A346875 -id:A346875 - cp's OEIS frontend

A262626 Visible parts of the perspective view of the stepped pyramid whose structure essentially arises after the 90-degree-zig-zag folding of the isosceles triangle A237593.

1, 1, 1, 3, 2, 2, 2, 2, 2, 1, 1, 2, 7, 3, 1, 1, 3, 3, 3, 3, 2, 2, 3, 12, 4, 1, 1, 1, 1, 4, 4, 4, 4, 2, 1, 1, 2, 4, 15, 5, 2, 1, 1, 2, 5, 5, 3, 5, 5, 2, 2, 2, 2, 5, 9, 9, 6, 2, 1, 1, 1, 1, 2, 6, 6, 6, 6, 3, 1, 1, 1, 1, 3, 6, 28, 7, 2, 2, 1, 1, 2, 2, 7, 7, 7, 7, 3, 2, 1, 1, 2, 3, 7, 12, 12, 8, 3, 1, 2, 2, 1, 3, 8, 8, 8, 8, 8, 3, 2, 1, 1

Offset: 1

Omar E. Pol, Sep 26 2015

Also the rows of both triangles A237270 and A237593 interleaved.
Also, irregular triangle read by rows in which T(n,k) is the area of the k-th region (from left to right in ascending diagonal) of the n-th symmetric set of regions (from the top to the bottom in descending diagonal) in the two-dimensional diagram of the perspective view of the infinite stepped pyramid described in A245092 (see the diagram in the Links section).
The diagram of the symmetric representation of sigma is also the top view of the pyramid, see Links section. For more information about the diagram see also A237593 and A237270.
The number of cubes at the n-th level is also A024916(n), the sum of all divisors of all positive integers <= n.
Note that this pyramid is also a quarter of the pyramid described in A244050. Both pyramids have infinitely many levels.
Odd-indexed rows are also the rows of the irregular triangle A237270.
Even-indexed rows are also the rows of the triangle A237593.
Lengths of the odd-indexed rows are in A237271.
Lengths of the even-indexed rows give 2*A003056.
Row sums of the odd-indexed rows gives A000203, the sum of divisors function.
Row sums of the even-indexed rows give the positive even numbers (see A005843).
Row sums give A245092.
From the front view of the stepped pyramid emerges a geometric pattern which is related to A001227, the number of odd divisors of the positive integers.
The connection with the odd divisors of the positive integers is as follows: A261697 --> A261699 --> A237048 --> A235791 --> A237591 --> A237593 --> A237270 --> this sequence.

			Irregular triangle begins:
  1;
  1, 1;
  3;
  2, 2;
  2, 2;
  2, 1, 1, 2;
  7;
  3, 1, 1, 3;
  3, 3;
  3, 2, 2, 3;
  12;
  4, 1, 1, 1, 1, 4;
  4, 4;
  4, 2, 1, 1, 2, 4;
  15;
  5, 2, 1, 1, 2, 5;
  5, 3, 5;
  5, 2, 2, 2, 2, 5;
  9, 9;
  6, 2, 1, 1, 1, 1, 2, 6;
  6, 6;
  6, 3, 1, 1, 1, 1, 3, 6;
  28;
  7, 2, 2, 1, 1, 2, 2, 7;
  7, 7;
  7, 3, 2, 1, 1, 2, 3, 7;
  12, 12;
  8, 3, 1, 2, 2, 1, 3, 8;
  8, 8, 8;
  8, 3, 2, 1, 1, 1, 1, 2, 3, 8;
  31;
  9, 3, 2, 1, 1, 1, 1, 2, 3, 9;
  ...
Illustration of the odd-indexed rows of triangle as the diagram of the symmetric representation of sigma which is also the top view of the stepped pyramid:
.
   n  A000203    A237270    _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
   1     1   =      1      |_| | | | | | | | | | | | | | | |
   2     3   =      3      |_ _|_| | | | | | | | | | | | | |
   3     4   =    2 + 2    |_ _|  _|_| | | | | | | | | | | |
   4     7   =      7      |_ _ _|    _|_| | | | | | | | | |
   5     6   =    3 + 3    |_ _ _|  _|  _ _|_| | | | | | | |
   6    12   =     12      |_ _ _ _|  _| |  _ _|_| | | | | |
   7     8   =    4 + 4    |_ _ _ _| |_ _|_|    _ _|_| | | |
   8    15   =     15      |_ _ _ _ _|  _|     |  _ _ _|_| |
   9    13   =  5 + 3 + 5  |_ _ _ _ _| |      _|_| |  _ _ _|
  10    18   =    9 + 9    |_ _ _ _ _ _|  _ _|    _| |
  11    12   =    6 + 6    |_ _ _ _ _ _| |  _|  _|  _|
  12    28   =     28      |_ _ _ _ _ _ _| |_ _|  _|
  13    14   =    7 + 7    |_ _ _ _ _ _ _| |  _ _|
  14    24   =   12 + 12   |_ _ _ _ _ _ _ _| |
  15    24   =  8 + 8 + 8  |_ _ _ _ _ _ _ _| |
  16    31   =     31      |_ _ _ _ _ _ _ _ _|
  ...
The above diagram arises from a simpler diagram as shown below.
Illustration of the even-indexed rows of triangle as the diagram of the deployed front view of the corner of the stepped pyramid:
.
.                                 A237593
Level                               _ _
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|
...
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).
The diagram represents the first 16 levels of the pyramid.
The diagram of the isosceles triangle and the diagram of the top view of the pyramid shows the connection between the partitions into consecutive parts and the sum of divisors function (see also A286000 and A286001). - _Omar E. Pol_, Aug 28 2018
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

