cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

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

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 2, 1, 3, 2, 2, 1, 2, 2, 3, 1, 2, 1, 2, 1, 4, 2, 2, 1, 3, 2, 4, 1, 2, 1, 2, 1, 4, 2, 3, 1, 2, 2, 4, 1, 2, 1, 2, 2, 3, 2, 2, 1, 3, 3, 4, 2, 2, 1, 4, 1, 4, 2, 2, 1, 2, 2, 5, 1, 4, 1, 2, 2, 4, 3, 2, 1, 2, 2, 4, 2, 3, 2, 2, 1, 5, 2, 2, 1, 4, 2, 4, 1, 2, 1
Offset: 1

Views

Author

Omar E. Pol, Feb 25 2014

Keywords

Comments

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

Examples

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

Crossrefs

Programs

  • Mathematica
    a237271[n_] := Length[a237270[n]] (* code defined in A237270 *)
    Map[a237271, Range[90]] (* data *)
    (* Hartmut F. W. Hoft, Jun 23 2014 *)
    a[n_] := Module[{d = Partition[Divisors[n], 2, 1]}, 1 + Count[d, ?(OddQ[#[[2]]] && #[[2]] >= 2*#[[1]] &)]]; Array[a, 100] (* _Amiram Eldar,  Dec 22 2024 *)
  • PARI
    fill(vcells, hga, hgb) = {ic = 1; for (i=1, #hgb, if (hga[i] < hgb[i], for (j=hga[i], hgb[i]-1, cell = vector(4); cell[1] = i - 1; cell[2] = j; vcells[ic] = cell; ic ++;););); vcells;}
    findfree(vcells) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[3] == 0) && (vcelli[4] == 0), return (i));); return (0);}
    findxy(vcells, x, y) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[1]==x) && (vcelli[2]==y) && (vcelli[3] == 0) && (vcelli[4] == 0), return (i));); return (0);}
    findtodo(vcells, iz) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[3] == iz) && (vcelli[4] == 0), return (i)); ); return (0);}
    zcount(vcells) = {nbz = 0; for (i=1, #vcells, nbz = max(nbz, vcells[i][3]);); nbz;}
    docell(vcells, ic, iz) = {x = vcells[ic][1]; y = vcells[ic][2]; if (icdo = findxy(vcells, x-1, y), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x+1, y), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x, y-1), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x, y+1), vcells[icdo][3] = iz); vcells[ic][4] = 1; vcells;}
    docells(vcells, ic, iz) = {vcells[ic][3] = iz; while (ic, vcells = docell(vcells, ic, iz); ic = findtodo(vcells, iz);); vcells;}
    nbzb(n, hga, hgb) = {vcells = vector(sigma(n)); vcells = fill(vcells, hga, hgb); iz = 1; while (ic = findfree(vcells), vcells = docells(vcells, ic, iz); iz++;); zcount(vcells);}
    lista(nn) = {hga = concat(heights(row237593(0), 0), 0); for (n=1, nn, hgb = heights(row237593(n), n); nbz = nbzb(n, hga, hgb); print1(nbz, ", "); hga = concat(hgb, 0););} \\ with heights() also defined in A237593; \\ Michel Marcus, Mar 28 2014
    
  • Python
    from sympy import divisors
    def a(n: int) -> int:
        divs = list(divisors(n))
        d = [divs[i:i+2] for i in range(len(divs) - 1)]
        s = sum(1 for pair in d if len(pair) == 2 and pair[1] % 2 == 1 and pair[1] >= 2 * pair[0])
        return s + 1
    print([a(n) for n in range(1, 80)])  # Peter Luschny, Aug 05 2025

Formula

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

A245092 The even numbers (A005843) and the values of sigma function (A000203) interleaved.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 6, 7, 8, 6, 10, 12, 12, 8, 14, 15, 16, 13, 18, 18, 20, 12, 22, 28, 24, 14, 26, 24, 28, 24, 30, 31, 32, 18, 34, 39, 36, 20, 38, 42, 40, 32, 42, 36, 44, 24, 46, 60, 48, 31, 50, 42, 52, 40, 54, 56, 56, 30, 58, 72, 60, 32, 62, 63, 64, 48
Offset: 0

