A335616 -id:A335616 - cp's OEIS frontend

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).

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.

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).

Original entry on oeis.org

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

Omar E. Pol, Feb 20 2022

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.
T(n,k) is also the number of staircases (or 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).
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.
Terms can be 0, 1 or 2.

			Triangle begins:
-----------------------
   n / k   1  2  3  4
-----------------------
   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:
                               _
                             _| |_
                           _|  _  |_
                         _|   | |   |_
                       _|    _| |_    |_
                     _|     |  _  |     |_
                   _|      _| | | |_      |_
                 _|       |   | |   |       |_
               _|        _|  _| |_  |_        |_
             _|         |   |  _  |   |         |_
           _|          _|   | | | |   |_          |_
         _|           |    _| | | |_    |           |_
       _|            _|   |   | |   |   |_            |_
     _|             |     |  _| |_  |     |             |_
   _|              _|    _| |  _  | |_    |_              |_
  |_ _ _ _ _ _ _ _|_ _ _|_ _|_|_|_|_ _|_ _ _|_ _ _ _ _ _ _ _|
  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 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:
                                _                                     _
                               |_|                                   | |
                               |_|                                   | |
                               |_|                                   | |
                               |_|                                   | |
                               |_|                                   | |
                               |_|                                   | |
                               |_|                                   | |
                          _ _ _|_|                              _ _ _|_|
                      _ _|_|      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.
.

Another (and more regular) version of A279387 and of A280940.
Row sums give A001227.
Row n has length A351846(n).
Cf. A000384, A082647, A196020, A235791, A236104, A237048, A237591, A237593, A262626, A335616, A346875, A347186, A347263, A347529.

A237593 Triangle read by rows in which row n lists the elements of the n-th row of A237591 followed by the same elements in reverse order.

Original entry on oeis.org

1, 1, 2, 2, 2, 1, 1, 2, 3, 1, 1, 3, 3, 2, 2, 3, 4, 1, 1, 1, 1, 4, 4, 2, 1, 1, 2, 4, 5, 2, 1, 1, 2, 5, 5, 2, 2, 2, 2, 5, 6, 2, 1, 1, 1, 1, 2, 6, 6, 3, 1, 1, 1, 1, 3, 6, 7, 2, 2, 1, 1, 2, 2, 7, 7, 3, 2, 1, 1, 2, 3, 7, 8, 3, 1, 2, 2, 1, 3, 8, 8, 3, 2, 1, 1, 1, 1, 2, 3, 8

Offset: 1

Omar E. Pol, Feb 22 2014

Row n is a palindromic composition of 2*n.
T(n,k) is also the length of the k-th segment in a Dyck path on the first quadrant of the square grid, connecting the x-axis with the y-axis, from (n, 0) to (0, n), starting with a segment in vertical direction, see example.
Conjecture 1: the area under the n-th Dyck path equals A024916(n), the sum of all divisors of all positive integers <= n.
If the conjecture is true then the n-th Dyck path represents the boundary segments after the alternating sum of the elements of the n-th row of A236104.
Conjecture 2: two adjacent Dyck paths never cross (checked by hand up to n = 128), hence the total area between the n-th Dyck path and the (n-1)-st Dyck path is equal to sigma(n) = A000203(n), the sum of divisors of n.
The connection between A196020 and A237271 is as follows: A196020 --> A236104 --> A235791 --> A237591 --> this sequence --> A239660 --> A237270 --> A237271.
PARI scripts area(n) and chkcross(n) have been written to check the 2 properties and have been run up to n=10000. - Michel Marcus, Mar 27 2014
Mathematica functions have been written that verified the 2 properties through n=30000. - Hartmut F. W. Hoft, Apr 07 2014
Comments from Franklin T. Adams-Watters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014: (Start)
The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.
You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.
Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n).
Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)
The symmetric representation of sigma, so defined, is row n of A237270. - Peter Munn, Jan 06 2025
It appears that, for the n-th set, the number of cells lying on the first diagonal is equal to A067742(n), the number of middle divisors of n. - Michel Marcus, Jun 21 2014
Checked Michel Marcus's conjecture with two Mathematica functions up to n=100000, for more information see A240542. - Hartmut F. W. Hoft, Jul 17 2014
A003056(n) is also the number of peaks of the Dyck path related to the n-th row of triangle. - Omar E. Pol, Nov 03 2015
The number of peaks of the Dyck path associated to the row A000396(n) of this triangle equals the n-th Mersenne prime A000668(n), hence Mersenne primes are visible in two ways at the pyramid described in A245092. - Omar E. Pol, Dec 19 2016
The limit as n approaches infinity (area under the Dyck path described in the n-th row of triangle divided by n^2) equals Pi^2/12 = zeta(2)/2. (Cf. A072691.) - Omar E. Pol, Dec 18 2021
The connection between the isosceles triangle and the stepped pyramid is due to the fact that this object can also be interpreted as a pop-up card. - Omar E. Pol, Nov 09 2022

