A347186 -id:A347186 - cp's OEIS frontend

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).

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.

A365434 Irregular triangle read by rows: T(n,k) is the number of cubes in the k-th level of the ziggurat of order n described in A347186, n >= 1, k >= 1.

Original entry on oeis.org

1, 3, 1, 4, 2, 7, 5, 3, 1, 6, 4, 2, 12, 9, 7, 5, 3, 1, 8, 6, 4, 2, 15, 13, 11, 9, 7, 5, 3, 1, 13, 9, 7, 5, 2, 18, 16, 14, 12, 10, 8, 6, 4, 2, 12, 10, 8, 6, 4, 2, 28, 24, 22, 20, 16, 14, 12, 9, 7, 5, 3, 1, 14, 12, 10, 8, 6, 4, 2, 24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 24, 19, 17, 15, 11, 9, 7, 3, 1, 1

Offset: 1

Omar E. Pol, Oct 19 2023

The values of n when the number of terms in row n is equal to n give A174973.
The values of n when the number of terms in row n is not equal to n give A238524.
If and only if n is a power of 2 then row n lists the first n odd numbers in decreasing order.
If and only if n is an odd prime then row n lists the first (n + 1)/2 positive even numbers in decreasing order.
If and only if n is an even perfect number then row n lists 2*n together with the first n - 1 odd numbers in decreasing order.

			Triangle begins:
   1;
   3,  1;
   4,  2;
   7,  5,  3,  1;
   6,  4,  2;
  12,  9,  7,  5,  3,  1;
   8,  6,  4,  2;
  15, 13, 11,  9,  7,  5,  3,  1;
  13,  9,  7,  5,  2;
  18, 16, 14, 12, 10,  8,  6,  4,  2;
  12, 10,  8,  6,  4,  2;
  28, 24, 22, 20, 16, 14, 12,  9,  7,  5,  3, 1;
  14, 12, 10,  8,  6,  4,  2;
  24, 22, 20, 18, 16, 14, 12, 10,  8,  6,  4, 2;
  24, 19, 17, 15, 11,  9,  7,  3,  1,  1;
  31, 29, 27, 25, 23, 21, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1;
  ...

Column 1 gives A000203.
Row sums give A347186.
Row lengths give A365433.
Cf. A000079, A000396, A065091, A174973, A196020, A235791, A236104, A237270, A237271, A237591, A237593, A238524, A245092, A262626, A347263, A347367, A365195.

A365195 Height of the first staircase of the ziggurat of order n described in A347186.

Original entry on oeis.org

1, 2, 2, 4, 3, 6, 4, 8, 5, 9, 6, 12, 7, 12, 8, 16, 9, 18, 10, 20, 11, 18, 12, 24, 13, 21, 14, 28, 15, 30, 16, 32, 17, 27, 18, 36, 19, 30, 20, 40, 21, 42, 22, 42, 23, 36, 24, 48, 25, 39, 26, 49, 27, 54, 28, 56, 29, 45, 30, 60, 31, 48, 32, 64, 33, 66, 34, 63, 35, 54, 36, 72, 37, 57, 38, 70, 39, 77, 40, 80

Offset: 1

Omar E. Pol, Aug 25 2023

nonn

a(n) is also the number of square cells in the first layer of the symmetric representation of sigma(n) from the border to, at most, the axis of symmetry of the diagram.
The values of a(n) where a(n) = n give A174973.
Since this is a supersequence of A174973 so all powers of 2 and all even perfect numbers are in the sequence.
From Omar E. Pol, Oct 22 2023: (Start)
The values of a(n) where a(n) is not equal to n give A238524.
If n is an odd prime then a(n) = (n + 1)/2.
Shares infinitely many terms with A365433 from which first differs at a(15). (End)

Cf. A000079, A000396, A065091, A174973, A196020, A235791, A236104, A237270, A237271, A237591, A237593, A238524, A262626, A279387, A347186, A347367, A365433, A365434.

a(n) <= n.

a(2*n-1) = n.

More terms from Omar E. Pol, Oct 22 2023

A365433 Height of the ziggurat of order n described in A347186.