Robert Price, Table of n, a(n) for n = 1..16048 (n=1..412 rows)
Omar E. Pol, An infinite stepped pyramid (A237593, A237270, A262626)
Omar E. Pol, Diagram of the isosceles triangle A237593 before the 90-degree-zig-zag folding (rows: 1..28)
Omar E. Pol, Perspective view of the stepped pyramid (levels: 1..16)
Index entries for sequences related to sigma(n)

Famous sequences that are visible in the stepped pyramid:
Cf. A000040 (prime numbers)......., for the characteristic shape see A346871.
Cf. A000079 (powers of 2)........., for the characteristic shape see A346872.
Cf. A000203 (sum of divisors)....., total area of the terraces in the n-th level.
Cf. A000217 (triangular numbers).., for the characteristic shape see A346873.
Cf. A000225 (Mersenne numbers)...., for a visualization see A346874.
Cf. A000384 (hexagonal numbers)..., for the characteristic shape see A346875.
Cf. A000396 (perfect numbers)....., for the characteristic shape see A346876.
Cf. A000668 (Mersenne primes)....., for a visualization see A346876.
Cf. A001097 (twin primes)........., for a visualization see A346871.
Cf. A001227 (# of odd divisors)..., number of subparts in the n-th level.
Cf. A002378 (oblong numbers)......, for a visualization see A346873.
Cf. A008586 (multiples of 4)......, perimeters of the successive levels.
Cf. A008588 (multiples of 6)......, for the characteristic shape see A224613.
Cf. A013661 (zeta(2))............., (area of the horizontal faces)/(n^2), n -> oo.
Cf. A014105 (second hexagonals)..., for the characteristic shape see A346864.
Cf. A067742 (# of middle divisors), # cells in the main diagonal in n-th level.
Other sequences that are visible in the stepped pyramid: A000096, A001065, A001359, A001747, A002939, A002943, A003056, A004125, A004277, A004526, A005279, A006512, A007606, A007607, A082647, A008438, A008578, A008864, A010814, A014106, A014107, A014132, A014574, A016945, A019434, A024206, A024916, A028552, A028982, A028983, A034856, A038550, A047836, A048050, A052928, A054735, A054844, A062731, A065091, A065475, A071561, A071562, A071904, A092506, A100484, A108605, A139256, A139257, A144396, A152677, A152678, A153485, A155085, A161680, A161983, A162917, A174905, A174973, A175254, A176810, A224880, A235791, A237270, A237271, A237591, A237593, A238005, A238524, A244049, A245092, A259176, A259177, A261348, A278972, A317302, A317303, A317304, A317305, A317307, A319529, A319796, A319801, A319802, A327329, A336305, (and several others).
Apart from zeta(2) other constants that are related to the stepped pyramid are A072691, A353908, A354238.
Cf. A054844, A131507, A196020, A236104, A237048, A239660, A244050, A259179, A261350, A261697, A261699, A262612, A280850, A286000, A286001, A296508.

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

A347529 Irregular triangle read by rows: T(n,k) is the sum of the subparts of the symmetric representation of sigma(n) that arise from the (2*k-1)-th double-staircase of the double-staircases diagram of n described in A335616, n >= 1, k >= 1, and the first element of column k is in row A000384(k).

Original entry on oeis.org

1, 3, 4, 7, 6, 11, 1, 8, 0, 15, 0, 10, 3, 18, 0, 12, 0, 23, 5, 14, 0, 24, 0, 16, 7, 1, 31, 0, 0, 18, 0, 0, 35, 4, 0, 20, 0, 0, 39, 0, 3, 22, 10, 0, 36, 0, 0, 24, 0, 0, 47, 13, 0, 26, 0, 5, 42, 0, 0, 28, 12, 0, 55, 0, 0, 1, 30, 0, 0, 0, 59, 6, 7, 0, 32, 0, 0, 0, 63, 0, 0, 0, 34, 14, 0, 0

Offset: 1

Omar E. Pol, Sep 05 2021

Conjecture 1: the number of nonzero terms in row n equals A082647(n).
Conjecture 2: column k lists positive integers interleaved with 2*k+2 zeros.
The k-th column of the triangle is related to the (2*k+1)-gonal numbers. For further information about this connection see A347186 and A347263.
If n is prime then the only nonzero term in row n is T(n,1) = 1 + n.
If n is a power of 2 then the only nonzero term in row n is T(n,1) = 2*n - 1.
If n is an even perfect number then there are two nonzero terms in row n, they are T(n,1) = 2*n - 1 and the last term in the row is 1.
If n is a hexagonal number then the last term in row n is 1.
Row n contains a subpart 1 if and only if n is a hexagonal number.
First differs from A279388 at a(10), or row 9 of triangle.