			Triangle begins:
   n
   1 |  1, 1;
   2 |  2, 2;
   3 |  2, 1, 1, 2;
   4 |  3, 1, 1, 3;
   5 |  3, 2, 2, 3;
   6 |  4, 1, 1, 1, 1, 4;
   7 |  4, 2, 1, 1, 2, 4;
   8 |  5, 2, 1, 1, 2, 5;
   9 |  5, 2, 2, 2, 2, 5;
  10 |  6, 2, 1, 1, 1, 1, 2, 6;
  11 |  6, 3, 1, 1, 1, 1, 3, 6;
  12 |  7, 2, 2, 1, 1, 2, 2, 7;
  13 |  7, 3, 2, 1, 1, 2, 3, 7;
  14 |  8, 3, 1, 2, 2, 1, 3, 8;
  15 |  8, 3, 2, 1, 1, 1, 1, 2, 3, 8;
  16 |  9, 3, 2, 1, 1, 1, 1, 2, 3, 9;
  17 |  9, 4, 2, 1, 1, 1, 1, 2, 4, 9;
  18 | 10, 3, 2, 2, 1, 1, 2, 2, 3, 10;
  19 | 10, 4, 2, 2, 1, 1, 2, 2, 4, 10;
  20 | 11, 4, 2, 1, 2, 2, 1, 2, 4, 11;
  21 | 11, 4, 3, 1, 1, 1, 1, 1, 1, 3, 4, 11;
  22 | 12, 4, 2, 2, 1, 1, 1, 1, 2, 2, 4, 12;
  23 | 12, 5, 2, 2, 1, 1, 1, 1, 2, 2, 5, 12;
  24 | 13, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 13;
  ...
Illustration of rows 8 and 9 interpreted as Dyck paths in the first quadrant and the illustration of the symmetric representation of sigma(9) = 5 + 3 + 5 = 13, see below:
.
y                       y
.                       .
.                       ._ _ _ _ _                _ _ _ _ _ 5
._ _ _ _ _              .         |              |_ _ _ _ _|
.         |             .         |_ _                     |_ _ 3
.         |_            .             |                    |_  |
.           |_ _        .             |_ _                   |_|_ _ 5
.               |       .                 |                      | |
.   Area = 56   |       .    Area = 69    |          Area = 13   | |
.               |       .                 |                      | |
.               |       .                 |                      | |
. . . . . . . . | . x   . . . . . . . . . | . x                  |_|
.
.    Fig. 1                    Fig. 2                  Fig. 3
.
Figure 1. For n = 8 the 8th row of triangle is [5, 2, 1, 1, 2, 5] and the area under the symmetric Dyck path is equal to A024916(8) = 56.
Figure 2. For n = 9 the 9th row of triangle is [5, 2, 2, 2, 2, 5] and the area under the symmetric Dyck path is equal to A024916(9) = 69.
Figure 3. The symmetric representation of sigma(9): between both symmetric Dyck paths there are three regions (or parts) of sizes [5, 3, 5].
The sum of divisors of 9 is 1 + 3 + 9 = A000203(9) = 13. On the other hand the difference between the areas under the Dyck paths equals the sum of the parts of the symmetric representation of sigma(9) = 69 - 56 = 5 + 3 + 5 = 13, equaling the sum of divisors of 9.
.
Illustration of initial terms as Dyck paths in the first quadrant:
(row n = 1..28)
.  _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
  |_ _ _ _ _ _ _ _ _ _ _ _ _ _  |
  |_ _ _ _ _ _ _ _ _ _ _ _ _ _| |
  |_ _ _ _ _ _ _ _ _ _ _ _ _  | |
  |_ _ _ _ _ _ _ _ _ _ _ _ _| | |
  |_ _ _ _ _ _ _ _ _ _ _ _  | | |_ _ _
  |_ _ _ _ _ _ _ _ _ _ _ _| | |_ _ _  |
  |_ _ _ _ _ _ _ _ _ _ _  | | |_ _  | |_
  |_ _ _ _ _ _ _ _ _ _ _| | |_ _ _| |_  |_
  |_ _ _ _ _ _ _ _ _ _  | |       |_ _|   |_
  |_ _ _ _ _ _ _ _ _ _| | |_ _    |_  |_ _  |_ _
  |_ _ _ _ _ _ _ _ _  | |_ _ _|     |_  | |_ _  |
  |_ _ _ _ _ _ _ _ _| | |_ _  |_      |_|_ _  | |
  |_ _ _ _ _ _ _ _  | |_ _  |_ _|_        | | | |_ _ _ _ _
  |_ _ _ _ _ _ _ _| |     |     | |_ _    | |_|_ _ _ _ _  |
  |_ _ _ _ _ _ _  | |_ _  |_    |_  | |   |_ _ _ _ _  | | |
  |_ _ _ _ _ _ _| |_ _  |_  |_ _  | | |_ _ _ _ _  | | | | |
  |_ _ _ _ _ _  | |_  |_  |_    | |_|_ _ _ _  | | | | | | |
  |_ _ _ _ _ _| |_ _|   |_  |   |_ _ _ _  | | | | | | | | |
  |_ _ _ _ _  |     |_ _  | |_ _ _ _  | | | | | | | | | | |
  |_ _ _ _ _| |_      | |_|_ _ _  | | | | | | | | | | | | |
  |_ _ _ _  |_ _|_    |_ _ _  | | | | | | | | | | | | | | |
  |_ _ _ _| |_  | |_ _ _  | | | | | | | | | | | | | | | | |
  |_ _ _  |_  |_|_ _  | | | | | | | | | | | | | | | | | | |
  |_ _ _|   |_ _  | | | | | | | | | | | | | | | | | | | | |
  |_ _  |_ _  | | | | | | | | | | | | | | | | | | | | | | |
  |_ _|_  | | | | | | | | | | | | | | | | | | | | | | | | |
  |_  | | | | | | | | | | | | | | | | | | | | | | | | | | |
  |_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|
.
n: 1 2 3 4 5 6 7 8 9 10..12..14..16..18..20..22..24..26..28
.
It appears that the total area (also the total number of cells) in the first n set of symmetric regions of the diagram is equal to A024916(n), the sum of all divisors of all positive integers <= n.
It appears that the total area (also the total number of cells) in the n-th set of symmetric regions of the diagram is equal to sigma(n) = A000203(n) (checked by hand up n = 128).
From _Omar E. Pol_, Aug 18 2015: (Start)
The above diagram is also the top view of the stepped pyramid described in A245092 and it is also the top view of the staircase described in A244580, in both cases the figure represents the first 28 levels of the structure. Note that the diagram contains (and arises from) a hidden pattern which is shown below.
.
Illustration of initial terms as an isosceles triangle:
Row                                 _ _
1                                 _|1|1|_
2                               _|2 _|_ 2|_
3                             _|2  |1|1|  2|_
4                           _|3   _|1|1|_   3|_
5                         _|3    |2 _|_ 2|    3|_
6                       _|4     _|1|1|1|1|_     4|_
7                     _|4      |2  |1|1|  2|      4|_
8                   _|5       _|2 _|1|1|_ 2|_       5|_
9                 _|5        |2  |2 _|_ 2|  2|        5|_
10              _|6         _|2  |1|1|1|1|  2|_         6|_
11            _|6          |3   _|1|1|1|1|_   3|          6|_
12          _|7           _|2  |2  |1|1|  2|  2|_           7|_
13        _|7            |3    |2 _|1|1|_ 2|    3|            7|_
14      _|8             _|3   _|1|2 _|_ 2|1|_   3|_             8|_
15    _|8              |3    |2  |1|1|1|1|  2|    3|              8|_
16   |9                |3    |2  |1|1|1|1|  2|    3|                9|
...
This diagram is the simpler representation of the sequence.
The number of horizontal line segments in the n-th level in each side of the diagram equals A001227(n), the number of odd divisors of n.
The number of horizontal line segments in the left side of the diagram plus the number of the horizontal line segment in the right side equals A054844(n).
The total number of vertical line segments in the n-th level of the diagram equals A131507(n).
Note that this symmetric pattern also emerges from the front view of the stepped pyramid described in A245092, which is related to sigma A000203, the sum-of-divisors function, and other related sequences. The diagram represents the first 16 levels of the pyramid. (End)

