A347186 Ziggurat sequence (see Comments lines for definition).

1, 4, 6, 16, 12, 37, 20, 64, 36, 90, 42, 161, 56, 156, 107, 256, 90, 334, 110, 408, 202, 342, 156, 697, 207, 462, 312, 785, 240, 976, 272, 1024, 446, 756, 441, 1586, 380, 930, 604, 1736, 462, 1841, 506, 1806, 1101, 1332, 600, 2921, 720, 1820, 992, 2450, 756, 2998, 1108, 3257

Offset: 1

Author

Omar E. Pol, Aug 21 2021

Keywords

• nonn

Comments

What do the parts and subparts of the symmetric representation of sigma(n) represent? (cf. A237270, A280851). This sequence gives an answer.

To calculate a(n) we must follow a geometric algorithm composed of three stages as follows:

Stage 1 (Construction):

On the infinite square grid we draw the diagram called "double-staircases" with n levels described in A335616.

Then we label the double-staircases from left to right starting from k = 1 to the central column of the diagram.

Note that k is also the difference in height between two steps of the ladder it would have if we drew it with more than one step.

Stage 2 (Debugging):

We remove all double-staircases that do not have at least one step at the level 1 of the diagram starting from the base.

Now the number of steps in the k-th double-staircase is equal to A196020(n,k).

Stage 3 (Annihilation):

From left to right, we remove each even-indexed double-staircase along with the steps of the nearest odd-indexed double-staircase that are just above it.

As a result of this geometric algorithm a diagram is obtained which can have only double-staircases, only simple-staircases, or both at the same time.

We call double-staircases those that have a step in the central column of the diagram. The rest are simple-staircases forming one or more symmetrical pairs equidistant from the central column of the diagram.

We call "parts" of the diagram to the staircases (and the cells below them) that are separated from each other by columns of zero height.

We call "subparts" of the diagram to the polygons formed by the cells that are under the staircases.

a(n) is the total area (or the total number of cells) under all the staircases, with multiplicity.

The connection with the polygonal numbers is as follows:

The area under a double-staircase labeled with the number k is equal to the m-th (k+2)-gonal number plus the (m-1)-th (k+2)-gonal number, where m is the number of steps on one side of the ladder from the base to the top.

If k = 1 then the area under the double-staircase is also equal to n^2 = A000290(n).

The area under a simple-staircase labeled with the number k is equal to the m-th (k+2)-gonal number, where m is the number of steps.

If n is a power of 2, or if n is an odd prime number, or if n is an even perfect number, or if n is a member of A246955, then the calculation of a(n) is easy (see the Formula section).

The connection with the symmetric representation of sigma(n) or "SRS(n)" is as follows:

The total number of steps in the diagram is equal to A000203(n), equaling the total area (or the number of cells) in the SRS(n).

The number of parts in the diagram is equal to A237271(n) equaling the number of parts in the SRS(n).

The number of double-staircases (also the number of steps in the central column in the diagram) is equal to A067742(n), equaling the number of central subparts in the SRS(n).

The number of simple-staircases is equal to A281009(n), equaling the total number of equidistant subparts in the SRS(n).

The total number of staircases is equal to A001227(n), equaling the number of subparts in the SRS(n).

The number of columns in the diagram is equal to 2*n - 1, equaling the number of "widths" in the SRS(n) (cf. A249351).

The number of steps in the successive parts of the diagram gives the n-th row of triangle A237270, equaling the successive parts in the SRS(n).

The number of steps in the successive staircases from left to right gives the n-th row of triangle A280851, equaling the successive subparts from left to right in the SRS(n).

The diagram is essentially the front view of a three-dimensional structure whose base is the symmetric representation of sigma(n), so a(n) is also the total number of cubic cells (or cubes) in the structure.

The number of polycubes in the structure is equal to A237271(n), equaling the number of parts in the SRS(n).

For some values of n the three-dimensional structure resembles a "ziggurat" which is a type of massive structure built in ancient Mesopotamia.

It appears that the geometric algorithm described here is equivalent to the numerical algorithm conjectured in A280850 in the sense that both allow the value of the subparts of the SRS(n) to be computed.

Both algorithms would also be equivalent to the geometric transformation of the isosceles triangle described in A237593 which, when folded, transforms into the vertical faces of the structure of the stepped pyramid described in A245092.

Note that the stage 2 (Debugging) is in correspondence with the formula A196020(n,k) = A237048(n,k)*A338721(n,k) which transform the triangle A338721 into the triangle A196020. - Omar E. Pol, Jun 23 2022