			Triangle begins:
---------------------------
   n / k   1   2   3   4
---------------------------
   1 |     1;
   2 |     3;
   3 |     4;
   4 |     7;
   5 |     6;
   6 |    11,  1;
   7 |     8,  0;
   8 |    15,  0;
   9 |    10,  3;
  10 |    18,  0;
  11 |    12,  0;
  12 |    23,  5;
  13 |    14,  0;
  14 |    24,  0;
  15 |    16,  7,  1;
  16 |    31,  0,  0;
  17 |    18,  0,  0;
  18 |    35,  4,  0;
  19 |    20,  0,  0;
  20 |    39,  0,  3;
  21 |    22, 10,  0;
  22 |    36,  0,  0;
  23 |    24,  0,  0;
  24 |    47, 13,  0;
  25 |    26,  0,  5;
  26 |    42,  0,  0;
  27 |    28, 12,  0;
  28 |    55,  0,  0,  1;
...
For n = 15 the calculation of the 15th row of the triangle (in accordance with the geometric algorithm described in A347186) is as follows:
Stage 1 (Construction):
We draw the diagram called "double-staircases" with 15 levels described in A335616.
Then we label the five double-staircases (j = 1..5) as shown below:
                               _
                             _| |_
                           _|  _  |_
                         _|   | |   |_
                       _|    _| |_    |_
                     _|     |  _  |     |_
                   _|      _| | | |_      |_
                 _|       |   | |   |       |_
               _|        _|  _| |_  |_        |_
             _|         |   |  _  |   |         |_
           _|          _|   | | | |   |_          |_
         _|           |    _| | | |_    |           |_
       _|            _|   |   | |   |   |_            |_
     _|             |     |  _| |_  |     |             |_
   _|              _|    _| |  _  | |_    |_              |_
  |_ _ _ _ _ _ _ _|_ _ _|_ _|_|_|_|_ _|_ _ _|_ _ _ _ _ _ _ _|
  1               2     3   4 5
.
Stage 2 (Debugging):
We remove the fourth double-staircase as it does not have at least one step at level 1 of the diagram starting from the base, as shown below:
                               _
                             _| |_
                           _|  _  |_
                         _|   | |   |_
                       _|    _| |_    |_
                     _|     |  _  |     |_
                   _|      _| | | |_      |_
                 _|       |   | |   |       |_
               _|        _|  _| |_  |_        |_
             _|         |   |     |   |         |_
           _|          _|   |     |   |_          |_
         _|           |    _|     |_    |           |_
       _|            _|   |         |   |_            |_
     _|             |     |         |     |             |_
   _|              _|    _|    _    |_    |_              |_
  |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _|
  1               2     3     5
.
Stage 3 (Annihilation):
We delete the second double-staircase and the steps of the first double-staircase that are just above the second double-staircase.
The new diagram has two double-staircases and two simple-staircases as shown below:
                               _
                              | |
                 _            | |            _
               _| |          _| |_          | |_
             _|   |         |     |         |   |_
           _|     |         |     |         |     |_
         _|       |        _|     |_        |       |_
       _|         |       |         |       |         |_
     _|           |       |         |       |           |_
   _|             |      _|    _    |_      |             |_
  |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _|
  1                     3     5
.
The diagram is called "ziggurat of 15".
The number of steps in the staircase labeled 1 is 8. There is a pair of these staircases, so T(15,1) = 2*8 = 16, since the symmetric representation of sigma(15) is also the base of the three dimensional version of the ziggurat .
The number of steps in the double-staircase labeled 3 is equal to 7, so T(15,2) = 7.
The number of steps in the double-staircase labeled 5 is equal to 1, so T(15,3) = 1.
Therefore the 15th row of triangle is [16, 7, 1].
The top view of the 3D-Ziggurat of 15 and the symmetric representation of sigma(15) with subparts look like this:
                                _                                     _
                               |_|                                   | |
                               |_|                                   | |
                               |_|                                   | |
                               |_|                                   | |
                               |_|                                   | |
                               |_|                                   | |
                               |_|                                   | |
                          _ _ _|_|                              _ _ _|_|
                      _ _|_|      36                        _ _| |      8
                     |_|_|_|                               |  _ _|
                    _|_|_|                                _| |_|
                   |_|_|  1                              |_ _|  1
                   |    34                               |    7
    _ _ _ _ _ _ _ _|                      _ _ _ _ _ _ _ _|
   |_|_|_|_|_|_|_|_|                     |_ _ _ _ _ _ _ _|
                    36                                    8
.
     Top view of the 3D-Ziggurat.        The symmetric representation of
     The ziggurat is formed by 3        of sigma(15) is formed by 3 parts.
   polycubes with 107 cubes             It has 4 subparts with 24 cells in
   in total. It has 4 staircases       total. It is the base of the ziggurat.
       with 24 steps in total.
.

Index entries for sequences related to sigma(n)

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

A347263 Irregular triangle read by rows: T(n,k) is the sum of the subparts of the ziggurat diagram of n (described in A347186) that arise from the (2*k-1)-th double-staircase of the double-staircases diagram of n (described in A335616), n >= 1, k >= 1, and the first element of column k is in row A000384(k).