Robert Price, Table of n, a(n) for n = 1..15008 (rows n = 1..412, flattened)
Michel Marcus, A colored version of the symmetric representation of sigma(n), multipage, n = 1..85
Omar E. Pol, An infinite stepped pyramid
Omar E. Pol, Illustration of initial terms of A000203 in the pyramid
Omar E. Pol, Illustration of initial terms of A001065 in the pyramid
Omar E. Pol, Illustration of initial terms of A048050 in the pyramid
Omar E. Pol, Illustration of initial terms of A067742 in the pyramid
Omar E. Pol, Illustration of initial terms of A224613, (black spiders)
Omar E. Pol, Illustration of initial terms as an isosceles triangle (rows: 1..28)
Omar E. Pol, The symmetric representation of sigma(n), n = 1..64, (version with six colors)
Index entries for sequences related to sigma(n)

Row n has length 2*A003056(n).
Row sums give A005843, n >= 1.
Column k starts in row A008805(k-1).
Column 1 = right border = A008619, n >= 1.
Bisections are in A259176, A259177.
For further information see A262626.
Cf. A000203, A000217, A001065, A001227, A024916, A048050, A054844, A067742, A072691, A131507, A196020, A221529, A224613, A235791, A236104, A237048, A237270, A237271, A237590, A237591, A239660, A239931-A239934, A244050, A244580, A245092, A249351, A261350, A261699, A262611, A262612, A279387, A280850, A280851, A286000, A286001, A296508, A335616, A340035.

row[n_]:=Floor[(Sqrt[8n+1]-1)/2]
s[n_,k_]:=Ceiling[(n+1)/k-(k+1)/2]-Ceiling[(n+1)/(k+1)-(k+2)/2]
f[n_,k_]:=If[k<=row[n],s[n,k],s[n,2 row[n]+1-k]]
TableForm[Table[f[n,k],{n,1,50},{k,1,2 row[n]}]] (* Hartmut F. W. Hoft, Apr 08 2014 *)

row(n) = {my(orow = row237591(n)); vector(2*#orow, i, if (i <= #orow, orow[i], orow[2*#orow-i+1]));}
area(n) = {my(rown = row(n)); surf = 0; h = n; odd = 1; for (i=1, #row, if (odd, surf += h*rown[i], h -= rown[i];); odd = !odd;); surf;}
heights(v, n) = {vh = vector(n); ivh = 1; h = n; odd = 1; for (i=1, #v, if (odd, for (j=1, v[i], vh[ivh] = h; ivh++), h -= v[i];); odd = !odd;); vh;}
isabove(hb, ha) = {for (i=1, #hb, if (hb[i] < ha[i], return (0));); return (1);}
chkcross(nn) = {hga = concat(heights(row(1), 1), 0); for (n=2, nn, hgb = heights(row(n), n); if (! isabove(hgb, hga), print("pb cross at n=", n)); hga = concat(hgb, 0););} \\ Michel Marcus, Mar 27 2014

from sympy import sqrt
import math
def row(n): return int(math.floor((sqrt(8*n + 1) - 1)/2))
def s(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2)) - int(math.ceil((n + 1)/(k + 1) - (k + 2)/2))
def T(n, k): return s(n, k) if k<=row(n) else s(n, 2*row(n) + 1 - k)
for n in range(1, 11): print([T(n, k) for k in range(1, 2*row(n) + 1)]) # Indranil Ghosh, Apr 21 2017

Let j(n)= floor((sqrt(8n+1)-1)/2) then T(n,k) = A237591(n,k), if k <= j(n); otherwise T(n,k) = A237591(n,2*j(n)+1-k). - Hartmut F. W. Hoft, Apr 07 2014 (corrected by Omar E. Pol, May 31 2015)

A minor edit to the definition. - N. J. A. Sloane, Jul 31 2025

A237270 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(n).

Original entry on oeis.org

1, 3, 2, 2, 7, 3, 3, 12, 4, 4, 15, 5, 3, 5, 9, 9, 6, 6, 28, 7, 7, 12, 12, 8, 8, 8, 31, 9, 9, 39, 10, 10, 42, 11, 5, 5, 11, 18, 18, 12, 12, 60, 13, 5, 13, 21, 21, 14, 6, 6, 14, 56, 15, 15, 72, 16, 16, 63, 17, 7, 7, 17, 27, 27, 18, 12, 18, 91, 19, 19, 30, 30, 20, 8, 8, 20, 90

Offset: 1

Omar E. Pol, Feb 19 2014

T(n,k) is the number of cells in the k-th region of the n-th set of regions in a diagram of the symmetry of sigma(n), see example.
Row n is a palindromic composition of sigma(n).
Row sums give A000203.
Row n has length A237271(n).
In the row 2n-1 of triangle both the first term and the last term are equal to n.
If n is an odd prime then row n is [m, m], where m = (1 + n)/2.
The connection with A196020 is as follows: A196020 --> A236104 --> A235791 --> A237591 --> A237593 --> A239660 --> this sequence.
For the boundary segments in an octant see A237591.
For the boundary segments in a quadrant see A237593.
For the boundary segments in the spiral see also A239660.
For the parts in every quadrant of the spiral see A239931, A239932, A239933, A239934.
We can find the spiral on the terraces of the stepped pyramid described in A244050. - Omar E. Pol, Dec 07 2016
T(n,k) is also the area of the k-th terrace, from left to right, at the n-th level, starting from the top, of the stepped pyramid described in A245092 (see Links section). - Omar E. Pol, Aug 14 2018