Views

Author

Omar E. Pol, Jul 15 2014

Keywords

Comments

Consider an irregular stepped pyramid with n steps. The base of the pyramid is equal to the symmetric representation of A024916(n), the sum of all divisors of all positive integers <= n. Two of the faces of the pyramid are the same as the representation of the n-th triangular numbers as a staircase. The total area of the pyramid is equal to 2*A024916(n) + A046092(n). The volume is equal to A175254(n). By definition a(2n-1) is A000203(n), the sum of divisors of n. Starting from the top a(2n-1) is also the total area of the horizontal part of the n-th step of the pyramid. By definition, a(2n) = A005843(n) = 2n. Starting from the top, a(2n) is also the total area of the irregular vertical part of the n-th step of the pyramid.
On the other hand the sequence also has a symmetric representation in two dimensions, see Example.
From Omar E. Pol, Dec 31 2016: (Start)
We can find the pyramid after the following sequences: A196020 --> A236104 --> A235791 --> A237591 --> A237593.
The structure of this infinite pyramid arises after the 90-degree-zig-zag folding of the diagram of the isosceles triangle A237593 (see the links).
The terraces at the m-th level of the pyramid are also the parts of the symmetric representation of sigma(m), m >= 1, hence the sum of the areas of the terraces at the m-th level equals A000203(m).
Note that the stepped pyramid is also one of the 3D-quadrants of the stepped pyramid described in A244050.
For more information about the pyramid see A237593 and all its related sequences. (End)

Examples

			Illustration of initial terms:
----------------------------------------------------------------------
a(n)                             Diagram
----------------------------------------------------------------------
0    _
1   |_|\ _
2    \ _| |\ _
3     |_ _| | |\ _
4      \ _ _|_| | |\ _
4       |_ _|  _| | | |\ _
6        \ _ _|  _| | | | |\ _
7         |_ _ _|  _|_| | | | |\ _
8          \ _ _ _|  _ _| | | | | |\ _
6           |_ _ _| |    _| | | | | | |\ _
10           \ _ _ _|  _|  _|_| | | | | | |\ _
12            |_ _ _ _|  _|  _ _| | | | | | | |\ _
12             \ _ _ _ _|  _|  _ _| | | | | | | | |\ _
8               |_ _ _ _| |  _|  _ _|_| | | | | | | | |\ _
14               \ _ _ _ _| |  _| |  _ _| | | | | | | | | |\ _
15                |_ _ _ _ _| |_ _| |  _ _| | | | | | | | | | |\ _
16                 \ _ _ _ _ _|  _ _|_|  _ _|_| | | | | | | | | | |\
13                  |_ _ _ _ _| |  _|  _|  _ _ _| | | | | | | | | | |
18                   \ _ _ _ _ _| |  _|  _|    _ _| | | | | | | | | |
18                    |_ _ _ _ _ _| |  _|     |  _ _|_| | | | | | | |
20                     \ _ _ _ _ _ _| |      _| |  _ _ _| | | | | | |
12                      |_ _ _ _ _ _| |  _ _|  _| |  _ _ _| | | | | |
22                       \ _ _ _ _ _ _| |  _ _|  _|_|  _ _ _|_| | | |
28                        |_ _ _ _ _ _ _| |  _ _|  _ _| |  _ _ _| | |
24                         \ _ _ _ _ _ _ _| |  _| |    _| |  _ _ _| |
14                          |_ _ _ _ _ _ _| | |  _|  _|  _| |  _ _ _|
26                           \ _ _ _ _ _ _ _| | |_ _|  _|  _| |
24                            |_ _ _ _ _ _ _ _| |  _ _|  _|  _|
28                             \ _ _ _ _ _ _ _ _| |  _ _|  _|
24                              |_ _ _ _ _ _ _ _| | |  _ _|
30                               \ _ _ _ _ _ _ _ _| | |
31                                |_ _ _ _ _ _ _ _ _| |
32                                 \ _ _ _ _ _ _ _ _ _|
...
a(n) is the total area of the n-th set of symmetric regions in the diagram.
.
From _Omar E. Pol_, Aug 21 2015: (Start)
The above structure contains a hidden pattern, simpler, as shown below:
Level                              _ _
1                                _| | |_
2                              _|  _|_  |_
3                            _|   | | |   |_
4                          _|    _| | |_    |_
5                        _|     |  _|_  |     |_
6                      _|      _| | | | |_      |_
7                    _|       |   | | |   |       |_
8                  _|        _|  _| | |_  |_        |_
9                _|         |   |  _|_  |   |         |_
10             _|          _|   | | | | |   |_          |_
11           _|           |    _| | | | |_    |           |_
12         _|            _|   |   | | |   |   |_            |_
13       _|             |     |  _| | |_  |     |             |_
14     _|              _|    _| |  _|_  | |_    |_              |_
15   _|               |     |   | | | | |   |     |               |_
16  |                 |     |   | | | | |   |     |                 |
...
The symmetric pattern emerges from the front view of the stepped pyramid.
Note that starting from this diagram A000203 is obtained as follows:
In the pyramid the area of the k-th vertical region in the n-th level on the front view is equal to A237593(n,k), and the sum of all areas of the vertical regions in the n-th level on the front view is equal to 2n.
The area of the k-th horizontal region in the n-th level is equal to A237270(n,k), and the sum of all areas of the horizontal regions in the n-th level is equal to sigma(n) = A000203(n). (End)
From _Omar E. Pol_, Dec 31 2016: (Start)
Illustration of the top view of the pyramid with 16 levels:
.
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      |_ _ _ _ _ _ _ _ _|
... (End)
		

