmetric representation of sigma(n), or T(n,k) = 0 if the mentioned subpart is already associated to a previous peak or if there is no subpart adjacent to the k-th peak, with n >= 1, k >= 1.

Original entry on oeis.org

1, 3, 2, 2, 7, 0, 3, 3, 11, 1, 0, 4, 0, 4, 15, 0, 0, 5, 3, 5, 9, 0, 9, 0, 6, 0, 0, 6, 23, 5, 0, 0, 7, 0, 0, 7, 12, 0, 12, 0, 8, 7, 1, 0, 8, 31, 0, 0, 0, 0, 9, 0, 0, 0, 9, 35, 2, 0, 2, 0, 10, 0, 0, 0, 10, 39, 0, 3, 0, 0, 11, 5, 0, 5, 0, 11, 18, 0, 0, 0, 18, 0, 12, 0, 0, 0, 0, 12, 47, 13, 0, 0, 0, 0, 13, 0, 5, 0, 0, 13

Offset: 1

Omar E. Pol, Feb 10 2018

Conjecture: row n is formed by the odd-indexed terms of the n-th row of triangle A280850 together with the even-indexed terms of the same row but listed in reverse order. Examples: the 15th row of A280850 is [8, 8, 7, 0, 1] so the 15th row of this triangle is [8, 7, 1, 0, 8]. The 75th row of A280850 is [38, 38, 21, 0, 3, 3, 0, 0, 0, 21, 0] so the 75h row of this triangle is [38, 21, 3, 0, 0, 0, 21, 0, 3, 0, 38].
For the definition of "subparts" see A279387.
For more information about the mentioned Dyck paths see A237593.
T(n,k) could be called the "charge" of the k-th peak of the largest Dyck path of the symmetric representation of sigma(n).
The number of zeros in row n is A238005(n). - Omar E. Pol, Sep 11 2021

			Triangle begins (rows 1..28):
   1;
   3;
   2,  2;
   7,  0;
   3,  3;
  11,  1,  0;
   4,  0,  4;
  15,  0,  0;
   5,  3,  5;
   9,  0,  9,  0;
   6,  0,  0,  6;
  23,  5,  0,  0;
   7,  0,  0,  7;
  12,  0, 12,  0;
   8,  7,  1,  0,  8;
  31,  0,  0,  0,  0;
   9,  0,  0,  0,  9;
  35,  2,  0,  2,  0;
  10,  0,  0,  0, 10;
  39,  0,  3,  0,  0;
  11,  5,  0,  5,  0, 11;
  18,  0,  0,  0, 18,  0;
  12,  0,  0,  0,  0, 12;
  47, 13,  0,  0,  0,  0;
  13,  0,  5,  0,  0, 13;
  21,  0,  0,  0  21,  0;
  14,  6,  0,  6,  0, 14;
  55,  0,  0,  1,  0,  0,  0;
  ...
For n = 15 we have that the 14th row of triangle A237593 is [8, 3, 1, 2, 2, 1, 3, 8] and the 15th row of the same triangle is [8, 3, 2, 1, 1, 1, 1, 2, 3, 8], so the diagram of the symmetric representation of sigma(15) is constructed in the third quadrant as shown below in Figure 1:
.    _                                  _
.   | |                                | |
.   | |                                | |
.   | |                                | |
. 8 | |                                | |
.   | |                                | |
.   | |                                | |
.   | |                                | |
.   |_|_ _ _                           |_|_ _ _
.         | |_ _                      8      | |_ _
.         |_    |                            |_ _  |
.           |_  |_                          7  |_| |_
.          8  |_ _|                           1  |_ _|
.                 |                             0    |
.                 |_ _ _ _ _ _ _ _                   |_ _ _ _ _ _ _ _
.                 |_ _ _ _ _ _ _ _|                  |_ _ _ _ _ _ _ _|
.                         8                         8
.
.   Figure 1. The symmetric            Figure 2. After the dissection
.   representation of sigma(15)        of the symmetric representation
.   has three parts of size 8          of sigma(15) into layers of
.   because every part contains        width 1 we can see four subparts,
.   8 cells, so the 15th row of        so the 15th row of this triangle is
.   triangle A237270 is [8, 8, 8].     [8, 7, 1, 0, 8]. See also below.
.
Illustration of first 50 terms (rows 1..16 of triangle) in an irregular spiral which can be find in the top view of the pyramid described in A244050:
.
.               12 _ _ _ _ _ _ _ _
.                 |  _ _ _ _ _ _ _|_ _ _ _ _ _ _ 7
.                 | |             |_ _ _ _ _ _ _|
.              0 _| |                           |
.               |_ _|9 _ _ _ _ _ _              |_ _ 0
.         12 _ _|     |  _ _ _ _ _|_ _ _ _ _ 5      |_ 0
.    0 _ _ _| |    0 _| |         |_ _ _ _ _|         |
.     |  _ _ _|  9 _|_ _|                   |_ _ 3    |_ _ _ 7
.     | |    0 _ _| |   11 _ _ _ _          |_  |         | |
.     | |     |  _ _|  1 _|  _ _ _|_ _ _ 3    |_|_ _ 5    | |
.     | |     | |    0 _|_| |     |_ _ _|         | |     | |
.     | |     | |     |  _ _|           |_ _ 3    | |     | |
.     | |     | |     | |    3 _ _        | |     | |     | |
.     | |     | |     | |     |  _|_ 1    | |     | |     | |
.    _|_|    _|_|    _|_|    _|_| |_|    _|_|    _|_|    _|_|    _
.   | |     | |     | |     | |         | |     | |     | |     | |
.   | |     | |     | |     |_|_ _     _| |     | |     | |     | |
.   | |     | |     | |    2  |_ _|_ _|  _|     | |     | |     | |
.   | |     | |     |_|_     2    |_ _ _|  0 _ _| |     | |     | |
.   | |     | |    4    |_               7 _|  _ _|0    | |     | |
.   | |     |_|_ _     0  |_ _ _ _        |  _|    _ _ _| |     | |
.   | |    6      |_      |_ _ _ _|_ _ _ _| |  0 _|  _ _ _|0    | |
.   |_|_ _ _     0  |_   4        |_ _ _ _ _|  _|  _| |    _ _ _| |
.  8      | |_ _   0  |                     15|  _|  _|   |  _ _ _|
.         |_ _  |     |_ _ _ _ _ _            | |_ _|  0 _| |      0
.        7  |_| |_    |_ _ _ _ _ _|_ _ _ _ _ _| |    5 _|  _|
.          1  |_ _|  6            |_ _ _ _ _ _ _|  _ _|  _|  0
.            0    |                             23|  _ _|  0
.                 |_ _ _ _ _ _ _ _                | |    0
.                 |_ _ _ _ _ _ _ _|_ _ _ _ _ _ _ _| |
.                8                |_ _ _ _ _ _ _ _ _|
.                                                    31
.
The diagram contains 30 subparts equaling A060831(16), the total number of partitions of all positive integers <= 16 into consecutive parts.
For the construction of the spiral see A239660.
From _Omar E. Pol_, Nov 26 2020: (Start)
Also consider the infinite double-staircases diagram defined in A335616 (see the theorem). For n = 15 the diagram with first 15 levels looks like this:
.
Level                         "Double-staircases" diagram
.                                          _
1                                        _|1|_
2                                      _|1 _ 1|_
3                                    _|1  |1|  1|_
4                                  _|1   _| |_   1|_
5                                _|1    |1 _ 1|    1|_
6                              _|1     _| |1| |_     1|_
7                            _|1      |1  | |  1|      1|_
8                          _|1       _|  _| |_  |_       1|_
9                        _|1        |1  |1 _ 1|  1|        1|_
10                     _|1         _|   | |1| |   |_         1|_
11                   _|1          |1   _| | | |_   1|          1|_
12                 _|1           _|   |1  | |  1|   |_           1|_
13               _|1            |1    |  _| |_  |    1|            1|_
14             _|1             _|    _| |1 _ 1| |_    |_             1|_
15            |1              |1    |1  | |1| |  1|    1|              1|
.
Starting from A196020 and after the algorithm described n A280850 and the conjecture applied to the above diagram we have a new diagram as shown below:
.
Level                             "Ziggurat" diagram
.                                          _
6                                         |1|
7                            _            | |            _
8                          _|1|          _| |_          |1|_
9                        _|1  |         |1   1|         |  1|_
10                     _|1    |         |     |         |    1|_
11                   _|1      |        _|     |_        |      1|_
12                 _|1        |       |1       1|       |        1|_
13               _|1          |       |         |       |          1|_
14             _|1            |      _|    _    |_      |            1|_
15            |1              |     |1    |1|    1|     |              1|
.
The 15th row
of A249351:   [1,1,1,1,1,1,1,1,0,0,0,1,1,1,2,1,1,1,0,0,0,1,1,1,1,1,1,1,1]
The 15th row
of A237270:   [              8,            8,            8              ]
The 15th row
of this seq:  [              8,      7,    1,    0,      8              ]
The 15th row
of A280851:   [              8,      7,    1,            8              ]
.
(End)