			Illustration of the first 27 terms as regions (or parts) of a spiral constructed with the first 15.5 rows of A239660:
.
.                  _ _ _ _ _ _ _ _
.                 |  _ _ _ _ _ _ _|_ _ _ _ _ _ _ 7
.                 | |             |_ _ _ _ _ _ _|
.             12 _| |                           |
.               |_ _|  _ _ _ _ _ _              |_ _
.         12 _ _|     |  _ _ _ _ _|_ _ _ _ _ 5      |_
.      _ _ _| |    9 _| |         |_ _ _ _ _|         |
.     |  _ _ _|  9 _|_ _|                   |_ _ 3    |_ _ _ 7
.     | |      _ _| |      _ _ _ _          |_  |         | |
.     | |     |  _ _| 12 _|  _ _ _|_ _ _ 3    |_|_ _ 5    | |
.     | |     | |      _|   |     |_ _ _|         | |     | |
.     | |     | |     |  _ _|           |_ _ 3    | |     | |
.     | |     | |     | |    3 _ _        | |     | |     | |
.     | |     | |     | |     |  _|_ 1    | |     | |     | |
.    _|_|    _|_|    _|_|    _|_| |_|    _|_|    _|_|    _|_|    _
.   | |     | |     | |     | |         | |     | |     | |     | |
.   | |     | |     | |     |_|_ _     _| |     | |     | |     | |
.   | |     | |     | |    2  |_ _|_ _|  _|     | |     | |     | |
.   | |     | |     |_|_     2    |_ _ _|7   _ _| |     | |     | |
.   | |     | |    4    |_                 _|  _ _|     | |     | |
.   | |     |_|_ _        |_ _ _ _        |  _|    _ _ _| |     | |
.   | |    6      |_      |_ _ _ _|_ _ _ _| | 15 _|    _ _|     | |
.   |_|_ _ _        |_   4        |_ _ _ _ _|  _|     |    _ _ _| |
.  8      | |_ _      |                       |      _|   |  _ _ _|
.         |_    |     |_ _ _ _ _ _            |  _ _|28  _| |
.           |_  |_    |_ _ _ _ _ _|_ _ _ _ _ _| |      _|  _|
.          8  |_ _|  6            |_ _ _ _ _ _ _|  _ _|  _|
.                 |                               |  _ _|  31
.                 |_ _ _ _ _ _ _ _                | |
.                 |_ _ _ _ _ _ _ _|_ _ _ _ _ _ _ _| |
.                8                |_ _ _ _ _ _ _ _ _|
.
.
[For two other drawings of the spiral see the links. - _N. J. A. Sloane_, Nov 16 2020]
If the sequence does not contain negative terms then its terms can be represented in a quadrant. For the construction of the diagram we use the symmetric Dyck paths of A237593 as shown below:
---------------------------------------------------------------
Triangle         Diagram of the symmetry of sigma (n = 1..24)
---------------------------------------------------------------
.              _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
1;            |_| | | | | | | | | | | | | | | | | | | | | | | |
3;            |_ _|_| | | | | | | | | | | | | | | | | | | | | |
2, 2;         |_ _|  _|_| | | | | | | | | | | | | | | | | | | |
7;            |_ _ _|    _|_| | | | | | | | | | | | | | | | | |
3, 3;         |_ _ _|  _|  _ _|_| | | | | | | | | | | | | | | |
12;           |_ _ _ _|  _| |  _ _|_| | | | | | | | | | | | | |
4, 4;         |_ _ _ _| |_ _|_|    _ _|_| | | | | | | | | | | |
15;           |_ _ _ _ _|  _|     |  _ _ _|_| | | | | | | | | |
5, 3, 5;      |_ _ _ _ _| |      _|_| |  _ _ _|_| | | | | | | |
9, 9;         |_ _ _ _ _ _|  _ _|    _| |    _ _ _|_| | | | | |
6, 6;         |_ _ _ _ _ _| |  _|  _|  _|   |  _ _ _ _|_| | | |
28;           |_ _ _ _ _ _ _| |_ _|  _|  _ _| | |  _ _ _ _|_| |
7, 7;         |_ _ _ _ _ _ _| |  _ _|  _|    _| | |    _ _ _ _|
12, 12;       |_ _ _ _ _ _ _ _| |     |     |  _|_|   |* * * *
8, 8, 8;      |_ _ _ _ _ _ _ _| |  _ _|  _ _|_|       |* * * *
31;           |_ _ _ _ _ _ _ _ _| |  _ _|  _|      _ _|* * * *
9, 9;         |_ _ _ _ _ _ _ _ _| | |_ _ _|      _|* * * * * *
39;           |_ _ _ _ _ _ _ _ _ _| |  _ _|    _|* * * * * * *
10, 10;       |_ _ _ _ _ _ _ _ _ _| | |       |* * * * * * * *
42;           |_ _ _ _ _ _ _ _ _ _ _| |  _ _ _|* * * * * * * *
11, 5, 5, 11; |_ _ _ _ _ _ _ _ _ _ _| | |* * * * * * * * * * *
18, 18;       |_ _ _ _ _ _ _ _ _ _ _ _| |* * * * * * * * * * *
12, 12;       |_ _ _ _ _ _ _ _ _ _ _ _| |* * * * * * * * * * *
60;           |_ _ _ _ _ _ _ _ _ _ _ _ _|* * * * * * * * * * *
...
The total number of cells in the first n set of symmetric regions of the diagram equals A024916(n), the sum of all divisors of all positive integers <= n, hence the total number of cells in the n-th set of symmetric regions of the diagram equals sigma(n) = A000203(n).
For n = 9 the 9th row of A237593 is [5, 2, 2, 2, 2, 5] and the 8th row of A237593 is [5, 2, 1, 1, 2, 5] therefore between both symmetric Dyck paths there are three regions (or parts) of sizes [5, 3, 5], so row 9 is [5, 3, 5].
The sum of divisors of 9 is 1 + 3 + 9 = A000203(9) = 13. On the other hand the sum of the parts of the symmetric representation of sigma(9) is 5 + 3 + 5 = 13, equaling the sum of divisors of 9.
For n = 24 the 24th row of A237593 is [13, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 13] and the 23rd row of A237593 is [12, 5, 2, 2, 1, 1, 1, 1, 2, 2, 5, 12] therefore between both symmetric Dyck paths there are only one region (or part) of size 60, so row 24 is 60.
The sum of divisors of 24 is 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = A000203(24) = 60. On the other hand the sum of the parts of the symmetric representation of sigma(24) is 60, equaling the sum of divisors of 24.
Note that the number of *'s in the diagram is 24^2 - A024916(24) = 576 - 491 = A004125(24) = 85.
From _Omar E. Pol_, Nov 22 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 triangle:  [              8,            8,            8              ]
The 15th row
of A296508:   [              8,      7,    1,    0,      8              ]
The 15th row
of A280851    [              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 A239387 and also A296508, A280851. (End)