Original entry on oeis.org

1, 2, 2, 4, 3, 6, 4, 8, 5, 9, 6, 12, 7, 12, 10, 16, 9, 18, 10, 20, 13, 18, 12, 24, 13, 21, 16, 28, 15, 30, 16, 32, 19, 27, 21, 36, 19, 30, 22, 40, 21, 42, 22, 42, 40, 36, 24, 48, 25, 39, 28, 49, 27, 54, 36, 56, 31, 45, 30, 60, 31, 48, 36, 64, 41, 66, 34, 63, 37, 56, 36, 72, 37, 57, 61, 70, 50, 77, 40, 80

Offset: 1

Omar E. Pol, Oct 18 2023

nonn

The values of a(n) where a(n) = n give A174973.
The values of a(n) where a(n) is not equal to n give A238524.
If n is an odd prime then a(n) = (n + 1)/2.
Since this is a supersequence of A174973 so all powers of 2 and all even perfect numbers are in the sequence.
Shares infinitely many terms with A365195 from which first differs at a(15).

Row lengths of A365434.

Cf. A000079, A000396, A065091, A174973, A196020, A235791, A236104, A237270, A237271, A237591, A237593, A238524, A245092, A262626, A347186, A347263, A347367, A365195.

a(2*n-1) >= n.

A356351 Partial sums of the ziggurat sequence A347186.

Original entry on oeis.org

1, 5, 11, 27, 39, 76, 96, 160, 196, 286, 328, 489, 545, 701, 808, 1064, 1154, 1488, 1598, 2006, 2208, 2550, 2706, 3403, 3610, 4072, 4384, 5169, 5409, 6385, 6657, 7681, 8127, 8883, 9324, 10910, 11290, 12220, 12824, 14560, 15022, 16863, 17369, 19175, 20276, 21608, 22208, 25129, 25849, 27669

Offset: 1

Omar E. Pol, Oct 15 2022

nonn

a(n) is the volume (or the number of cubes) in a polycube whose base is the symmetric representation of A024916(n) which is formed with the first n 3D-Ziggurats described in A347186.

a(n) is also the total number of cubes in a three-dimensional spiral formed with the first n 3D-Ziggurats described in A347186 (see example). The base of the 3D-spiral is the spiral formed with the symmetric representation of sigma of the first n positive integers as shown in the example section of A239660.

			For n = 16 the figure shows the top view of a three-dimensional spiral formed with the first 16 3D-Ziggurats described in A347186. There are four 3D-Ziggurats in every quadrant:
.
                  _ _ _ _ _ _ _ _
                 |_|_|_|_|_|_|_|_|_ _ _ _ _ _ _
                 |_|             |_|_|_|_|_|_|_|
                _|_|                           |
               |_|_|  _ _ _ _ _ _              |_ _
            _ _|     |_|_|_|_|_|_|_ _ _ _ _        |_
      _ _ _|_|      _|_|         |_|_|_|_|_|         |
     |_|_|_|_|    _|_|_|                   |_ _      |_ _ _
     |_|      _ _|_|      _ _ _ _          |_|_|         |_|
     |_|     |_|_|_|    _|_|_|_|_|_ _ _      |_|_ _      |_|
     |_|     |_|      _|_|_|     |_|_|_|         |_|     |_|
     |_|     |_|     |_|_|_|           |_ _      |_|     |_|
     |_|     |_|     |_|      _ _        |_|     |_|     |_|
     |_|     |_|     |_|     |_|_|_      |_|     |_|     |_|
    _|_|    _|_|    _|_|    _|_| |_|    _|_|    _|_|    _|_|    _
   |_|     |_|     |_|     |_|         |_|     |_|     |_|     |_|
   |_|     |_|     |_|     |_|_ _     _|_|     |_|     |_|     |_|
   |_|     |_|     |_|       |_|_|_ _|_|_|     |_|     |_|     |_|
   |_|     |_|     |_|_          |_|_|_|    _ _|_|     |_|     |_|
   |_|     |_|         |_                 _|_|_|_|     |_|     |_|
   |_|     |_|_ _        |_ _ _ _        |_|_|    _ _ _|_|     |_|
   |_|           |_      |_|_|_|_|_ _ _ _|_|    _|_|_|_|_|     |_|
   |_|_ _ _        |_            |_|_|_|_|_|  _|_|_|_|    _ _ _|_|
         |_|_ _      |                       |_|_|_|_|   |_|_|_|_|
         |_|_|_|     |_ _ _ _ _ _            |_|_|_|    _|_|
           |_|_|_    |_|_|_|_|_|_|_ _ _ _ _ _|_|      _|_|_|
             |_|_|               |_|_|_|_|_|_|_|  _ _|_|_|
                 |                               |_|_|_|
                 |_ _ _ _ _ _ _ _                |_|
                 |_|_|_|_|_|_|_|_|_ _ _ _ _ _ _ _|_|
                                 |_|_|_|_|_|_|_|_|_|
.
The number of square cells in the top view of the n-th 3D-Ziggurat equals A000203(n).
The total number of square cells in the top view of the 3D-Spiral with the first n 3D-Ziggurats equals A024916(n).
In the above figure the total number of square cells equals A024916(16) = 220.
a(16) = 1064 is the total number of cubes in the 3D-Spiral with the first 16 3D-Ziggurats.