Original entry on oeis.org

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

Offset: 1

Omar E. Pol, Sep 05 2021

Conjecture 1: the number of nonzero terms in row n equals A082647(n).
Conjecture 2: column k lists positive integers interleaved with 2*k+2 zeros.
The subparts of the ziggurat diagram are the polygons formed by the cells that are under the staircases.
The connection of the subparts of the ziggurat diagram with the polygonal numbers is as follows:
The area under a double-staircase labeled with the number j is equal to the m-th (j+2)-gonal number plus the (m-1)-th (j+2)-gonal number, where m is the number of steps on one side of the ladder from the base to the top.
The area under a simple-staircase labeled with the number j is equal to the m-th (j+2)-gonal number, where m is the number of steps.
So the k-th column of the triangle is related to the (2*k+1)-gonal numbers, for example:
For the calculation of column 1 we use triangular numbers A000217.
For the calculation of column 2 we use pentagonal numbers A000326.
For the calculation of column 3 we use heptagonal numbers A000566.
For the calculation of column 4 we use enneagonal numbers A001106.
And so on.
More generally, for the calculation of column k we use the (2*k+1)-gonal numbers.
For further information about the ziggurat diagram see A347186.

			Triangle begins:
   n / k   1    2    3    4
------------------------------
   1 |     1;
   2 |     4;
   3 |     6;
   4 |    16;
   5 |    12;
   6 |    36,   1;
   7 |    20,   0;
   8 |    64,   0;
   9 |    30,   6;
  10 |    90,   0;
  11 |    42,   0;
  12 |   144,  17;
  13 |    56,   0;
  14 |   156,   0;
  15 |    72,  34,   1;
  16 |   256,   0,   0;
  17 |    90,   0,   0;
  18 |   324,  10,   0;
  19 |   110,   0    0;
  20 |   400,   0,   8;
  21 |   132,  70,   0;
  22 |   342,   0,   0;
  23 |   156,   0,   0;
  24 |   576, 121,   0;
  25 |   182,   0,  25;
  26 |   462,   0,   0;
  27 |   210, 102,   0;
  28 |   784,   0,   0,   1;
...
For n = 15 the calculation of the 15th row of the triangle (in accordance with the geometric algorithm described in A347186) is as follows:
Stage 1 (Construction):
We draw the diagram called "double-staircases" with 15 levels described in A335616.
Then we label the five double-staircases (j = 1..5) as shown below:
                               _
                             _| |_
                           _|  _  |_
                         _|   | |   |_
                       _|    _| |_    |_
                     _|     |  _  |     |_
                   _|      _| | | |_      |_
                 _|       |   | |   |       |_
               _|        _|  _| |_  |_        |_
             _|         |   |  _  |   |         |_
           _|          _|   | | | |   |_          |_
         _|           |    _| | | |_    |           |_
       _|            _|   |   | |   |   |_            |_
     _|             |     |  _| |_  |     |             |_
   _|              _|    _| |  _  | |_    |_              |_
  |_ _ _ _ _ _ _ _|_ _ _|_ _|_|_|_|_ _|_ _ _|_ _ _ _ _ _ _ _|
  1               2     3   4 5
.
Stage 2 (Debugging):
We remove the fourth double-staircase as it does not have at least one step at level 1 of the diagram starting from the base, as shown below:
                               _
                             _| |_
                           _|  _  |_
                         _|   | |   |_
                       _|    _| |_    |_
                     _|     |  _  |     |_
                   _|      _| | | |_      |_
                 _|       |   | |   |       |_
               _|        _|  _| |_  |_        |_
             _|         |   |     |   |         |_
           _|          _|   |     |   |_          |_
         _|           |    _|     |_    |           |_
       _|            _|   |         |   |_            |_
     _|             |     |         |     |             |_
   _|              _|    _|    _    |_    |_              |_
  |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _|
  1               2     3     5
.
Stage 3 (Annihilation):
We delete the second double-staircase and the steps of the first double-staircase that are just above the second double-staircase.
The new diagram has two double-staircases and two simple-staircases as shown below:
                               _
                              | |
                 _            | |            _
               _| |          _| |_          | |_
             _|   |         |     |         |   |_
           _|     |         |     |         |     |_
         _|       |        _|     |_        |       |_
       _|         |       |         |       |         |_
     _|           |       |         |       |           |_
   _|             |      _|    _    |_      |             |_
  |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _|
  1                     3     5
.
The diagram is called "ziggurat of 15".
Now we calculate the area (or the number of cells) under the staircases with multiplicity using polygonal numbers as shown below:
The area under the staircase labeled 1 is equal to A000217(8) = 36. There is a pair of these staircases, so T(15,1) = 2*36 = 72.
The area under the double-staircase labeled 3 is equal to A000326(4) + A000326(3) = 22 + 12 = 34, so T(15,2) = 34.
The area under the double-staircase labeled 5 is equal to A000566(1) + A000566(0) = 1 + 0 = 1, so T(15,3) = 1.
Therefore the 15th row of the triangle is [72, 34, 1].

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

A346873 Triangle read by rows in which row n lists the row A000217(n) of A237591, n >= 1.