Robert Price, Table of n, a(n) for n = 1..15542 (rows n = 1..5000, flattened)
Hartmut F. W. Hoft, Sample visual documentation for Mathematica code
Michel Marcus, A colored version of the symmetric representation of sigma(n), multipage, n = 1..85
Omar E. Pol, An infinite stepped pyramid (A237593, A237270, A262626)
Omar E. Pol, Perspective view of the stepped pyramid (16 levels)
Omar E. Pol, Perspective view of the stepped pyramid into four quadrants (11 levels). This is formed by combing four copies of the pyramid back-to-back (cf. A244050).
N. J. A. Sloane, Another drawing of the spiral
N. J. A. Sloane, Spiral showing only the outer boundary
Index entries for sequences related to sigma(n)

Cf. A000203, A004125, A023196, A024916, A153485, A196020, A221529, A231347, A235791, A235796, A236104, A236112, A236540, A237046, A237048, A237271, A237590, A237591, A237593, A239050, A239660, A239663, A239665, A239931, A239932, A239933, A239934, A240020, A240062, A244050, A245092, A249351, A262626, A280850, A280851, A296508, A335616, A340035, A384149.

T[n_,k_] := Ceiling[(n + 1)/k - (k + 1)/2] (* from A235791 *)
path[n_] := Module[{c = Floor[(Sqrt[8n + 1] - 1)/2], h, r, d, rd, k, p = {{0, n}}}, h = Map[T[n, #] - T[n, # + 1] &, Range[c]]; r = Join[h, Reverse[h]]; d = Flatten[Table[{{1, 0}, {0, -1}}, {c}], 1];
rd = Transpose[{r, d}]; For[k = 1, k <= 2c, k++, p = Join[p, Map[Last[p] + rd[[k, 2]] * # &, Range[rd[[k, 1]]]]]]; p]
segments[n_] := SplitBy[Map[Min, Drop[Drop[path[n], 1], -1] - path[n - 1]], # == 0 &]
a237270[n_] := Select[Map[Apply[Plus, #] &, segments[n]], # != 0 &]
Flatten[Map[a237270, Range[40]]] (* data *)
(* Hartmut F. W. Hoft, Jun 23 2014 *)

T(n, k) = (A384149(n, k) + A384149(n, m+1-k))/2, where m = A237271(n) is the row length. (conjectured) - Peter Munn, Jun 01 2025

Drawing of the spiral extended by Omar E. Pol, Nov 22 2020

A235791 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k copies of every positive integer in nondecreasing order, and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

1, 2, 3, 1, 4, 1, 5, 2, 6, 2, 1, 7, 3, 1, 8, 3, 1, 9, 4, 2, 10, 4, 2, 1, 11, 5, 2, 1, 12, 5, 3, 1, 13, 6, 3, 1, 14, 6, 3, 2, 15, 7, 4, 2, 1, 16, 7, 4, 2, 1, 17, 8, 4, 2, 1, 18, 8, 5, 3, 1, 19, 9, 5, 3, 1, 20, 9, 5, 3, 2, 21, 10, 6, 3, 2, 1, 22, 10, 6, 4, 2, 1, 23, 11, 6, 4, 2, 1, 24, 11, 7, 4, 2, 1

Offset: 1

Omar E. Pol, Jan 23 2014

The alternating sum of the squares of the elements of the n-th row equals the sum of all divisors of all positive integers <= n, i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*(T(n,k))^2 = A024916(n).
Row n has length A003056(n) hence the first element of column k is in row A000217(k).
For more information see A236104.
The sum of row n gives A060831(n), the sum of the number of odd divisors of all positive integers <= n. - Omar E. Pol, Mar 01 2014. [An equivalent assertion is that the sum of row n of A237048 is the number of odd divisors of n, and this was proved by Hartmut F. W. Hoft in a comment in A237048. - N. J. A. Sloane, Dec 07 2020]
Comments from Franklin T. Adams-Watters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014: (Start)
The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.
You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.
Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n).
Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)
From Hartmut F. W. Hoft, Apr 07 2014: (Start)
Mathematica function has been written to check the first property up to n = 20000.
T(n,(sqrt(8n+1)-1)/2+1) = 0 for all n >= 1, which is useful for formulas for A237591 and A237593. (End)
Alternating row sums give A240542. - Omar E. Pol, Apr 16 2014
Conjecture: T(n,k) is also the total number of partitions of all positive integers <= n into exactly k consecutive parts, i.e., the partial column sum of A285898, or in accordance with the triangles of the same family: the partial column sum of A237048. - Omar E. Pol, Apr 28 2017, Nov 24 2020
The above conjecture is true. The proof will be added soon (it uses the generating function for the columns). - N. J. A. Sloane, Nov 24 2020
T(n,k) is also the total length of all line segments between the k-th vertex and the central vertex of the largest Dyck path of the symmetric representation of sigma(n). In other words: T(n,k) is the sum of the last (A003056(n)-k+1) terms of the n-th row of A237591. - Omar E. Pol, Sep 07 2021
T(n,k) is also the Manhattan distance between the k-th vertex and the central vertex of the Dyck path described in the n-th row of the triangle A237593. - Omar E. Pol, Jan 11 2023