Cf. A000203, A024916, A196020, A235791, A236104, A237270, A237591, A237593, A239660, A239931, A239932, A239933, A239934, A347186, A296508, A299778, A347186, A347263, A347367, A347529, A351819.

A237271 Number of parts in the symmetric representation of sigma(n).

Original entry on oeis.org

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

Offset: 1

Omar E. Pol, Feb 25 2014

nonn

The diagram of the symmetry of sigma has been via A196020 --> A236104 --> A235791 --> A237591 --> A237593.
For more information see A237270.
a(n) is also the number of terraces at n-th level (starting from the top) of the stepped pyramid described in A245092. - Omar E. Pol, Apr 20 2016
a(n) is also the number of subparts in the first layer of the symmetric representation of sigma(n). For the definion of "subpart" see A279387. - Omar E. Pol, Dec 08 2016
Note that the number of subparts in the symmetric representation of sigma(n) equals A001227(n), the number of odd divisors of n. (See the second example). - Omar E. Pol, Dec 20 2016
From Hartmut F. W. Hoft, Dec 26 2016: (Start)
Using odd prime number 3, observe that the 1's in the 3^k-th row of the irregular triangle of A237048 are at index positions
     3^0 < 2*3^0 < 3^1 < 2*3^1 < ... < 2*3^((k-1)/2) < 3^(k/2) < ...
  the last being 2*3^((k-1)/2) when k is odd and 3^(k/2) when k is even. Since odd and even index positions alternate, each pair (3^i, 2*3^i) specifies one part in the symmetric representation with a center part present when k is even. A straightforward count establishes that the symmetric representation of 3^k, k>=0, has k+1 parts. Since this argument is valid for any odd prime, every positive integer occurs infinitely many times in the sequence. (End)
a(n) = number of runs of consecutive nonzero terms in row n of A262045. - N. J. A. Sloane, Jan 18 2021
Indices of odd terms give A071562. Indices of even terms give A071561. - Omar E. Pol, Feb 01 2021
a(n) is also the number of prisms in the three-dimensional version of the symmetric representation of k*sigma(n) where k is the height of the prisms, with k >= 1. - Omar E. Pol, Jul 01 2021
With a(1) = 0; a(n) is also the number of parts in the symmetric representation of A001065(n), the sum of aliquot parts of n. - Omar E. Pol, Aug 04 2021
The parity of this sequence is also the characteristic function of numbers that have middle divisors. - Omar E. Pol, Sep 30 2021
a(n) is also the number of polycubes in the 3D-version of the ziggurat of order n described in A347186. - Omar E. Pol, Jun 11 2024
Conjecture 1: a(n) is the number of odd divisors of n except the "e" odd divisors described in A005279. Thus a(n) is the length of the n-th row of A379288. - Omar E. Pol, Dec 21 2024
The conjecture 1 was checked up n = 10000 by Amiram Eldar. - Omar E. Pol, Dec 22 2024
The conjecture 1 is true. For a proof see A379288. - Hartmut F. W. Hoft, Jan 21 2025
From Omar E. Pol, Jul 31 2025: (Start)
Conjecture 2: a(n) is the number of 2-dense sublists of divisors of n.
We call "2-dense sublists of divisors of n" to the maximal sublists of divisors of n whose terms increase by a factor of at most 2.
In a 2-dense sublist of divisors of n the terms are in increasing order and two adjacent terms are the same two adjacent terms in the list of divisors of n.
Example: for n = 10 the list of divisors of 10 is [1, 2, 5, 10]. There are two 2-dense sublists of divisors of 10, they are [1, 2], [5, 10], so a(10) = 2.
The conjecture 2 is essentially the same as the second conjecture in the Comments of A384149. See also Peter Munn's formula in A237270.
The indices where a(n) = 1 give A174973 (2-dense numbers). See the proof there. (End)
Conjecture 3: a(n) is the number of divisors p of n such that p is greater than twice the adjacent previous divisor of n. The divisors p give the n-th row of A379288. - Omar E. Pol, Aug 02 2025