Index entries for sequences related to sigma(n)

Row sums give A000203.
Row n has length A003056(n).
Column k starts in row A000217(k).
Nonzero terms give A280851.
The number of nonzero terms in row n is A001227(n).
The triangle with n rows contain A060831(n) nonzero terms.
Cf. A024916, A196020, A235791, A236104, A237048, A237270, A237591, A237593, A238005, A239657, A239660, A239931-A239934, A240542, A244050, A245092, A250068, A250070, A261699, A262626, A279387, A279388, A279391, A280850.

A280850 Irregular triangle read by rows in which row n is constructed with an algorithm using the n-th row of triangle A196020 (see Comments for precise definition).

Original entry on oeis.org

1, 3, 2, 2, 7, 0, 3, 3, 11, 0, 1, 4, 4, 0, 15, 0, 0, 5, 5, 3, 9, 0, 0, 9, 6, 6, 0, 0, 23, 0, 5, 0, 7, 7, 0, 0, 12, 0, 0, 12, 8, 8, 7, 0, 1, 31, 0, 0, 0, 0, 9, 9, 0, 0, 0, 35, 0, 2, 2, 0, 10, 10, 0, 0, 0, 39, 0, 0, 0, 3, 11, 11, 5, 0, 0, 5, 18, 0, 0, 18, 0, 0, 12, 12, 0, 0, 0, 0, 47, 0, 13, 0, 0, 0

Offset: 1

Omar E. Pol, Jan 09 2017

For the construction of the n-th row of this triangle start with a copy of the n-th row of the triangle A196020.
Then replace each element of the m-th pair of positive integers (x, y) with the value (x - y)/2, where "y" is the m-th even-indexed term of the row, and "x" is its previous nearest odd-indexed term not used in another pair in the same row, if such a pair exist. Otherwise T(n,k) = A196020(n,k). (See example).
Observation 1: at least for the first 28 rows of the triangle the nonzero terms in the n-th row are also the subparts of the symmetric representation of sigma(n), assuming the ordering of the subparts in the same row does not matter.
Question 1: are always the nonzero terms of the n-th row the same as all the subparts of the symmetric representation of sigma(n)? If not, what is the index of the row in which appears the first counterexample?
Note that the "subparts" are the regions that arise after the dissection of the symmetric representation of sigma(n) into successive layers of width 1.
For more information about "subparts" see A279387 and A237593.
About the question 1, it appears that the n-th row of the triangle A280851 and the n-th row of this triangle contain the same nonzero numbers, though in different order; checked through n = 250000. - Hartmut F. W. Hoft, Jan 31 2018
From Omar E. Pol, Feb 02 2018: (Start)
Observation 2: at least for the first 28 rows of the triangle we have that in the n-th row the odd-indexed terms, from left to right, together with the even-indexed terms, from right to left, form a finite sequence in which the nonzero terms are the same as the n-th row of triangle A280851, which lists the subparts of the symmetric representation of sigma(n).
Question 2: Are always the same for all rows? If not, what is the index of the row in which appears the first counterexample? (End)
Conjecture: the odd-indexed terms of the n-th row together with the even-indexed terms of the same row but listed in reverse order give the n-th row of triangle A296508 (this is the same conjecture from A296508). - Omar E. Pol, Apr 20 2018