			Triangle begins:
   1;
   2;
   3,  1;
   4,  1;
   5,  2;
   6,  2,  1;
   7,  3,  1;
   8,  3,  1;
   9,  4,  2;
  10,  4,  2,  1;
  11,  5,  2,  1;
  12,  5,  3,  1;
  13,  6,  3,  1;
  14,  6,  3,  2;
  15,  7,  4,  2,  1;
  16,  7,  4,  2,  1;
  17,  8,  4,  2,  1;
  18,  8,  5,  3,  1;
  19,  9,  5,  3,  1;
  20,  9,  5,  3,  2;
  21, 10,  6,  3,  2,  1;
  22, 10,  6,  4,  2,  1;
  23, 11,  6,  4,  2,  1;
  24, 11,  7,  4,  2,  1;
  25, 12,  7,  4,  3,  1;
  26, 12,  7,  5,  3,  1;
  27, 13,  8,  5,  3,  2;
  28, 13,  8,  5,  3,  2,  1;
  ...
For n = 10 the 10th row of triangle is 10, 4, 2, 1, so we have that 10^2 - 4^2 + 2^2 - 1^2 = 100 - 16 + 4 - 1 = 87, the same as A024916(10) = 87, the sum of all divisors of all positive integers <= 10.
From _Omar E. Pol_, Nov 19 2015: (Start)
Illustration of initial terms in the third quadrant:
.                                                            y
Row                                                         _|
1                                                         _|1|
2                                                       _|2 _|
3                                                     _|3  |1|
4                                                   _|4   _|1|
5                                                 _|5    |2 _|
6                                               _|6     _|2|1|
7                                             _|7      |3  |1|
8                                           _|8       _|3 _|1|
9                                         _|9        |4  |2 _|
10                                      _|10        _|4  |2|1|
11                                    _|11         |5   _|2|1|
12                                  _|12          _|5  |3  |1|
13                                _|13           |6    |3 _|1|
14                              _|14            _|6   _|3|2 _|
15                            _|15             |7    |4  |2|1|
16                          _|16              _|7    |4  |2|1|
17                        _|17               |8     _|4 _|2|1|
18                      _|18                _|8    |5  |3  |1|
19                    _|19                 |9      |5  |3 _|1|
20                  _|20                  _|9     _|5  |3|2 _|
21                _|21                   |10     |6   _|3|2|1|
22              _|22                    _|10     |6  |4  |2|1|
23            _|23                     |11      _|6  |4  |2|1|
24          _|24                      _|11     |7    |4 _|2|1|
25        _|25                       |12       |7   _|4|3  |1|
26      _|26                        _|12      _|7  |5  |3 _|1|
27    _|27                         |13       |8    |5  |3|2 _|
28   |28                           |13       |8    |5  |3|2|1|
...
T(n,k) is also the number of cells between the k-th vertical line segment (from left to right) and the y-axis in the n-th row of the structure.
Note that the number of horizontal line segments in the n-th row of the structure equals A001227(n), the number of odd divisors of n.
Also the diagram represents the left part of the front view of the pyramid described in A245092. (End)
For more information about the diagram see A286001. - _Omar E. Pol_, Dec 19 2020
From _Omar E. Pol_, Sep 08 2021: (Start)
For n = 12 the symmetric representation of sigma(12) in the fourth quadrant is as shown below:
                            _
                           | |
                           | |
                           | |
                           | |
                           | |
                      _ _ _| |
                    _|    _ _|
                  _|     |
                 |      _|
                 |  _ _|
      _ _ _ _ _ _| |3   1
     |_ _ _ _ _ _ _|
    12              5
.
For n = 12 and k = 1 the total length of all line segments between the first vertex and the central vertex of the largest Dyck path is equal to 12, so T(12,1) = 12.
For n = 12 and k = 2 the total length of all line segments between the second vertex and the central vertex of the largest Dyck path is equal to 5, so T(12,2) = 5.
For n = 12 and k = 3 the total length of all line segments between the third vertex and the central vertex of the largest Dyck path is equal to 3, so T(12,3) = 3.
For n = 12 and k = 4 the total length of all line segments between the fourth vertex and the central vertex of the largest Dyck path is equal to 1, so T(12,4) = 1.
Hence the 12th row of triangle is [12, 5, 3, 1]. (End)

G. C. Greubel, Table of n, a(n) for the first 150 rows, flattened

Columns 1..3: A000027, A008619, A008620.
Operations on rows: A003056 (number of terms), A237591 (differences between terms), A060831 (sums), A339577 (products), A240542 (alternating sums), A236104 (squares), A339576 (sums of squares), A024916 (alternating sums of squares), A237048 (differences between rows), A042974 (right border).
Cf. A000203, A000217, A001227, A196020, A211343, A228813, A231345, A231347, A235794, A236106, A236112, A237270, A237271, A237593, A239660, A245092, A261699, A262626, A286000, A286001, A280850, A280851, A296508, A335616.

row[n_] := Floor[(Sqrt[8*n + 1] - 1)/2]; f[n_, k_] := Ceiling[(n + 1)/k - (k + 1)/2]; Table[f[n, k], {n, 1, 150}, {k, 1, row[n]}] // Flatten (* Hartmut F. W. Hoft, Apr 07 2014 *)

row(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i); \\ Michel Marcus, Mar 27 2014

from sympy import sqrt
import math
def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2))
for n in range(1, 21): print([T(n, k) for k in range(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)]) # Indranil Ghosh, Apr 25 2017

T(n,k) = ceiling((n+1)/k - (k+1)/2) for 1 <= n, 1 <= k <= floor((sqrt(8n+1)-1)/2) = A003056(n). - Hartmut F. W. Hoft, Apr 07 2014
G.f. for column k (k >= 1): x^(k*(k+1)/2)/( (1-x)*(1-x^k) ). - N. J. A. Sloane, Nov 24 2020
T(n,k) = Sum_{j=1..n} A237048(j,k). - Omar E. Pol, May 18 2017
T(n,k) = sqrt(A236104(n,k)). - Omar E. Pol, Feb 14 2018
Sigma(n) = Sum_{k=1..A003056(n)} (-1)^(k-1) * (T(n,k)^2 - T(n-1,k)^2), assuming that T(k*(k+1)/2-1,k) = 0. - Omar E. Pol, Oct 10 2018
a(s(n,k)) = T(n,k), n >= 1, 1 <= k <= r = floor((sqrt(8*n + 1) - 1)/2), where s(n,k) = r*n - r*(r+1)*(r+2)/6 + k translates position (row n, column k) in the triangle of this sequence to its position in the sequence. - Hartmut F. W. Hoft, Feb 24 2021

A196020 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the odd numbers interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

1, 3, 5, 1, 7, 0, 9, 3, 11, 0, 1, 13, 5, 0, 15, 0, 0, 17, 7, 3, 19, 0, 0, 1, 21, 9, 0, 0, 23, 0, 5, 0, 25, 11, 0, 0, 27, 0, 0, 3, 29, 13, 7, 0, 1, 31, 0, 0, 0, 0, 33, 15, 0, 0, 0, 35, 0, 9, 5, 0, 37, 17, 0, 0, 0, 39, 0, 0, 0, 3, 41, 19, 11, 0, 0, 1, 43, 0, 0, 7, 0, 0, 45, 21, 0, 0, 0, 0, 47, 0, 13, 0, 0, 0