Original entry on oeis.org

1, 2, 1, 4, 1, 1, 6, 2, 1, 1, 8, 3, 2, 1, 1, 11, 4, 3, 1, 1, 1, 15, 5, 3, 2, 1, 1, 1, 19, 6, 4, 2, 2, 1, 1, 1, 23, 8, 5, 2, 2, 2, 1, 1, 1, 28, 10, 5, 3, 3, 2, 1, 1, 1, 1, 34, 11, 6, 4, 3, 2, 2, 1, 1, 1, 1, 40, 13, 7, 5, 3, 2, 2, 2, 1, 1, 1, 1, 46, 16, 8, 5, 4, 2, 3

Offset: 1

Omar E. Pol, Aug 06 2021

The characteristic shape of the symmetric representation of sigma(A000217(n)) consists in that in the main diagonal of the diagram the smallest Dyck path has a valley and the largest Dyck path has a peak, or vice versa, the smallest Dyck path has a peak and the largest Dyck path has valley.
So knowing this characteristic shape we can know if a number is a triangular number (or not) just by looking at the diagram, even ignoring the concept of triangular number.
Therefore we can see a geometric pattern of the distribution of the triangular numbers in the stepped pyramid described in A245092.
T(n,k) is also the length of the k-th line segment of the largest Dyck path of the symmetric representation of sigma(A000217(n)), from the border to the center, hence the sum of the n-th row of triangle is equal to A000217(n).
T(n,k) is also the difference between the total number of partitions of all positive integers <= n-th triangular number into exactly k consecutive parts, and the total number of partitions of all positive integers <= n-th triangular number into exactly k + 1 consecutive parts.

			Triangle begins:
   1;
   2,  1;
   4,  1, 1;
   6,  2, 1, 1;
   8,  3, 2, 1, 1;
  11,  4, 3, 1, 1, 1;
  15,  5, 3, 2, 1, 1, 1;
  19,  6, 4, 2, 2, 1, 1, 1;
  23,  8, 5, 2, 2, 2, 1, 1, 1;
  28, 10, 5, 3, 3, 2, 1, 1, 1, 1;
  34, 11, 6, 4, 3, 2, 2, 1, 1, 1, 1;
  40, 13, 7, 5, 3, 2, 2, 2, 1, 1, 1, 1;
  46, 16, 8, 5, 4, 2, 3, 1, 2, 1, 1, 1, 1;
...
Illustration of initial terms:
Column T gives the triangular numbers (A000217).
Column S gives A074285, the sum of the divisors of the triangular numbers which equals the area (and the number of cells) of the associated diagram.
-------------------------------------------------------------------------
  n    T    S   Diagram
-------------------------------------------------------------------------
                 _   _     _       _         _           _             _
  1    1    1   |_| | |   | |     | |       | |         | |           | |
               1 _ _|_|   | |     | |       | |         | |           | |
  2    3    4   |_ _|  _ _| |     | |       | |         | |           | |
                  2  1|    _|     | |       | |         | |           | |
                 _ _ _|  _|    _ _| |       | |         | |           | |
  3    6   12   |_ _ _ _| 1   |  _ _|       | |         | |           | |
                    4    1 _ _|_|           | |         | |           | |
                          |  _|1       _ _ _|_|         | |           | |
                 _ _ _ _ _| | 1    _ _| |               | |           | |
  4   10   18   |_ _ _ _ _ _|2    |    _|               | |           | |
                      6          _|  _|          _ _ _ _|_|           | |
                                |_ _|1 1        | |                   | |
                                | 2            _| |                   | |
                 _ _ _ _ _ _ _ _|4            |  _|          _ _ _ _ _| |
  3   15   24   |_ _ _ _ _ _ _ _|          _ _|_|           |  _ _ _ _ _|
                        8              _ _|  _|1            | |
                                      |_ _ _|1 1         _ _| |
                                      |  3           _ _|  _ _|
                                      |4            |    _|
                 _ _ _ _ _ _ _ _ _ _ _|            _|  _|
  4   21   32   |_ _ _ _ _ _ _ _ _ _ _|      _ _ _|  _|1 1
                          11                |  _ _ _|2
                                            | |  3
                                            | |
                                            | |5
                 _ _ _ _ _ _ _ _ _ _ _ _ _ _| |
  5   28   56   |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
                              15
.

Row sums give A000217, n >= 1.
Column 1 gives A039823.
For the characteristic shape of sigma(A000040(n)) see A346871.
For the characteristic shape of sigma(A000079(n)) see A346872.
For the visualization of Mersenne numbers A000225 see A346874.
For the characteristic shape of sigma(A000384(n)) see A346875.
For the characteristic shape of sigma(A000396(n)) see A346876.
For the characteristic shape of sigma(A008588(n)) see A224613.
Cf. A000203, A074285, A237591, A237593, A245092, A249351, A262626.

T(n,k) = A237591(A000217(n),k). - Omar E. Pol, Feb 06 2023

Name corrected by Omar E. Pol, Feb 06 2023

A346876 Irregular triangle read by rows in which row n is the "n-th even perfect number" row of A237591, n >= 1.