			Triangle begins (rows 1..28):
   1;
   3;
   2,  2;
   7,  0;
   3,  3;
  11,  0,  1;
   4,  4,  0;
  15,  0,  0;
   5,  5,  3;
   9,  0,  0,  9;
   6,  6,  0,  0;
  23,  0,  5,  0;
   7,  7,  0,  0;
  12,  0,  0, 12;
   8,  8,  7,  0,  1;
  31,  0,  0,  0,  0;
   9,  9,  0,  0,  0;
  35,  0,  2,  2,  0;
  10, 10,  0,  0,  0;
  39,  0,  0,  0,  3;
  11, 11,  5,  0,  0,  5;
  18,  0,  0, 18,  0,  0;
  12, 12,  0,  0,  0,  0;
  47,  0, 13,  0,  0,  0;
  13, 13,  0,  0,  5,  0;
  21,  0,  0, 21,  0,  0;
  14, 14,  6,  0,  0,  6;
  55,  0,  0,  0,  0,  0,  1;
  ...
An example of the algorithm.
For n = 75, the construction of the 75th row of this triangle is as shown below:
.
75th row of A196020:             [149,  73, 47, 0, 25, 19, 0, 0, 0,  5, 0]
.
Odd-indexed terms:                149       47     25      0     0      0
Even-indexed terms:                     73      0      19     0      5
.
First even-indexed nonzero term:        73
First pair:                       149   73
.                                   *----*
Difference: 149 - 73 =                76
76/2 = 38                           *----*
New first pair:                    38   38
.
Second even-indexed nonzero term:                      19
Second pair:                                       25  19
.                                                   *---*
Difference: 25 - 19 =                                 6
6/2 = 3                                             *---*
New second pair:                                    3   3
.
Third even-indexed nonzero term:                                     5
Third pair:                                  47                      5
.                                             *----------------------*
Difference: 47 - 5 =                                     42
42/2 = 21                                     *----------------------*
New third pair:                              21                     21
.
So the 75th row
of this triangle is                [38,  38, 21, 0, 3,  3, 0, 0, 0, 21, 0]
.
On the other hand, the 75th row of A237593 is [38, 13, 7, 4, 3, 3, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 3, 3, 4, 7, 13, 38], and the 74th row of the same triangle is [38, 13, 6, 5, 3, 2, 2, 1, 2, 1, 1, 1, 1, 2, 1, 2, 2, 3, 5, 6, 13, 38], therefore between both symmetric Dyck paths (described in A237593 and A279387) there are six subparts: [38, 38, 21, 21, 3, 3]. (The diagram of the symmetric representation of sigma(75) is too large to include.) At least in this case the nonzero terms of the 75th row of the triangle coincide with the subparts of the symmetric representation of sigma(75). The ordering of the elements does not matter.
Continuing with the original example, in the 75th row of this triangle we have that the odd-indexed terms, from left to right, together with the even-indexed terms, from right to left, form the finite sequence [38, 21, 3, 0, 0, 0, 21, 0, 3, 0, 38] which is the 75th row of a triangle. At least in this case the nonzero terms coincide with the 75th row of triangle A280851: [38, 21, 3, 21, 3, 38], which lists the six subparts of the symmetric representation of sigma(75) in order of appearance from left to right. - _Omar E. Pol_, Feb 02 2018
In accordance with the conjecture from the Comments section, the finite sequence [38, 21, 3, 0, 0, 0, 21, 0, 3, 0, 38] mentioned above should be the 75th row of triangle A296508. - _Omar E. Pol_, Apr 20 2018

Index entries for sequences related to sigma(n)

Row sums give A000203.
The number of positive terms in row n is A001227(n).
Row n has length A003056(n).
Column k starts in row A000217(k).
Cf. A196020, A235791, A236104, A237048, A237270, A237591, A237593, A239657, A239660, A244050, A245092, A250068, A250070, A261699, A262626, A279387, A279388, A279391, A280851, A296508.

(* functions row[], line[] and their support are defined in A196020 *)
(* maintain a stack of odd indices with nonzero entries for matching *)
a280850[n_] := Module[{a=line[n], r=row[n], stack={1}, i, j, b}, For[i=2, i<=r, i++, If[a[[i]]!=0, If[OddQ[i], AppendTo[stack, i], j=Last[stack]; b=(a[[j]]-a[[i]])/2; a[[i]]=b; a[[j]]=b; stack=Drop[stack, -1]]]]; a]
Flatten[Map[a280850,Range[24]]] (* data *)
TableForm[Map[a280850, Range[28]], TableDepth->2] (* triangle in Example *)
(* Hartmut F. W. Hoft, Jan 31 2018 *)

Name edited by Omar E. Pol, Nov 11 2018

A280851 Irregular triangle read by rows in which row n lists the subparts of the symmetric representation of sigma(n), ordered by order of appearance in the structure, from left to right.