Offset: 1

Omar E. Pol, Feb 02 2013

Gives an identity for sigma(n): alternating sum of row n equals the sum of divisors of n. For proof see Max Alekseyev link.
Row n has length A003056(n) hence column k starts in row A000217(k).
The number of positive terms in row n is A001227(n), the number of odd divisors of n.
If n = 2^j then the only positive integer in row n is T(n,1) = 2^(j+1) - 1.
If n is an odd prime then the only two positive integers in row n are T(n,1) = 2n - 1 and T(n,2) = n - 2.
If T(n,k) = 3 then T(n+1,k+1) = 1, the first element of the column k+1.
The partial sums of column k give the column k of A236104.
The connection with the symmetric representation of sigma is as follows: A236104 --> A235791 --> A237591 --> A237593 --> A239660 --> A237270.
Alternating sum of row n equals the number of units cubes that protrude from the n-th level of the stepped pyramid described in A245092. - Omar E. Pol, Oct 28 2015
Conjecture: T(n,k) is the difference between the square of the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the square of the total number of partitions of all positive integers < n into exactly k consecutive parts. - Omar E. Pol, Feb 14 2018
From Omar E. Pol, Nov 24 2020: (Start)
T(n,k) is also the number of steps in the first n levels of the k-th double-staircase that has at least one step in the n-th level of the "Double- staircases" diagram, otherwise T(n,k) = 0, (see the Example section).
For the connection with A280851 see also the algorithm of A280850 and the conjecture of A296508. (End)
The number of zeros in the n-th row equals A238005(n). - Omar E. Pol, Sep 11 2021
Apart from the alternating row sums and the sum of divisors function A000203 another connection with Euler's pentagonal theorem is that in the irregular triangle of A238442 the k-th column starts in the row that is the k-th generalized pentagonal number A001318(k) while here the k-th column starts in the row that is the k-th generalized hexagonal number A000217(k). Both A001318 and A000217 are successive members of the same family: the generalized polygonal numbers. - Omar E. Pol, Sep 23 2021
Other triangle with the same row lengths and alternating row sums equals sigma(n) is A252117. - Omar E. Pol, May 03 2022

			Triangle begins:
   1;
   3;
   5,  1;
   7,  0;
   9,  3;
  11,  0,  1;
  13,  5,  0;
  15,  0,  0;
  17,  7,  3;
  19,  0,  0,  1;
  21,  9,  0,  0;
  23,  0,  5,  0;
  25, 11,  0,  0;
  27,  0,  0,  3;
  29, 13,  7,  0,  1;
  31,  0,  0,  0,  0;
  33, 15,  0,  0,  0;
  35,  0,  9,  5,  0;
  37, 17,  0,  0,  0;
  39,  0,  0,  0,  3;
  41, 19, 11,  0,  0,  1;
  43,  0,  0,  7,  0,  0;
  45, 21,  0,  0,  0,  0;
  47,  0, 13,  0,  0,  0;
  49, 23,  0,  0,  5,  0;
  51,  0,  0,  9,  0,  0;
  53, 25, 15,  0,  0,  3;
  55,  0,  0,  0,  0,  0,  1;
  ...
For n = 15 the divisors of 15 are 1, 3, 5, 15, so the sum of divisors of 15 is 1 + 3 + 5 + 15 = 24. On the other hand, the 15th row of the triangle is 29, 13, 7, 0, 1, so the alternating row sum is 29 - 13 + 7 - 0 + 1 = 24, equaling the sum of divisors of 15.
If n is even then the alternating sum of the n-th row is simpler to evaluate than the sum of divisors of n. For example the sum of divisors of 24 is 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60, and the alternating sum of the 24th row of triangle is 47 - 0 + 13 - 0 + 0 - 0 = 60.
From _Omar E. Pol_, Nov 24 2020: (Start)
For an illustration of the rows of triangle consider the infinite "double-staircases" diagram defined in A335616 (see also the theorem there).
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|
.
The first largest double-staircase has 29 horizontal steps, the second double-staircase has 13 steps, the third double-staircase has 7 steps, and the fifth double-staircases has only one step. Note that the fourth double-staircase does not count because it does not have horizontal steps in the 15th level, so the 15th row of triangle is [29, 13, 7, 0, 1].
For a connection with the "Ziggurat" diagram and the parts and subparts of the symmetric representation of sigma(15) see also A237270. (End)

G. C. Greubel, Table of n, a(n) for the first 200 rows, flattened
Max Alekseyev, Proof of the alternating sum property of A196020, SeqFan Mailing List, Nov 17 2013.
Paul D. Hanna, About an identity for sigma, SeqFan Mailing List, Nov 18 2013.
Index entries for sequences related to sigma(n)

Columns 1-2: A005408, A193356.
Compare with A027750, A238442, A237270, A237273, A252117, A280851, A296508.
Cf. A000203, A000217, A001227, A001318, A003056, A211343, A212119, A228813, A231345, A231347, A235791, A235794, A236104, A236106, A236112, A237048, A237271, A237591, A237593, A238005, A239660, A244050, A245092, A261699, A262626, A286000, A286001, A280850, A335616, A338721.

T_row := proc(n) local T;
T := (n, k) -> if modp(n-k/2, k) = 0 and n >= k*(k+1)/2 then 2*n/k-k else 0 fi;
seq(T(n,k), k=1..floor((sqrt(8*n+1)-1)/2)) end:
seq(print(T_row(n)),n=1..24); # Peter Luschny, Oct 27 2015

T[n_, k_] := If[Mod[n - k*(k+1)/2, k] == 0 ,2*n/k - k, 0]
row[n_] := Floor[(Sqrt[8n+1]-1)/2]
line[n_] := Map[T[n, #]&, Range[row[n]]]
a196020[m_, n_] := Map[line, Range[m, n]]
Flatten[a196020[1,22]] (* data *)
(* Hartmut F. W. Hoft, Oct 26 2015 *)
A196020row = Function[n,Table[If[Divisible[Numerator[n-k/2],k] && CoprimeQ[ Denominator[n- k/2], k],2*n/k-k,0],{k,1,Floor[(Sqrt[8 n+1]-1)/2]}]]
Flatten[Table[A196020row[n], {n,1,24}]] (* Peter Luschny, Oct 28 2015 *)

def T(n,k):
    q = (2*n-k)/2
    b = k.divides(q.numerator()) and gcd(k,q.denominator()) == 1
    return 2*n/k - k if b else 0
for n in (1..24): [T(n, k) for k in (1..floor((sqrt(8*n+1)-1)/2))] # Peter Luschny, Oct 28 2015