Original entry on oeis.org

4, 1, 1, 15, 5, 3, 2, 1, 1, 1, 249, 83, 42, 25, 17, 13, 9, 7, 6, 5, 5, 3, 4, 2, 3, 2, 2, 2, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4065, 1355, 678, 407, 271, 194, 146, 113, 91, 75, 62, 52, 45, 40, 34, 30, 27, 25, 22, 19, 19, 16, 15, 14, 13, 12, 12, 10, 10, 9, 9, 8, 8, 7

Offset: 1

Omar E. Pol, Aug 06 2021

The characteristic shape of the symmetric representation of sigma(A000396(n)) consists in that the diagram has only one region (or part) and that region has whidth 1 except in the main diagonal where the width is 2.
So knowing this characteristic shape we can know if a number is an even perfect number (or not) just by looking at the diagram, even ignoring the concept of even perfect number (see the examples).
Therefore we can see a geometric pattern of the distribution of the even perfect numbers in the stepped pyramid described in A245092.
For the definition of "width" see A249351.
T(n,k) is the length of the k-th line segment of the largest Dyck path of the symmetric representation of sigma(A000396(n)), from the border to the center, hence the sum of the n-th row of triangle is equal to A000396(n) assuming there are no odd perfect numbers.
T(n,k) is also the difference between the total number of partitions of all positive integers <= n-th even perfect number into exactly k consecutive parts, and the total number of partitions of all positive integers <= n-th perfect number into exactly k + 1 consecutive parts.

			Triangle begins:
    4, 1, 1;
   15, 5, 3, 2, 1, 1,1;
  249,83,42,25,17,13,9,7,6,5,5,3,4,2,3,2,2,2,2,2,1,2,1,2,1,1,1,1,1,1,1;
...
Illustration of initial terms:
Column P gives the even perfect numbers (A000396 assuming there are no odd perfect numbers).
Column S gives A139256, the sum of the divisors of the even perfect numbers equals the area (and the number of cells) of the associated diagram.
-------------------------------------------------------------------------
  n    P   S    Diagram:   1                                           2
-------------------------------------------------------------------------
                           _                                           _
                          | |                                         | |
                          | |                                         | |
                       _ _| |                                         | |
                      |    _|                                         | |
                 _ _ _|  _|                                           | |
  1    6   12   |_ _ _ _| 1                                           | |
                    4    1                                            | |
                                                                      | |
                                                                      | |
                                                                      | |
                                                                      | |
                                                                      | |
                                                                      | |
                                                             _ _ _ _ _| |
                                                            |  _ _ _ _ _|
                                                            | |
                                                         _ _| |
                                                     _ _|  _ _|
                                                    |    _|
                                                   _|  _|
                                                  |  _|1 1
                                             _ _ _| | 1
                                            |  _ _ _|2
                                            | |  3
                                            | |
                                            | |5
                 _ _ _ _ _ _ _ _ _ _ _ _ _ _| |
  2   28   56   |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
                              15
.
For n = 3, P = 496, the diagram is too large to include here. To draw that diagram note that the lengths of the line segments of the smallest Dyck path are [248, 83, 42, 25, 17, 13, 9, 7, 6, 5, 5, 3, 4, 2, 3, 2, 2, 2, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 2, 2, 3, 2, 4, 3, 5, 5, 6, 7, 9, 13, 17, 25, 42, 83, 248] and the lengths of the line segments of the largest Dyck path are [249, 83, 42, 25, 17, 13, 9, 7, 6, 5, 5, 3, 4, 2, 3, 2, 2, 2, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 2, 2, 3, 2, 4, 3, 5, 5, 6, 7, 9, 13, 17, 25, 42, 83, 249].

Michel Marcus, Table of n, a(n) for n = 1..8359 (rows 1..5).

Row sums give A000396.
Row lengths give A000668.
For the characteristic shape of sigma(A000040(n)) see A346871.
For the characteristic shape of sigma(A000079(n)) see A346872.
For the characteristic shape of sigma(A000217(n)) see A346873.
For the visualization of Mersenne numbers A000225 see A346874.
For the characteristic shape of sigma(A000384(n)) see A346875.
For the characteristic shape of sigma(A008588(n)) see A224613.
Cf. A000203, A139256, A237591, A237593, A245092, A249351, A262626.

row235791(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i);
row(n) = {my(orow = concat(row235791(n), 0)); vector(#orow -1, i, orow[i] - orow[i+1]); } \\ A237591
tabf(nn) = {for (n=1, nn, my(p=prime(n)); if (isprime(2^n-1), print(row(2^(n-1)*(2^n-1)));););}
tabf(7) \\ Michel Marcus, Aug 31 2021

More terms from Michel Marcus, Aug 31 2021

Name edited by Michel Marcus, Jun 16 2023

A346871 Irregular triangle read by rows in which row n lists the row A000040(n) of A237591, n >= 1.

Original entry on oeis.org

2, 2, 1, 3, 2, 4, 2, 1, 6, 3, 1, 1, 7, 3, 2, 1, 9, 4, 2, 1, 1, 10, 4, 2, 2, 1, 12, 5, 2, 2, 1, 1, 15, 6, 3, 2, 1, 1, 1, 16, 6, 3, 2, 2, 1, 1, 19, 7, 4, 2, 2, 1, 1, 1, 21, 8, 4, 2, 2, 2, 1, 1, 22, 8, 4, 3, 2, 1, 2, 1, 24, 9, 4, 3, 2, 2, 1, 1, 1, 27, 10, 5, 3, 2, 2, 1, 2, 1

Offset: 1

Omar E. Pol, Aug 06 2021

The characteristic shape of the symmetric representation of sigma(prime(n)) consists in that the diagram contains exactly two regions (or parts) and each region is a rectangle (or bar), except for the first prime number (the 2) whose symmetric representation of sigma(2) consists of only one region which contains three cells.
So knowing this characteristic shape we can know if a number is prime (or not) just by looking at the diagram, even ignoring the concept of prime number.
Therefore we can see a geometric pattern of the exact distribution of prime numbers in the stepped pyramid described in A245092.
T(n,k) is the length of the k-th line segment of the largest Dyck path of the symmetric representation of sigma(prime(n)), from the border to the center, hence the sum of the n-th row of triangle is equal to A000040(n).
T(n,k) is also the difference between the total number of partitions of all positive integers <= n-th prime into exactly k consecutive parts, and the total number of partitions of all positive integers <= n-th prime into exactly k + 1 consecutive parts.