Original entry on oeis.org

1, 3, 2, 2, 7, 3, 3, 11, 1, 4, 4, 15, 5, 3, 5, 9, 9, 6, 6, 23, 5, 7, 7, 12, 12, 8, 7, 1, 8, 31, 9, 9, 35, 2, 2, 10, 10, 39, 3, 11, 5, 5, 11, 18, 18, 12, 12, 47, 13, 13, 5, 13, 21, 21, 14, 6, 6, 14, 55, 1, 15, 15, 59, 3, 7, 3, 16, 16, 63, 17, 7, 7, 17, 27, 27, 18, 9, 3, 18, 71, 10, 10, 19, 19, 30, 30

Offset: 1

Omar E. Pol, Jan 09 2017

The terms in the n-th row are the same as the terms in the n-th row of triangle A279391, but in some rows the terms appear in distinct order.
First differs from A279391 at a(28) = T(15,3).
Also nonzero terms of A296508. - Omar E. Pol, Feb 11 2018

			Triangle begins (rows 1..16):
   1;
   3;
   2,  2;
   7;
   3,  3;
  11,  1;
   4,  4;
  15;
   5,  3,  5;
   9,  9;
   6,  6;
  23,  5;
   7,  7;
  12, 12;
   8,  7,  1,  8;
  31;
...
For n = 12 we have that the 11th row of triangle A237593 is [6, 3, 1, 1, 1, 1, 3, 6] and the 12th row of the same triangle is [7, 2, 2, 1, 1, 2, 2, 7], so the diagram of the symmetric representation of sigma(12) = 28 is constructed as shown below in Figure 1:
.                          _                                    _
.                         | |                                  | |
.                         | |                                  | |
.                         | |                                  | |
.                         | |                                  | |
.                         | |                                  | |
.                    _ _ _| |                             _ _ _| |
.                  _|    _ _|                           _|  _ _ _|
.                _|     |                             _|  _| |
.               |      _|                            |  _|  _|
.               |  _ _|                              | |_ _|
.    _ _ _ _ _ _| |     28                _ _ _ _ _ _| |    5
.   |_ _ _ _ _ _ _|                      |_ _ _ _ _ _ _|
.                                                       23
.
.   Figure 1. The symmetric            Figure 2. After the dissection
.   representation of sigma(12)        of the symmetric representation
.   has only one part which            of sigma(12) into layers of
.   contains 28 cells, so              width 1 we can see two subparts
.   the 12th row of the                that contain 23 and 5 cells
.   triangle A237270 is [28].          respectively, so the 12th row of
.                                      this triangle is [23, 5].
.
For n = 15 we have that the 14th row of triangle A237593 is [8, 3, 1, 2, 2, 1, 3, 8] and the 15th row of the same triangle is [8, 3, 2, 1, 1, 1, 1, 2, 3, 8], so the diagram of the symmetric representation of sigma(15) = 24 is constructed as shown below in Figure 3:
.                                _                                  _
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                          _ _ _|_|                           _ _ _|_|
.                      _ _| |      8                      _ _| |      8
.                     |    _|                            |  _ _|
.                    _|  _|                             _| |_|
.                   |_ _|  8                           |_ _|  1
.                   |                                  |    7
.    _ _ _ _ _ _ _ _|                   _ _ _ _ _ _ _ _|
.   |_ _ _ _ _ _ _ _|                  |_ _ _ _ _ _ _ _|
.                    8                                  8
.
.   Figure 3. The symmetric            Figure 4. After the dissection
.   representation of sigma(15)        of the symmetric representation
.   has three parts of size 8          of sigma(15) into layers of
.   because every part contains        width 1 we can see four "subparts".
.   8 cells, so the 15th row of        The first layer has three subparts:
.   triangle A237270 is [8, 8, 8].     [8, 7, 8]. The second layer has
.                                      only one subpart of size 1. The
.                                      15th row of this triangle is
.                                      [8, 7, 1, 8].
.
From _Hartmut F. W. Hoft_, Jan 31 2018: (Start)
The subparts of 36 whose symmetric representation of sigma has maximum width 2 are 71, 10, and 10.
The (size, width level) pairs of the six subparts of the symmetric representation of sigma(63) which consists of five parts are (32,1), (12,1), (11,1), (5,2), (12,1), and (32,1).
The subparts of perfect number 496 are 991, the length of its entire Dyck path, and 1 at the diagonal.
Number 10080, the smallest number whose symmetric representation of sigma has maximum width 10 (see A250070), has 12 subparts; its (size, width level) pairs are (20159,1), (6717,2), (4027,3), (2873,4), (2231,5), (1329,6), (939,7), (541,8), (403,9), (3,10), (87,10), and (3,10). The size of the first subpart is the length of the entire Dyck path so that the symmetric representation consists of a single part. The first subpart at the 10th level occurs at coordinates (6926,7055) ... (6929,7055). (End)
From _Omar E. Pol_, Dec 26 2020: (Start)
Also consider the infinite double-staircases diagram defined in A335616 (see the theorem).
For n = 15 the diagram with first 15 levels looks like this:
.
Level                         "Double-staircases" diagram
.                                          _
1                                        _|1|_
2                                      _|1 _ 1|_
3                                    _|1  |1|  1|_
4                                  _|1   _| |_   1|_
5                                _|1    |1 _ 1|    1|_
6                              _|1     _| |1| |_     1|_
7                            _|1      |1  | |  1|      1|_
8                          _|1       _|  _| |_  |_       1|_
9                        _|1        |1  |1 _ 1|  1|        1|_
10                     _|1         _|   | |1| |   |_         1|_
11                   _|1          |1   _| | | |_   1|          1|_
12                 _|1           _|   |1  | |  1|   |_           1|_
13               _|1            |1    |  _| |_  |    1|            1|_
14             _|1             _|    _| |1 _ 1| |_    |_             1|_
15            |1              |1    |1  | |1| |  1|    1|              1|
.
Starting from A196020 and after the algorithm described in A280850 and A296508 applied to the above diagram we have a new diagram as shown below:
.
Level                             "Ziggurat" diagram
.                                          _
6                                         |1|
7                            _            | |            _
8                          _|1|          _| |_          |1|_
9                        _|1  |         |1   1|         |  1|_
10                     _|1    |         |     |         |    1|_
11                   _|1      |        _|     |_        |      1|_
12                 _|1        |       |1       1|       |        1|_
13               _|1          |       |         |       |          1|_
14             _|1            |      _|    _    |_      |            1|_
15            |1              |     |1    |1|    1|     |              1|
.
The 15th row
of A249351 :  [1,1,1,1,1,1,1,1,0,0,0,1,1,1,2,1,1,1,0,0,0,1,1,1,1,1,1,1,1]
The 15th row
of A237270:   [              8,            8,            8              ]
The 15th row
of A296508:   [              8,      7,    1,    0,      8              ]
The 15th row
of triangle   [              8,      7,    1,            8              ]
.
More generally, for n >= 1, it appears there is the same correspondence between the original diagram of the symmetric representation of sigma(n) and the "Ziggurat" diagram of n.
For the definition of subparts see A279387 and also A296508. (End)

