A338721 -id:A338721 - cp's OEIS frontend

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

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

Offset: 1

Omar E. Pol, Feb 02 2013

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

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

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

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

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

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

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

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

A335616 a(n) is twice the number of partitions of n into consecutive parts, minus the number of partitions of n into consecutive parts that contain 1 as a part.

Original entry on oeis.org

1, 2, 3, 2, 4, 3, 4, 2, 6, 3, 4, 4, 4, 4, 7, 2, 4, 6, 4, 4, 7, 4, 4, 4, 6, 4, 8, 3, 4, 8, 4, 2, 8, 4, 8, 5, 4, 4, 8, 4, 4, 8, 4, 4, 11, 4, 4, 4, 6, 6, 8, 4, 4, 8, 7, 4, 8, 4, 4, 8, 4, 4, 12, 2, 8, 7, 4, 4, 8, 8, 4, 6, 4, 4, 12, 4, 8, 7, 4, 4, 10, 4, 4, 8, 8, 4, 8, 4, 4, 12, 7, 4, 8, 4, 8, 4, 4, 6, 12, 6

Offset: 1

Omar E. Pol, Oct 02 2020

a(n) is twice the number of partitions of n into consecutive parts, minus the number of partitions of n into distinct parts such that the greatest part equals the number of all parts.
For a visualization of the sequence, consider a diagram formed for infinitely many double staircases as shown in the Example section.
a(n) is the number of horizontal line segments (or steps) that are only in the n-th level of the structure, starting from the top.
The total length of all vertical line segments that are adjacent and below the steps of the n-th level of the structure equals twice the total number of parts in all partitions of n into consecutive parts.
Note that in the n-th double staircase the top step is located in the level A000217(n), n >= 1, every horizontal line segment has length 1, and every vertical line segment has length n.
a(n) is also the number of horizontal line segments of length 1 or 2 in the n-th level of the similar diagram used to represent the sequence A237593 and other isosceles triangles related to A237593.
a(n) is odd if and only if n is a nonzero triangular number (A000217).
Double-staircases theorem of the sum of divisors: the total number of steps from level n to the top of all the odd-indexed double staircases that have at least one step in the level n, minus the total number of steps from the level n to the top of all the even-indexed double staircases that have at least one step in the level n equals sigma(n) = A000203(n).
The above theorem shows a symmetry of sigma in accordance with the symmetric Dyck Paths described in A237593 and with the pyramid described in A245092.
For the connection with the partitions into consecutive integers see also A196020, since we can see here that A196020(n,k) is also the number of steps in the first n levels of the k-th double staircase that has at least one step in the n-th level of the diagram, otherwise A196020(n,k) = 0. Also, it is the width of the mentioned staircase in n-th level of the diagram.
It appears that odd primes (A065091) are also the levels where there are steps in the staircases 1 and 2, but no step from other staircases.
It appears that powers of 2 (A000079) are also the levels where there are only one or two steps in total.
This sequence could be related to several other sequences (see the Crossrefs section of A262626).

			Illustration of initial terms:
n   a(n)                               Diagram
                                          _
1     1                                 _|1|_
2     2                               _|1 _ 1|_
3     3                             _|1  |1|  1|_
4     2                           _|1   _| |_   1|_
5     4                         _|1    |1 _ 1|    1|_
6     3                       _|1     _| |1| |_     1|_
7     4                     _|1      |1  | |  1|      1|_
8     2                   _|1       _|  _| |_  |_       1|_
9     6                 _|1        |1  |1 _ 1|  1|        1|_
10    3               _|1         _|   | |1| |   |_         1|_
11    4             _|1          |1   _| | | |_   1|          1|_
12    4           _|1           _|   |1  | |  1|   |_           1|_
13    4         _|1            |1    |  _| |_  |    1|            1|_
14    4       _|1             _|    _| |1 _ 1| |_    |_             1|_
15    7     _|1              |1    |1  | |1| |  1|    1|              1|_
16    2    |1                |     |   | | | |   |     |                1|
...
For n = 6 (above), the total number of steps in all double staircases that have at least one step in the 6th level of the structure is equal to 3, since there are two steps in the first double staircase, there are no steps in the second double staircase, and there is only one step in the third double staircase, so a(3) = 2 + 0 + 1 = 3.
From the theorem (see comments) for n = 6, let s(k) = A196020(6,k) be the total number of steps from level n to the top, in the k-th double staircase that has at least a step in the 6th level of the structure, otherwise s(k) = 0. We have that s(1) = 11, s(2) = 0 and s(3) = 1. So the alternating sum is 11 - 0 + 1 = 12, which equals sigma(6) = 1 + 2 + 3 + 6 = 12.
Note that to evaluate sigma(n), it is sufficient to have only the n-th level of the diagram, since the width of the base level of a double staircase equals the number of its steps. See below:
For n = 6 the 6th level of the above diagram looks like this:
                                _         _         _
                               |1      | |1| |      1|
.
Width of the 1st staircase:    |<-------- 11 ------->|
.
Width of the 3rd staircase:          --->|1|<---
.
The width of the first double staircase is 11, the width of the second double staircase does not count, and the width of the third double staircase is 1, so the alternating sum is 11 - 0 + 1 = 12 = sigma(6).
For n = 15 the alternating sum is 29 - 13 + 7 - 0 + 1 = 24 = sigma(15).
For n = 16 the alternating sum is 31 -  0 + 0 - 0 + 0 = 31 = sigma(16).
For more information about these alternating sums see A196020.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Row sums of A339275.
Partial sums give A338722.
Cf. A000079, A000203, A000217, A001227, A054843, A054844, A065091, A136107. A196020, A204217, A236104, A235791, A237048, A237593, A237593, A245092, A249351, A262611, A262626, A279693, A279733, A281010, A281011, A286001, A299765, A338721.