			Illustration of initial terms (n = 1..12):
---------------------------------------------------------
n   A000203  A237270    a(n)            Diagram
---------------------------------------------------------
.                               _ _ _ _ _ _ _ _ _ _ _ _
1       1      1         1     |_| | | | | | | | | | | |
2       3      3         1     |_ _|_| | | | | | | | | |
3       4      2+2       2     |_ _|  _|_| | | | | | | |
4       7      7         1     |_ _ _|    _|_| | | | | |
5       6      3+3       2     |_ _ _|  _|  _ _|_| | | |
6      12      12        1     |_ _ _ _|  _| |  _ _|_| |
7       8      4+4       2     |_ _ _ _| |_ _|_|    _ _|
8      15      15        1     |_ _ _ _ _|  _|     |
9      13      5+3+5     3     |_ _ _ _ _| |      _|
10     18      9+9       2     |_ _ _ _ _ _|  _ _|
11     12      6+6       2     |_ _ _ _ _ _| |
12     28      28        1     |_ _ _ _ _ _ _|
...
For n = 9 the sum of divisors of 9 is 1+3+9 = A000203(9) = 13. On the other hand the 9th set of symmetric regions of the diagram is formed by three regions (or parts) with 5, 3 and 5 cells, so the total number of cells is 5+3+5 = 13, equaling the sum of divisors of 9. There are three parts: [5, 3, 5], so a(9) = 3.
From _Omar E. Pol_, Dec 21 2016: (Start)
Illustration of the diagram of subparts (n = 1..12):
---------------------------------------------------------
n   A000203  A279391  A001227           Diagram
---------------------------------------------------------
.                               _ _ _ _ _ _ _ _ _ _ _ _
1       1      1         1     |_| | | | | | | | | | | |
2       3      3         1     |_ _|_| | | | | | | | | |
3       4      2+2       2     |_ _|  _|_| | | | | | | |
4       7      7         1     |_ _ _|  _ _|_| | | | | |
5       6      3+3       2     |_ _ _| |_|  _ _|_| | | |
6      12      11+1      2     |_ _ _ _|  _| |  _ _|_| |
7       8      4+4       2     |_ _ _ _| |_ _|_|  _ _ _|
8      15      15        1     |_ _ _ _ _|  _|  _| |
9      13      5+3+5     3     |_ _ _ _ _| |  _|  _|
10     18      9+9       2     |_ _ _ _ _ _| |_ _|
11     12      6+6       2     |_ _ _ _ _ _| |
12     28      23+5      2     |_ _ _ _ _ _ _|
...
For n = 6 the symmetric representation of sigma(6) has two subparts: [11, 1], so A000203(6) = 12 and A001227(6) = 2.
For n = 12 the symmetric representation of sigma(12) has two subparts: [23, 5], so A000203(12) = 28 and A001227(12) = 2. (End)
From _Hartmut F. W. Hoft_, Dec 26 2016: (Start)
Two examples of the general argument in the Comments section:
Rows 27 in A237048 and A249223 (4 parts)
i:  1  2 3 4 5 6 7 8 9 . . 12
27: 1  1 1 0 0 1                           1's in A237048 for odd divisors
    1 27 3     9                           odd divisors represented
27: 1  0 1 1 1 0 0 1 1 1 0 1               blocks forming parts in A249223
Rows 81 in A237048 and A249223 (5 parts)
i:  1  2 3 4 5 6 7 8 9 . . 12. . . 16. . . 20. . . 24
81: 1  1 1 0 0 1 0 0 1 0 0 0                          1's in A237048 f.o.d
    1 81 3    27     9                                odd div. represented
81: 1  0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 1  blocks fp in A249223
(End)

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..5000 from Michel Marcus)

Row lengths of A237270 and of A379288.
Column 1 of A279387.
Partial sums give A237590.
Parity gives A347950.
Cf. A000203, A000265, A001065, A001227, A005279, A024916, A060831, A061345, A067742, A071561, A071562, A175254, A196020, A221529, A235791, A236104, A237048, A237591, A237593, A239657, A244050, A244971, A245092, A249223, A250068, A261699, A262045, A262612, A262626, A274824, A279387, A279693, A319073, A340583, A340846, A342344, A347186, A379288.
Cf. A027750, A174973 (2-dense numbers), A239663, A240062, A243982, A379379, A380580, A384149, A384222, A384225, A384226, A384230, A384930.