Hartmut F. W. Hoft, Table of n, a(n) for n = 1..987
Hartmut F. W. Hoft, A proof that the symmetric representation of sigma equals sigma
Hartmut F. W. Hoft, Numbers having a subpart of size 1 in the symmetric representation of sigma are the hexagonal numbers, A proof of _Omar E. Pol_'s conjecture dated Aug 28 2021 in A000384.
Index entries for sequences related to sigma(n)

Row sums give A000203.
The length of row n equals A001227(n).
Hence, if n is odd the length of row n equals A000005(n).
For the definition of "subparts" see A279387.
For the triangle of sums of subparts see A279388.
Cf. A000384, A001227, A196020, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A239657, A240542, A244050, A245092, A249351, A250068, A250070, A261699, A262626, A279391, A280850, A296508, A335616, A346875.

row[n_] := Floor[(Sqrt[8n+1]-1)/2]
f[n_] := Map[Ceiling[(n+1)/#-(#+1)/2] - Ceiling[(n+1)/(#+1)-(#+2)/2]&, Range[row[n]]]
a237593[n_] := Module[{a=f[n]}, Join[a, Reverse[a]]]
g[n_] := Map[If[Mod[n - #*(#+1)/2, #]==0, (-1)^(#+1), 0]&, Range[row[n]]]
a262045[n_] := Module[{a=Accumulate[g[n]]}, Join[a, Reverse[a]]]
findStart[list_] := Module[{i=1}, While[list[[i]]==0, i++]; i]
a280851[n_] := Module[{lenL=a237593[n], widL=a262045[n], r=row[n], subs={}, acc, start, i}, While[!AllTrue[widL, #==0&], start=findStart[widL]; acc=lenL[[start]]; widL[[start]]-=1; i=start+1; While[i<=2*r && acc!=0, If[widL[[i]]==0, If[start<=r2*r && acc!=0, If[start<=r2] (* triangle *) (* Hartmut F. W. Hoft, Jan 31 2018 *)

Name clarified by Hartmut F. W. Hoft and Omar E. Pol, Jan 31 2018

A250068 Maximum width of any region in the symmetric representation of sigma(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2

Offset: 1

Hartmut F. W. Hoft, Nov 11 2014

nonn

Since the width of the single region of the symmetric representation of sigma( 2^ceiling((p-1)*(log_2 3) - 1) * 3^(p-1) ), for prime number p, at the diagonal equals p, this sequence contains an increasing subsequence (see A250071).

a(n) is also the number of layers of width 1 in the symmetric representation of sigma(n). For more information see A001227. - Omar E. Pol, Dec 13 2016

			a(6) = 2 since the sequence of widths at each unit step in the symmetric representation of sigma(6) = 12 is 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1. For visual examples see A237270, A237593 and sequences referenced in these.

N. J. A. Sloane, Table of n, a(n) for n = 1..10000

Cf. A000203, A001227, A237048, A237270, A237271, A237591, A237593, A241008, A241010, A246955, A247687, A249223, A249351 (widths), A279387, A279388, A279391.

See A250070, A250071, A340506 for records.

(* function a2[ ] is defined in A249223 *)
a250068[n_]:=Max[a2[n]]
a250068[{m_,n_}]:=Map[a250068,Range[m,n]]
a250068[{1,100}](* data *)

t237048(n,k) = if (k % 2, (n % k) == 0, ((n - k/2) % k) == 0);
kmax(n) = (sqrt(1+8*n)-1)/2;
t249223(n,k) = sum(j=1, k, (-1)^(j+1)*t237048(n,j));
a(n) = my(wm = t249223(n, 1)); for (k=2, kmax(n), wm = max(wm, t249223(n, k))); wm; \\ Michel Marcus, Sep 20 2015

a(n) = max_{k=1..floor((sqrt(8*n+1) - 1)/2)} (Sum_{j=1..k}(-1)^(j+1)*A237048(n, j)), for n >= 1.

A279391 Irregular triangle read by rows in which row n lists the subparts of the successive layers of the symmetric representation of sigma(n).

Original entry on oeis.org

1, 3, 2, 2, 7, 3, 3, 11, 1, 4, 4, 15, 5, 3, 5, 9, 9, 6, 6, 23, 5, 7, 7, 12, 12, 8, 7, 8, 1, 31, 9, 9, 35, 2, 2, 10, 10, 39, 3, 11, 5, 5, 11, 18, 18, 12, 12, 47, 13, 13, 5, 13, 21, 21, 14, 6, 6, 14, 55, 1, 15, 15, 59, 3, 7, 3, 16, 16, 63, 17, 7, 7, 17, 27, 27, 18, 9, 18, 3, 71, 10, 10, 19, 19, 30, 30

Offset: 1

Omar E. Pol, Dec 12 2016

Note that the terms in the n-th row are the same as the terms in the n-th row of triangle A280851, but in some rows the terms appear in distinct order. First differs from A280851 at a(28) = T(15,3). - Omar E. Pol, Apr 24 2018

Row n in the triangle is a sequence of A250068(n) symmetric sections, each section consisting of the sizes of the subparts on that level in the symmetric representation of sigma of n - from the top down in the images below or left to right as drawn in A237593. - Hartmut F. W. Hoft, Sep 05 2021