A000203(n) = Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k).
T(n,k) = 2*A211343(n,k) - 1, if A211343(n,k) >= 1 otherwise T(n,k) = 0.
If n==k/2 (mod k) and n>=k(k+1)/2, then T(n,k) = 2*n/k - k; otherwise T(n,k) = 0. - Max Alekseyev, Nov 18 2013
T(n,k) = A236104(n,k) - A236104(n-1,k), assuming that A236104(k*(k+1)/2-1,k) = 0. - Omar E. Pol, Oct 14 2018
T(n,k) = A237048(n,k)*A338721(n,k). - Omar E. Pol, Feb 22 2022

A249351 Triangle read by rows in which row n lists the widths of the symmetric representation of sigma(n).

Original entry on oeis.org

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

Offset: 1

Omar E. Pol, Oct 26 2014

Here T(n,k) is defined to be the "k-th width" of the symmetric representation of sigma(n), with n>=1 and 1<=k<=2n-1. Explanation: consider the diagram of the symmetric representation of sigma(n) described in A236104, A237593 and other related sequences. Imagine that the diagram for sigma(n) contains 2n-1 equidistant segments which are parallel to the main diagonal [(0,0),(n,n)] of the quadrant. The segments are located on the diagonal of the cells. The distance between two parallel segment is equal to sqrt(2)/2. T(n,k) is the length of the k-th segment divided by sqrt(2). Note that the triangle contains nonnegative terms because for some n the value of some widths is equal to zero. For an illustration of some widths see Hartmut F. W. Hoft's contribution in the Links section of A237270.
Row n has length 2*n-1.
Row sums give A000203.
If n is a power of 2 then all terms of row n are 1's.
If n is an even perfect number then all terms of row n are 1's except the middle term which is 2.
If n is an odd prime then row n lists (n+1)/2 1's, n-2 zeros, (n+1)/2 1's.
The number of blocks of positive terms in row n gives A237271(n).
The sum of the k-th block of positive terms in row n gives A237270(n,k).
It appears that the middle diagonal is also A067742 (which was conjectured by Michel Marcus in the entry A237593 and checked with two Mathematica functions up to n = 100000 by Hartmut F. W. Hoft).
It appears that the trapezoidal numbers (A165513) are also the numbers k > 1 with the property that some of the noncentral widths of the symmetric representation of sigma(k) are not equal to 1. - Omar E. Pol, Mar 04 2023

			Triangle begins:
  1;
  1,1,1;
  1,1,0,1,1;
  1,1,1,1,1,1,1;
  1,1,1,0,0,0,1,1,1;
  1,1,1,1,1,2,1,1,1,1,1;
  1,1,1,1,0,0,0,0,0,1,1,1,1;
  1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
  1,1,1,1,1,0,0,1,1,1,0,0,1,1,1,1,1;
  1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1;
  1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1;
  1,1,1,1,1,1,1,1,1,2,2,2,2,2,1,1,1,1,1,1,1,1,1;
  ...
---------------------------------------------------------------------------
.        Written as an isosceles triangle              Diagram of
.              the sequence begins:               the symmetry of sigma
---------------------------------------------------------------------------
.                                                _ _ _ _ _ _ _ _ _ _ _ _
.                      1;                       |_| | | | | | | | | | | |
.                    1,1,1;                     |_ _|_| | | | | | | | | |
.                  1,1,0,1,1;                   |_ _|  _|_| | | | | | | |
.                1,1,1,1,1,1,1;                 |_ _ _|    _|_| | | | | |
.              1,1,1,0,0,0,1,1,1;               |_ _ _|  _|  _ _|_| | | |
.            1,1,1,1,1,2,1,1,1,1,1;             |_ _ _ _|  _| |  _ _|_| |
.          1,1,1,1,0,0,0,0,0,1,1,1,1;           |_ _ _ _| |_ _|_|    _ _|
.        1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;         |_ _ _ _ _|  _|     |
.      1,1,1,1,1,0,0,1,1,1,0,0,1,1,1,1,1;       |_ _ _ _ _| |      _|
.    1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1;     |_ _ _ _ _ _|  _ _|
.  1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1;   |_ _ _ _ _ _| |
.1,1,1,1,1,1,1,1,1,2,2,2,2,2,1,1,1,1,1,1,1,1,1; |_ _ _ _ _ _ _|
...
From _Omar E. Pol_, Nov 22 2020: (Start)
Also consider the infinite double-staircases diagram defined in A335616.
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 this seq:  [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 A280851    [              8,      7,    1,            8              ]
.
The number of horizontal steps (or 1's) in the successive columns of the above diagram gives the 15th row of this triangle.
For more information about the parts of the symmetric representation of sigma(n) see A237270. For more information about the subparts see A239387, A296508, A280851.
More generally, it appears there is the same correspondence between the original diagram of the symmetric representation of sigma(n) and the "Ziggurat" diagram of n. (End)

Index entries for sequences related to sigma(n)

Cf. A000203, A003056, A067742, A071562, A165513, A196020, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A238443, A239660, A239932-A239934, A240542, A241008, A241010, A245092, A245685, A246955, A246956, A247687, A249223, A250068, A250070, A250071, A262626, A280850, A280851, A296508, A235616, A347186.

(* function segments are defined in A237270 *)
a249351[n_] := Flatten[Map[segments, Range[n]]]
a249351[10] (* Hartmut F. W. Hoft, Jul 20 2022 *)

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.

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

cp's OEIS Frontend

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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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).

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A237593 Triangle read by rows in which row n lists the elements of the n-th row of A237591 followed by the same elements in reverse order.

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A237270 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(n).

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A235791 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k copies of every positive integer in nondecreasing order, and the first element of column k is in row k(k+1)/2.

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A196020 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the odd numbers interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

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A249351 Triangle read by rows in which row n lists the widths of the symmetric representation of sigma(n).

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