N:= 100:
S := convert(series( add( x^(n*(n+1)/2)*(1 + x^n)/(1 - x^n), n = 1..floor(sqrt(2*N)) ), x, N+1 ), polynom):
seq(coeff(S, x, n), n = 1..N); # Peter Bala, Jan 20 2021

A335616[n_]:=2DivisorSigma[0,n/2^IntegerExponent[n,2]]-Boole[IntegerQ[(Sqrt[8n+1]-1)/2]];Array[A335616,100] (* Paolo Xausa, Sep 03 2023 *)

a(n) = 2*A001227(n) - A010054(n).
a(n) = A054844(n) - A010054(n).
a(n) = 2*A136107(n) + A010054(n). - Omar E. Pol, Nov 27 2020
G.f.: Sum_{n >= 1} x^(n*(n+1)/2)*(1 + x^n)/(1 - x^n). Cf. A000005 with g.f. Sum_{n >= 1} x^(n^2)*(1 + x^n)/(1 - x^n). - Peter Bala, Jan 20 2021

Simpler definition from Omar E. Pol, Nov 27 2020

A347186 Ziggurat sequence (see Comments lines for definition).

Original entry on oeis.org

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

Omar E. Pol, Aug 21 2021

nonn

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

			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.
.

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

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.

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.

A339275 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the terms of A040000: 1, 2, 2, 2, ... interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

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

Offset: 1

Omar E. Pol, Dec 01 2020

T(n,k) is also the number of horizontal line segments in the n-th level of the k-th largest double-staircase of the diagram defined in A335616 (see example).

The partial sums of column k give the k-th column of A338721.

			Triangle begins (rows 1..28):
1;
2;
2,  1;
2,  0;
2,  2;
2,  0,  1;
2,  2,  0;
2,  0,  0;
2,  2,  2;
2,  0,  0,  1;
2,  2,  0,  0;
2,  0,  2,  0;
2,  2,  0,  0;
2,  0,  0,  2;
2,  2,  2,  0,  1;
2,  0,  0,  0,  0;
2,  2,  0,  0,  0;
2,  0,  2,  2,  0;
2,  2,  0,  0,  0;
2,  0,  0,  0,  2;
2,  2,  2,  0,  0,  1;
2,  0,  0,  2,  0,  0;
2,  2,  0,  0,  0,  0;
2,  0,  2,  0,  0,  0;
2,  2,  0,  0,  2,  0;
2,  0,  0,  2,  0,  0;
2,  2,  2,  0,  0,  2;
2,  0,  0,  0,  0,  0,  1;
...
For an illustration of the rows of triangle consider the infinite "double-staircases" diagram defined in A335616.
The first 15 levels of the structure looks like this:
.
Level                         "Double-staircases" diagram
n                                          _
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|
.
For n = 15, in the 15th level of the diagram we have that the first largest double-staircase has two horizontal steps, the second double-staircase has two steps, the third double-staircase has two steps, there are no steps in the fourth double-stairce and the fifth double-staircase has only one step, so the 15th row of triangle is [2, 2, 2, 0, 1].

Column 1 is A040000.
Row sums give A335616.
Row n has length A003056(n).
Column k starts in row A000217(k).
The number of positive terms in row n is A001227(n).
Cf. A196020, A236104, A237048, A237270, A237591, A237593, A249351, A280850, A296508, A299484, A338721.

cp's OEIS Frontend

A338722 Row sums in triangle A338721.

Views

Author

Keywords

Comments

Links

Crossrefs

Extensions

A338723 Alternating row sums in triangle A338721.

Views

Author

Keywords

Links

Crossrefs

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Sage

Formula

A335616 a(n) is twice the number of partitions of n into consecutive parts, minus the number of partitions of n into consecutive parts that contain 1 as a part.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A347186 Ziggurat sequence (see Comments lines for definition).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A339275 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the terms of A040000: 1, 2, 2, 2, ... interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

Views

Author

Keywords

Comments

Examples

Crossrefs