			Triangle begins (first 15 rows):
   [1];
   [3];
   [2, 2];
   [7];
   [3, 3];
   [11], [1];
   [4, 4];
   [15];
   [5, 3, 5];
   [9, 9];
   [6, 6];
   [23], [5];
   [7, 7];
   [12, 12];
   [8, 7, 8], [1];
  ...
For n = 12 we have that the 11th row of triangle A237593 is [6, 3, 1, 1, 1, 1, 3, 6] and the 12th row of the same triangle is [7, 2, 2, 1, 1, 2, 2, 7], so the diagram of the symmetric representation of sigma(12) = 28 is constructed as shown below in Figure 1:
.                          _                                    _
.                         | |                                  | |
.                         | |                                  | |
.                         | |                                  | |
.                         | |                                  | |
.                         | |                                  | |
.                    _ _ _| |                             _ _ _| |
.                  _|    _ _|                           _|  _ _ _|
.                _|     |                             _|  _| |
.               |      _|                            |  _|  _|
.               |  _ _|                              | |_ _|
.    _ _ _ _ _ _| |    28                 _ _ _ _ _ _| |    5
.   |_ _ _ _ _ _ _|                      |_ _ _ _ _ _ _|
.                                                       23
.
.   Figure 1. The symmetric            Figure 2. After the dissection
.   representation of sigma(12)        of the symmetric representation
.   has only one part which            of sigma(12) into layers of
.   contains 28 cells, so              width 1 we can see two "subparts"
.   the 12th row of the                that contain 23 and 5 cells
.   triangle A237270 is [28].          respectively, so the 12th row of
.                                      this triangle is [23], [5].
.
For n = 15 we have that the 14th row of triangle A237593 is [8, 3, 1, 2, 2, 1, 3, 8] and the 15th row of the same triangle is [8, 3, 2, 1, 1, 1, 1, 2, 3, 8], so the diagram of the symmetric representation of sigma(15) = 24 is constructed as shown below in Figure 3:
.                                _                                  _
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                          _ _ _|_|                           _ _ _|_|
.                      _ _| |      8                      _ _| |      8
.                     |    _|                            |  _ _|
.                    _|  _|                             _| |_|
.                   |_ _|  8                           |_ _|  1
.                   |                                  |    7
.    _ _ _ _ _ _ _ _|                   _ _ _ _ _ _ _ _|
.   |_ _ _ _ _ _ _ _|                  |_ _ _ _ _ _ _ _|
.                    8                                  8
.
.   Figure 3. The symmetric            Figure 4. After the dissection
.   representation of sigma(15)        of the symmetric representation
.   has three parts of size 8          of sigma(15) into layers of
.   because every part contains        width 1 we can see four "subparts".
.   8 cells, so the 15th row of        The first layer has three subparts:
.   triangle A237270 is [8, 8, 8].     8, 7, 8. The second layer has
.                                      only one subpart of size 1, so
.                                      the 15th row of this triangle is
.                                      [8, 7, 8], [1].
.
The smallest even number with 3 levels is 60; its row of subparts is: [119], [37], [6, 6]. The smallest odd number with 3 levels is 315; its row of subparts is:  [158, 207, 158], [11, 26, 5, 9, 5, 26, 11], [4, 4]. - _Hartmut F. W. Hoft_, Sep 05 2021

Index entries for sequences related to sigma(n)

The length of row n equals A001227(n).
If n is odd the length of row n equals A000005(n).
Row sums give A000203.
For the definition of "subparts" see A279387.
For the triangle of sums of subparts see A279388.
Cf. A001227, A005279, A196020, A235791, A236104, A237048, A237270, A237591, A237593, A239657, A244050, A245092, A250068, A250070, A261699, A262626, A280851, A296508.

(* support functions are defined in aA237593 and A262045 *)
subP[level_] := Module[{s=Map[Apply[Plus, #]&, Select[level, First[#]!=0&]]}, If[OddQ[Length[s]], s[[(Length[s]+1)/2]]-=1]; s]
a279391[n_] := Module[{widL=a262045[n], lenL=a237593[n], srs, subs}, srs=Transpose[Map[PadRight[If[widL[[#]]>0, Table[1, widL[[#]]], {0}], Max[widL]]&, Range[Length[lenL]]]]; subs=Map[SplitBy[lenL srs[[#]], #!=0&]&, Range[Max[widL]]]; Flatten[Map[subP, subs]]]
Flatten[Map[a279391, Range[38]]] (* Hartmut F. W. Hoft, Sep 05 2021 *)

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).

Original entry on oeis.org

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

Omar E. Pol, Sep 05 2021

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.

			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.
.

Index entries for sequences related to sigma(n)

Another (and more regular) version of A279388.
Row sums give A000203.
Row n has length A351846(n).
Cf. A347263 (analog for the ziggurat diagram).
Cf. A000040, A000079, A000384, A000396, A082647, A196020, A235791, A236104, A237591, A237270, A237271, A237593, A279387, A262626, A335616, A346875, A347186.

A347263 Irregular triangle read by rows: T(n,k) is the sum of the 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), n >= 1, k >= 1, and the first element of column k is in row A000384(k).

Original entry on oeis.org

1, 4, 6, 16, 12, 36, 1, 20, 0, 64, 0, 30, 6, 90, 0, 42, 0, 144, 17, 56, 0, 156, 0, 72, 34, 1, 256, 0, 0, 90, 0, 0, 324, 10, 0, 110, 0, 0, 400, 0, 8, 132, 70, 0, 342, 0, 0, 156, 0, 0, 576, 121, 0, 182, 0, 25, 462, 0, 0, 210, 102, 0, 784, 0, 0, 1, 240, 0, 0, 0, 900, 24, 52, 0, 272, 0, 0, 0

Offset: 1

Omar E. Pol, Sep 05 2021

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 subparts of the ziggurat diagram are the polygons formed by the cells that are under the staircases.
The connection of the subparts of the ziggurat diagram with the polygonal numbers is as follows:
The area under a double-staircase labeled with the number j is equal to the m-th (j+2)-gonal number plus the (m-1)-th (j+2)-gonal number, where m is the number of steps on one side of the ladder from the base to the top.
The area under a simple-staircase labeled with the number j is equal to the m-th (j+2)-gonal number, where m is the number of steps.
So the k-th column of the triangle is related to the (2*k+1)-gonal numbers, for example:
For the calculation of column 1 we use triangular numbers A000217.
For the calculation of column 2 we use pentagonal numbers A000326.
For the calculation of column 3 we use heptagonal numbers A000566.
For the calculation of column 4 we use enneagonal numbers A001106.
And so on.
More generally, for the calculation of column k we use the (2*k+1)-gonal numbers.
For further information about the ziggurat diagram see A347186.