a237271[n_] := Length[a237270[n]] (* code defined in A237270 *)
Map[a237271, Range[90]] (* data *)
(* Hartmut F. W. Hoft, Jun 23 2014 *)
a[n_] := Module[{d = Partition[Divisors[n], 2, 1]}, 1 + Count[d, ?(OddQ[#[[2]]] && #[[2]] >= 2*#[[1]] &)]]; Array[a, 100] (* _Amiram Eldar,  Dec 22 2024 *)

fill(vcells, hga, hgb) = {ic = 1; for (i=1, #hgb, if (hga[i] < hgb[i], for (j=hga[i], hgb[i]-1, cell = vector(4); cell[1] = i - 1; cell[2] = j; vcells[ic] = cell; ic ++;););); vcells;}
findfree(vcells) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[3] == 0) && (vcelli[4] == 0), return (i));); return (0);}
findxy(vcells, x, y) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[1]==x) && (vcelli[2]==y) && (vcelli[3] == 0) && (vcelli[4] == 0), return (i));); return (0);}
findtodo(vcells, iz) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[3] == iz) && (vcelli[4] == 0), return (i)); ); return (0);}
zcount(vcells) = {nbz = 0; for (i=1, #vcells, nbz = max(nbz, vcells[i][3]);); nbz;}
docell(vcells, ic, iz) = {x = vcells[ic][1]; y = vcells[ic][2]; if (icdo = findxy(vcells, x-1, y), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x+1, y), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x, y-1), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x, y+1), vcells[icdo][3] = iz); vcells[ic][4] = 1; vcells;}
docells(vcells, ic, iz) = {vcells[ic][3] = iz; while (ic, vcells = docell(vcells, ic, iz); ic = findtodo(vcells, iz);); vcells;}
nbzb(n, hga, hgb) = {vcells = vector(sigma(n)); vcells = fill(vcells, hga, hgb); iz = 1; while (ic = findfree(vcells), vcells = docells(vcells, ic, iz); iz++;); zcount(vcells);}
lista(nn) = {hga = concat(heights(row237593(0), 0), 0); for (n=1, nn, hgb = heights(row237593(n), n); nbz = nbzb(n, hga, hgb); print1(nbz, ", "); hga = concat(hgb, 0););} \\ with heights() also defined in A237593; \\ Michel Marcus, Mar 28 2014

from sympy import divisors
def a(n: int) -> int:
    divs = list(divisors(n))
    d = [divs[i:i+2] for i in range(len(divs) - 1)]
    s = sum(1 for pair in d if len(pair) == 2 and pair[1] % 2 == 1 and pair[1] >= 2 * pair[0])
    return s + 1
print([a(n) for n in range(1, 80)])  # Peter Luschny, Aug 05 2025

a(n) = A001227(n) - A239657(n). - Omar E. Pol, Mar 23 2014
a(p^k) = k + 1, where p is an odd prime and k >= 0. - Hartmut F. W. Hoft, Dec 26 2016
Theorem: a(n) <= number of odd divisors of n (cf. A001227). The differences are in A239657. - N. J. A. Sloane, Jan 19 2021
a(n) = A340846(n) - A340833(n) + 1 (Euler's formula). - Omar E. Pol, Feb 01 2021
a(n) = A000005(n) - A243982(n). - Omar E. Pol, Aug 02 2025

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 *)

A239657 Number of odd divisors m of n such that there is a divisor d of n with d < m < 2*d.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 3, 0, 0, 0, 0, 1, 2, 0, 0, 0, 1, 0, 3, 0, 0, 3, 0, 0, 1, 0, 0, 0, 0, 0, 3, 0, 1, 0, 0, 0, 3, 0, 0, 1, 0, 0, 3, 0, 0, 0, 1, 0, 2, 0, 0, 2, 0, 1, 2, 0, 1, 0, 0, 0, 3, 0, 0, 0, 1, 0, 5, 1, 0, 0, 0, 0, 1, 0, 0, 1, 2, 0, 2, 0, 1, 4, 0, 0, 3, 0, 1, 0, 1, 0, 2, 0, 0, 1, 0, 0, 3, 0, 0, 0, 0, 0, 5, 0, 0