			Triangle begins:
   2;
   2, 1;
   3, 2;
   4, 2, 1;
   6, 3, 1, 1;
   7, 3, 2, 1;
   9, 4, 2, 1, 1;
  10, 4, 2, 2, 1;
  12, 5, 2, 2, 1, 1;
  15, 6, 3, 2, 1, 1, 1;
  16, 6, 3, 2, 2, 1, 1;
  19, 7, 4, 2, 2, 1, 1, 1;
  21, 8, 4, 2, 2, 2, 1, 1;
  22, 8, 4, 3, 2, 1, 2, 1;
  24, 9, 4, 3, 2, 2, 1, 1, 1;
...
Illustration of initial terms:
Row 1:    _
        _| |
       |_ _|
         2                         Semilength = 2
.
Row 2:      _
           | |
        _ _|_|
       |_ _|1                      Semilength = 3
         2
.
Row 3:          _
               | |
               | |
              _|_|
        _ _ _|                     Semilength = 5
       |_ _ _|2
          3
.
Row 4:              _
                   | |
                   | |
                   | |
                  _|_|
                _|
        _ _ _ _| 1                 Semilength = 7
       |_ _ _ _|2
           4
.
Row 5:                         _
                              | |
                              | |
                              | |
                              | |
                              | |
                           _ _|_|
                         _|
                       _|1         Semilength = 11
                      |1
           _ _ _ _ _ _|
          |_ _ _ _ _ _|3
                6
.
The area (also the number of cells) of the successive diagrams gives A008864.

Row sums give A000040.
For the characteristic shape of sigma(A000079(n)) see A346872.
For the characteristic shape of sigma(A000217(n)) see A346873.
For the visualization of Mersenne numbers A000225 see A346874.
For the characteristic shape of sigma(A000384(n)) see A346875.
For the characteristic shape of sigma(A000396(n)) see A346876.
For the characteristic shape of sigma(A008588(n)) see A224613.
Cf. A000203, A008864, A018252, A237591, A237593, A245092, A249351, A262626.

A346872 Irregular triangle read by rows in which row n lists the row 2^(n-1) of A237591, n >= 1.

Original entry on oeis.org

1, 2, 3, 1, 5, 2, 1, 9, 3, 2, 1, 1, 17, 6, 3, 2, 2, 1, 1, 33, 11, 6, 4, 2, 2, 2, 1, 2, 1, 65, 22, 11, 7, 5, 3, 3, 2, 2, 2, 1, 2, 1, 1, 1, 129, 43, 22, 13, 9, 7, 5, 4, 3, 3, 3, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 257, 86, 43, 26, 18, 12, 10, 8, 6, 5, 4, 4, 3, 3, 3, 2, 3

Offset: 1

Omar E. Pol, Aug 06 2021

The characteristic shape of the symmetric representation of sigma(2^(n-1)) consists in that the diagram contains exactly one region (or part) and that region has width 1.
So knowing this characteristic shape we can know if a number is power of 2 or not just by looking at the diagram, even ignoring the concept of power of 2.
Therefore we can see a geometric pattern of the distribution of the powers of 2 in the stepped pyramid described in A245092.
For the definition of "width" see A249351.
T(n,k) is the length of the k-th line segment of the largest Dyck path of the symmetric representation of sigma(2^(n-1)), from the border to the center, hence the sum of the n-th row of triangle is equal to A000079(n-1).
T(n,k) is also the difference between the total number of partitions of all positive integers <= 2^(n-1) into exactly k consecutive parts, and the total number of partitions of all positive integers <= 2^(n-1) into exactly k + 1 consecutive parts.

			Triangle begins:
    1;
    2;
    3,  1;
    5,  2,  1;
    9,  3,  2,  1, 1;
   17,  6,  3,  2, 2, 1, 1;
   33, 11,  6,  4, 2, 2, 2, 1, 2, 1;
   65, 22, 11,  7, 5, 3, 3, 2, 2, 2, 1, 2, 1, 1, 1;
  129, 43, 22, 13, 9, 7, 5, 4, 3, 3, 3, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1;
...
Illustration of initial terms:
.
Row 1:  _
       |_|                              Semilength = 1
        1
Row 2:    _
        _| |
       |_ _|
         2                              Semilength = 2
.
Row 3:        _
             | |
            _| |
        _ _|  _|
       |_ _ _|1                         Semilength = 4
          3
.
Row 4:                _
                     | |
                     | |
                     | |
                  _ _| |
                _|  _ _|
               |  _|
        _ _ _ _| | 1                    Semilength = 8
       |_ _ _ _ _|2
            5
.
Row 5:                                _
                                     | |
                                     | |
                                     | |
                                     | |
                                     | |
                                     | |
                                     | |
                                _ _ _| |
                               |  _ _ _|
                              _| |
                            _|  _|
                        _ _|  _|        Semilength = 16
                       |  _ _|1 1
                       | | 2
        _ _ _ _ _ _ _ _| |3
       |_ _ _ _ _ _ _ _ _|
                9
.
The area (also the number of cells) of the successive diagrams gives the nonzero Mersenne numbers A000225.