			Triangle begins:
   n / k   1    2    3    4
------------------------------
   1 |     1;
   2 |     4;
   3 |     6;
   4 |    16;
   5 |    12;
   6 |    36,   1;
   7 |    20,   0;
   8 |    64,   0;
   9 |    30,   6;
  10 |    90,   0;
  11 |    42,   0;
  12 |   144,  17;
  13 |    56,   0;
  14 |   156,   0;
  15 |    72,  34,   1;
  16 |   256,   0,   0;
  17 |    90,   0,   0;
  18 |   324,  10,   0;
  19 |   110,   0    0;
  20 |   400,   0,   8;
  21 |   132,  70,   0;
  22 |   342,   0,   0;
  23 |   156,   0,   0;
  24 |   576, 121,   0;
  25 |   182,   0,  25;
  26 |   462,   0,   0;
  27 |   210, 102,   0;
  28 |   784,   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".
Now we calculate the area (or the number of cells) under the staircases with multiplicity using polygonal numbers as shown below:
The area under the staircase labeled 1 is equal to A000217(8) = 36. There is a pair of these staircases, so T(15,1) = 2*36 = 72.
The area under the double-staircase labeled 3 is equal to A000326(4) + A000326(3) = 22 + 12 = 34, so T(15,2) = 34.
The area under the double-staircase labeled 5 is equal to A000566(1) + A000566(0) = 1 + 0 = 1, so T(15,3) = 1.
Therefore the 15th row of the triangle is [72, 34, 1].

Row sums give A347186.
Row n has length A351846(n).
Cf. A347529 (analog for the symmetric representation of sigma).
Cf. A000217, A000326, A000384, A000566, A001106, A082647, A139600, A139601, A237270, A237271, A237593, A279387, A279388, A279391, A335616, A346875.

A280940 Irregular triangle read by rows: T(n,k) = number of subparts in the k-th part of the symmetric representation of sigma(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Omar E. Pol, Jan 11 2017

The "subparts" of the symmetric representation of sigma(n) are the regions that arise after the dissection of the symmetric representation of sigma(n) into successive layers of width 1.
The number of subparts in the symmetric representation of sigma(n) equals the number of odd divisors of n.
For more information about "subparts" see A279387, A279388 and A279391.
Note that we can find the symmetric representation of sigma(n) as the terraces at the n-th level (starting from the top) of the stepped pyramid described in A245092.

			Triangle begins (n = 1..21):
1;
1;
1, 1;
1;
1, 1;
2;
1, 1;
1;
1, 1, 1;
1, 1;
1, 1;
2;
1, 1;
1, 1;
1, 2, 1;
1;
1, 1;
3;
1, 1;
2;
1, 1, 1, 1;
...
For n = 12 we have that the 11th row of triangle A237593 is [6, 3, 1, 1, 1, 1, 3, 6] and the 12th row of the same triangle is [7, 2, 2, 1, 1, 2, 2, 7], so the diagram of the symmetric representation of sigma(12) = 28 is constructed as shown below in Figure 1:
.                          _                                    _
.                         | |                                  | |
.                         | |                                  | |
.                         | |                                  | |
.                         | |                                  | |
.                         | |                                  | |
.                    _ _ _| |                             _ _ _| |
.              28  _|    _ _|                       23  _|  _ _ _|
.                _|     |                             _|  _| |
.               |      _|                            |  _|  _|
.               |  _ _|                              | |_ _|
.    _ _ _ _ _ _| |                       _ _ _ _ _ _| |      5
.   |_ _ _ _ _ _ _|                      |_ _ _ _ _ _ _|
.
.   Figure 1. The symmetric            Figure 2. After the dissection
.   representation of sigma(12)        of the symmetric representation
.   has only one part which            of sigma(12) into layers of
.   contains 28 cells, so              width 1 we can see two "subparts"
.   A237271(12) = 1, and               that contain 23 and 5 cells
.   A000203(12) = 28.                  respectively, so the 12th row of
.                                      this triangle is [2], and the
.                                      row sum is A001227(12) = 2, equaling
.                                      the number of odd divisors of 12.
.
For n = 15 we have that the 14th row of triangle A237593 is [8, 3, 1, 2, 2, 1, 3, 8] and the 15th row of the same triangle is [8, 3, 2, 1, 1, 1, 1, 2, 3, 8], so the diagram of the symmetric representation of sigma(15) = 24 is constructed as shown below in Figure 3:
.                                _                                  _
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                               | |                                | |
.                           8   | |                            8   | |
.                               | |                                | |
.                               | |                                | |
.                          _ _ _|_|                           _ _ _|_|
.                   8  _ _| |                          7  _ _| |
.                     |    _|                            |  _ _|
.                    _|  _|                             _| |_|
.                   |_ _|                              |_ _|  1
.           8       |                          8       |
.    _ _ _ _ _ _ _ _|                   _ _ _ _ _ _ _ _|
.   |_ _ _ _ _ _ _ _|                  |_ _ _ _ _ _ _ _|
.
.   Figure 3. The symmetric            Figure 4. After the dissection
.   representation of sigma(15)        of the symmetric representation
.   has three parts of size 8          of sigma(15) into layers of
.   because every part contains        width 1 we can see four "subparts".
.   8 cells, so A237271(15) = 3,       The first and third part contains
.   and A000203(15) = 8+8+8 = 24.      one subpart each. The second part contains
.                                      two subparts, so the 15th row of this
.                                      triangle is [1, 2, 1], and the row
.                                      sum is A001227(15) = 4, equaling the
.                                      number of odd divisors of 15.
.