Offset: 1

Omar E. Pol, Mar 23 2014

nonn

The original name was: Number of odd divisors of n minus the number of parts in the symmetric representation of sigma(n).
Observation: at least the indices of the first 42 positive elements coincide with A005279: 6, 12, 15, 18, 20, 24..., checked (by hand) up to n = 2^7.
The observation is true for the indices of all positive elements. Hence the indices of the zeros give A174905. - Omar E. Pol, Jan 06 2017
a(n) is the number of subparts minus the number of parts in the symmetric representation of sigma(n). For the definition of "subpart" see A279387. - Omar E. Pol, Sep 26 2018
a(n) is the number of subparts of the symmetric representation of sigma(n) that are not in the first layer. - Omar E. Pol, Jan 26 2025

			Illustration of the symmetric representation of sigma(15) = 24 in the third quadrant:
.      _
.     | |
.     | |
.     | |
.     | |
.     | |
.     | |
.     | |
.     |_|_ _ _
.    8      | |_ _
.           |_    |
.             |_  |_
.            8  |_ _|
.                   |
.                   |_ _ _ _ _ _ _ _
.                   |_ _ _ _ _ _ _ _|
.                 8
.
For n = 15 the divisors of 15 are 1, 3, 5, 15, so the number of odd divisors of 15 is equal to 4. On the other hand the parts of the symmetric representation of sigma(15) are [8, 8, 8], there are three parts, so a(15) = 4 - 3 = 1.
From _Omar E. Pol_, Sep 26 2018: (Start)
Also the number of odd divisors of 15 equals the number of partitions of 15 into consecutive parts and equals the number of subparts in the symmetric representation of sigma(15). Then we have that the number of subparts minus the number of parts is  4 - 3 = 1, so a(15) = 1.
.      _
.     | |
.     | |
.     | |
.     | |
.     | |
.     | |
.     | |
.     |_|_ _ _
.    8      | |_ _
.           |_ _  |
.          7  |_| |_
.            1  |_ _|
.                   |
.                   |_ _ _ _ _ _ _ _
.                   |_ _ _ _ _ _ _ _|
.                 8
.
The above diagram shows the symmetric representation of sigma(15) with its four subparts: [8, 7, 1, 8]. (End)
From _Omar E. Pol_, Mar 30 2025: (Start)
The above diagram also shows that in the first layer there are three parts (having sizes [8, 7, 8]). Also there is another part that is not in the first layer, so a(15) = 1.
On the other hand for n = 15 there is only one odd divisor m of 15 such that  d < m < 2*d and d divides 15. That odd divisor is 5 as shown below, so a(15) = 1.
   d  <  m  <  2*d
--------------------
   1            2
   3     5      6
   5           10
  15           30
.
For n = 18 there are two odd divisors m of 18 such that  d < m < 2*d and d divides 18. Those odd divisors are 3 and 9 as shown below, so a(18) = 2.
   d  <  m  <  2*d
--------------------
   1            2
   2     3      4
   3            6
   6     9     12
   9           18
  18           36
.
(End)

Antti Karttunen, Table of n, a(n) for n = 1..5000 (computed from the b-file of A237271 provided by Michel Marcus)

Cf. A000203, A001227, A005279, A174905, A196020, A236104, A235791, A237048, A237270, A237271, A237591, A237593, A245092, A262626, A279387, A279391, A262626, A286000, A286001, A296508, A347186, A383147.

a[n_] := Module[{d = Partition[Divisors[n], 2, 1]}, Count[d, ?(OddQ[#[[2]]] && #[[2]] < 2*#[[1]] &)]]; Array[a, 100] (* _Amiram Eldar, Apr 01 2025 *)

a(n) = A001227(n) - A237271(n).

New Name from Omar E. Pol, Jan 26 2025

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.

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.

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A365434 Irregular triangle read by rows: T(n,k) is the number of cubes in the k-th level of the ziggurat of order n described in A347186, n >= 1, k >= 1.

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A365195 Height of the first staircase of the ziggurat of order n described in A347186.

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A365433 Height of the ziggurat of order n described in A347186.

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A356351 Partial sums of the ziggurat sequence A347186.

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A237271 Number of parts in the symmetric representation of sigma(n).

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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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A239657 Number of odd divisors m of n such that there is a divisor d of n with d < m < 2*d.

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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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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).