Row sums give A000079.
Column 1 gives A094373.
For the characteristic shape of sigma(A000040(n)) see A346871.
For the characteristic shape of sigma(A000217(n)) see A346873.
For the visualization of Mersenne numbers A000225 see A346874.
For the characteristic shape of sigma(A000384(n)) see A346875.
For the characteristic shape of sigma(A000396(n)) see A346876.
For the characteristic shape of sigma(A008588(n)) see A224613.
Cf. A000203, A000225, A237591, A237593, A245092, A249351, A262626.

A346874 Irregular triangle read by rows in which row n lists the row 2^n - 1 of A237591, n >= 1.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 8, 3, 2, 1, 1, 16, 6, 3, 2, 2, 1, 1, 32, 11, 6, 4, 2, 2, 2, 1, 2, 1, 64, 22, 11, 7, 5, 3, 3, 2, 2, 2, 1, 2, 1, 1, 1, 128, 43, 22, 13, 9, 7, 5, 4, 3, 3, 3, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 256, 86, 43, 26, 18, 12, 10, 8, 6, 5, 4, 4, 3, 3, 3, 2, 3

Offset: 1

Omar E. Pol, Aug 06 2021

The Mersenne number A000225(n) does not has a characteristic shape of its symmetric representation of sigma(A000225(n)). On the other hand, we can find that number in two ways in the symmetric representation of the powers of 2 as follows: the Mersenne numbers are the semilength of the smallest Dyck path and also they equals the area (or the number of cells) of the region of the diagram (see examples).
Therefore we can see a geometric pattern of the distribution of the Mersenne numbers in the stepped pyramid described in A245092.
T(n,k) is the length of the k-th line segment of the largest Dyck path of the symmetric representation of sigma(A000225(n)), from the border to the center, hence the sum of the n-th row of triangle is equal to A000225(n).
T(n,k) is also the difference between the total number of partitions of all positive integers <= Mersenne number A000225(n) into k consecutive parts, and the total number of partitions of all positive integers <= Mersenne number A000225(n) into k + 1 consecutive parts.

			Triangle begins:
    1;
    2,  1;
    4,  2,  1;
    8,  3,  2,  1, 1;
   16,  6,  3,  2, 2, 1, 1;
   32, 11,  6,  4, 2, 2, 2, 1, 2, 1;
   64, 22, 11,  7, 5, 3, 3, 2, 2, 2, 1, 2, 1, 1, 1;
  128, 43, 22, 13, 9, 7, 5, 4, 3, 3, 3, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1;
...
Illustration of initial terms:
.
Row 1:
       0_                               Semilength = 0    Area = 1
       |_|
Row 2:
          _
       1_| |                            Semilength = 1    Area = 3
       |_ _|
.
Row 3:        _
             | |
         1  _| |
       2_ _|  _|                        Semilength = 3    Area = 7
       |_ _ _|
.
Row 4:                _
                     | |
                     | |
                     | |
                  _ _| |
              1 _|  _ _|
          4   2|  _|                    Semilength = 7    Area = 15
        _ _ _ _| |
       |_ _ _ _ _|
.
Row 5:                                _
                                     | |
                                     | |
                                     | |
                                     | |
                                     | |
                                     | |
                                     | |
                                _ _ _| |
                               |  _ _ _|
                              _| |
                         1 1_|  _|
                      2 _ _|  _|        Semilength = 15   Area = 31
                       |  _ _|
               8      3| |
        _ _ _ _ _ _ _ _| |
       |_ _ _ _ _ _ _ _ _|
.

Row sums give A000225, n >= 1.
Column 1 gives A000079.
For the characteristic shape of sigma(A000040(n)) see A346871.
For the characteristic shape of sigma(A000079(n)) see A346872.
For the characteristic shape of sigma(A000217(n)) see A346873.
For the characteristic shape of sigma(A000384(n)) see A346875.
For the characteristic shape of sigma(A000396(n)) see A346876.
For the characteristic shape of sigma(A008588(n)) see A224613.
Cf. A000203, A237591, A237593, A245092, A249351, A262626.

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.

cp's OEIS Frontend

A262626 Visible parts of the perspective view of the stepped pyramid whose structure essentially arises after the 90-degree-zig-zag folding of the isosceles triangle A237593.

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

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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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A346873 Triangle read by rows in which row n lists the row A000217(n) of A237591, n >= 1.

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A346876 Irregular triangle read by rows in which row n is the "n-th even perfect number" row of A237591, n >= 1.

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A346871 Irregular triangle read by rows in which row n lists the row A000040(n) of A237591, n >= 1.

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A346872 Irregular triangle read by rows in which row n lists the row 2^(n-1) of A237591, n >= 1.

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A346874 Irregular triangle read by rows in which row n lists the row 2^n - 1 of A237591, n >= 1.

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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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