Row sums give A001227 (number of odd divisors of n).
Row lengths is A237271.
Cf. A000203, A196020, A235791, A236104, A237048, A237270, A237591, A237593, A244050, A245092, A249351, A250068, A262626, A279387, A279388, A279391.

A281010 Triangle read by rows in which row 2n-1 lists the widths of the symmetric representation of sigma(n), and row 2n lists a finite sequence S together with -1, with the property that the partial sums of S give the row 2n-1.

Original entry on oeis.org

1, 1, -1, 1, 1, 1, 1, 0, 0, -1, 1, 1, 0, 1, 1, 1, 0, -1, 1, 0, -1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, -1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, -1, 0, 0, 1, 0, 0, -1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, -1, 0, 0, 0, 0, -1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, -1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1

Offset: 1

Omar E. Pol, Jan 12 2017

The row 2n-1 lists the widths of the terraces at the n-th level (starting from the top) of the pyramid described in A245092.
The sum of the areas of these terraces equals A000203(n): the sum of the divisors of n.
The k-th element of row 2n is associated to the k-th vertical cells at the n-th level of the pyramid.
The row 2n shows where the subparts (or subregions) of the terraces starting and ending, in accordance with the values 1 or -1.
The number of subparts in the n-th terrace equals A001227(n): the number of odd divisors of n.
If n is odd then the number of subparts in the n-th terrace is also A000005(n): the number of divisors of n.

			Triangle begins:
1;
1,-1;
1, 1, 1;
1, 0, 0,-1;
1, 1, 0, 1, 1;
1, 0,-1, 1, 0;-1;
1, 1, 1, 1, 1, 1, 1;
1, 0, 0, 0, 0, 0, 0,-1;
1, 1, 1, 0, 0, 0, 1, 1, 1;
1, 0, 0,-1, 0, 0, 1, 0, 0,-1;
1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1;
1, 0, 0, 0, 0, 1,-1, 0, 0, 0, 0,-1;
1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1;
1, 0, 0, 0,-1, 0, 0, 0, 0, 1, 0, 0, 0,-1;
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1;
1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1;
1, 0, 0, 0, 0,-1, 0, 1, 0, 0,-1, 0, 1, 0, 0, 0, 0,-1;
1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1;
1, 0, 0, 0, 0, 0, 0, 0, 0,-1, 1, 0, 0, 0, 0, 0, 0, 0, 0,-1;
1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1;
1, 0, 0, 0, 0, 0,-1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,-1;
1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1;
1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,-1, 0, 0, 0, 0, 0, 0, 0, 0,-1;
...
Written as an isosceles triangle the sequence begins:
.
.                                        1;
.                                      1, -1;
.                                    1,  1,  1;
.                                  1,  0,  0, -1;
.                                1,  1,  0,  1,  1;
.                              1,  0, -1,  1,  0, -1;
.                            1,  1,  1,  1,  1,  1,  1;
.                          1,  0,  0,  0,  0,  0,  0, -1;
.                        1,  1,  1,  0,  0,  0,  1,  1,  1;
.                      1,  0,  0, -1,  0,  0,  1,  0,  0, -1;
.                    1,  1,  1,  1,  1,  2,  1,  1,  1,  1,  1;
.                  1,  0,  0,  0,  0,  1, -1,  0,  0,  0,  0, -1;
.                1,  1,  1,  1,  0,  0,  0,  0,  0,  1,  1,  1,  1;
.              1,  0,  0,  0, -1,  0,  0,  0,  0,  1,  0,  0,  0, -1;
.            1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1;
.          1,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0, -1;
.        1,  1,  1,  1,  1,  0,  0,  1,  1,  1,  0,  0,  1,  1,  1,  1,  1;
.      1,  0,  0,  0,  0, -1,  0,  1,  0,  0, -1,  0,  1,  0,  0,  0,  0, -1;
.    1,  1,  1,  1,  1,  1,  1,  1,  1,  0,  1,  1,  1,  1,  1,  1,  1,  1,  1;
.  1,  0,  0,  0,  0,  0,  0,  0,  0, -1,  1,  0,  0,  0,  0,  0,  0,  0,  0, -1;
...

Omar E. Pol, Diagram of the isosceles triangle A237593 before the 90-degree-zig-zag folding (rows: 1..28)
Omar E. Pol, Perspective view of the pyramid (first 16 levels)

The sum of row 2n-1 is A000203(n).
The sum of row 2n is A000004(n) = 0.
The number of positive terms in row 2n is A001227(n).
The number of nonzero terms in row 2n is A054844(n).
Middle diagonal (or central column of the isosceles triangle) gives A067742.
Row 2n-1 is also the n-th row of A249351.
Row 2n is also the n-th row of A281011.
Row 2n-1 lists the partial sums of the terms, except the last term, of the row 2n.
Cf. A000005, A001227, A196020, A235048, A236104, A237048, A237591, A237593, A244050, A245092, A262626, A279387, A279388, A279391.

cp's OEIS Frontend

A279387 Irregular triangle read by rows: suppose the symmetric representation of sigma(n) consists of m = A250068(n) layers of width 1, arranged in increasing order; then T(n,k) (n >= 1, 1 <= k <= m) is the number of subparts in the k-th layer.

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A280850 Irregular triangle read by rows in which row n is constructed with an algorithm using the n-th row of triangle A196020 (see Comments for precise definition).

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A280851 Irregular triangle read by rows in which row n lists the subparts of the symmetric representation of sigma(n), ordered by order of appearance in the structure, from left to right.

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A250068 Maximum width of any region in the symmetric representation of sigma(n).

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A279391 Irregular triangle read by rows in which row n lists the subparts of the successive layers of the symmetric representation of sigma(n).

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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).

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A280940 Irregular triangle read by rows: T(n,k) = number of subparts in the k-th part of the symmetric representation of sigma(n).

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