Examples

			Illustration of the geometric algorithm and the initial terms (n = 1..6):
-------------------------------------------------------------------------------
          Stage 1                  Stage 2                  Stage 3
      (Construction)             (Debugging)            (Annihilation)
-------------------------------------------------------------------------------
     Double-staircases           Diagram of                Ziggurat
n         diagram                  A196020                  diagram        a(n)
-------------------------------------------------------------------------------
             _                        _                        _
1           |_|                      |_|                      |_|            1
            1                        1                        1
.
             _                        _                        _
           _| |_                    _| |_                    _| |_
2         |_ _ _|                  |_ _ _|                  |_ _ _|          4
          1                        1                        1
.
             _                        _
           _| |_                    _| |_                    _   _
         _|  _  |_                _|  _  |_                _| | | |_
3       |_ _|_|_ _|              |_ _|_|_ _|              |_ _|_|_ _|        6
        1   2                    1   2                    1
.
             _                        _                        _
           _| |_                    _| |_                    _| |_
         _|  _  |_                _|     |_                _|     |_
       _|   | |   |_            _|         |_            _|         |_
4     |_ _ _|_|_ _ _|          |_ _ _ _ _ _ _|          |_ _ _ _ _ _ _|     16
      1     2                  1                        1
.
             _                        _
           _| |_                    _| |_
         _|  _  |_                _|  _  |_                _       _
       _|   | |   |_            _|   | |   |_            _| |     | |_
     _|    _| |_    |_        _|    _| |_    |_        _|   |     |   |_
5   |_ _ _|_ _ _|_ _ _|      |_ _ _|_ _ _|_ _ _|      |_ _ _|_ _ _|_ _ _|   12
    1     2                 1      2                  1
.
             _                        _                        _
           _| |_                    _| |_                    _| |_
         _|  _  |_                _|     |_                _|     |_
       _|   | |   |_            _|         |_            _|         |_
     _|    _| |_    |_        _|             |_        _|             |_
   _|     |  _  |     |_    _|        _        |_    _|        _        |_
6 |_ _ _ _|_|_|_|_ _ _ _|  |_ _ _ _ _|_|_ _ _ _ _|  |_ _ _ _ _|_|_ _ _ _ _| 37
  1       2 3              1         3              1         3
.
For n = 7..14 the examples are omitted.
For n = 15 the illustration of the geometric algorithm 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 (k = 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
.
Note that the number of steps in the successive double-staircases gives [29, 13, 7, 0, 1], the same as the 15th row of triangle A196020 (whose alternate sums equals sigma(15) = A000203(15) = 24).
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.
As a result of this geometric algorithm a new diagram is obtained which in this case has two double-staircases and two simple-staircases as shown below:
                               _
                              | |
                 _            | |            _
               _| |          _| |_          | |_
             _|   |         |     |         |   |_
           _|     |         |     |         |     |_
         _|       |        _|     |_        |       |_
       _|         |       |         |       |         |_
     _|           |       |         |       |           |_
   _|             |      _|    _    |_      |             |_
  |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _|
  1                     3     5
.
The diagram is called here "ziggurat of order 15".
Now we calculate the total area (or the total 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 are a pair of this staircases, so the total area of this pair is equal to 2*36 = 72.
The area under the double-staircase labeled 3 is equal to A000326(4) + A000326(3) = 22 + 12 = 34.
The area under the double-staircase labeled 5 is equal to A000566(1) + A000566(0) = 1 + 0 = 1.
Therefore the total area is a(15) = 72 + 34 + 1 = 107.
The connection with the symmetric representation of sigma(15) or "SRS(15)" is as follows:
The total number of steps is equal to A000203(15) = 24, equaling the total area (or number of cells) in the SRS(15).
The number of parts in the diagram is equal to A237271(15) = 3 equaling the number of parts in the SRS(15).
The number of double-staircases (also the number of steps in the central column in the diagram) is equal to A067742(15) = 2, equaling the number of central subparts in the SRS(15).
The number of simple-staircases is equal to A281009(15) = 2, equaling the total number of equidistant subparts in the SRS(15).
The total number of staircases is qual to A001227(15) = 4, equaling the number of subparts in the SRS(15).
The number of columns in the diagram is equal to 2*15 - 1 = 29 equaling the number of "widths" in the SRS(15) (cf. A249351).
The number of steps in the successive parts of the diagram are [8, 8, 8], the same as the 15th row of triangle A237270, matching the successive parts in the SRS(15).
The number of steps in the successive staircases from left to right are respectively [8, 7, 1, 8], the same as the 15th row of triangle A280851, matching the successive subparts in the SRS(15).
a(15) = 107 is also the number of cubic cells in the three-dimensional version of the structure whose base is the SRS(15).
The number of polycubes in the structure is equal to A237271(15) = 3, equaling the number of parts in the SRS(15).
The top view of the 3D-Ziggurat of order 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 a(15) = 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.
.
		

Links

• Omar E. Pol, Illustration of a(1)..a(8), 3D-Ziggurats
• Omar E. Pol, Illustration of a(9)..a(13), 3D-Ziggurats
• Omar E. Pol, Illustration of a(14)..a(15), 3D-Ziggurats

Crossrefs

Row sums of A347367 and of A347263.

Cf. A356351 (partial sums).

Cf. A279387 (definition of subpart).

Cf. A000079, A000203, A000217, A000290, A000326, A000396, A001227, A002378, A065091, A067742, A131576, A196020, A235791, A236104, A237270, A237271, A237591, A237593, A245092, A246955, A249351, A262626, A280850, A280851, A281009, A296508, A296512, A296513, A335616, A338721, A347273, A347361, A347529, A351819.

Formula

a(2^(n-1)) = n^2 = A000290(n).

a(P) = 1 + P^2, if P is an even perfect number.

a(p) = 2*A000217((p+1)/2) = A002378((p+1)/2), if p is an odd prime.

a(k) = 2*A000217(A000203(k)/2), if k is a member of A246955.