Crossrefs

Programs

  • Mathematica
    Table[If[EvenQ@ n, n, DivisorSigma[1, (n + 1)/2]], {n, 0, 65}] (* or *)
    Transpose@ {Range[0, #, 2], DivisorSigma[1, #] & /@ Range[#/2 + 1]} &@ 65 // Flatten (* Michael De Vlieger, Dec 31 2016 *)
    With[{nn=70},Riffle[Range[0,nn,2],DivisorSigma[1,Range[nn/2]]]] (* Harvey P. Dale, Aug 05 2024 *)

Formula

a(2*n-1) + a(2n) = A224880(n).

A244371 Number of toothpicks added at n-th stage in the toothpick structure of the symmetric representation of sigma in the four quadrants.

Original entry on oeis.org

8, 16, 24, 32, 32, 48, 40, 64, 64, 80, 56, 96, 64, 104, 104, 128, 80, 144, 88, 160, 144, 152, 104, 192, 136, 176, 176, 224, 128, 240, 136, 256
Offset: 1

Views

Author

Omar E. Pol, Jun 26 2014

Keywords

Comments

Partial sums give A244370.

Crossrefs

Formula

a(n) = 4*A244363(n) = 8*A244361(n). - Omar E. Pol, Oct 16 2014

Extensions

a(13)-a(28) from Omar E. Pol, Oct 18 2014
a(29)-a(32) from Omar E. Pol, May 04 2023

A244370 Total number of toothpicks after n-th stage in the toothpick structure of the symmetric representation of sigma in the four quadrants.

Original entry on oeis.org

8, 24, 48, 80, 112, 160, 200, 264, 328, 408, 464, 560, 624, 728, 832, 960, 1040, 1184, 1272, 1432, 1576, 1728, 1832, 2024, 2160, 2336, 2512, 2736
Offset: 1

Views

Author

Omar E. Pol, Jun 26 2014

Keywords

Comments

Partial sums of A244371.
If we use toothpicks of length 1/2, so the area of the central square is equal to 1. The total area of the structure after n-th stage is equal to A024916(n), the sum of all divisors of all positive integers <= n, hence the total area of the n-th set of symmetric regions added at n-th stage is equal to sigma(n) = A000203(n), the sum of divisors of n.
If we use toothpicks of length 1, so the number of cells (and the area) of the central square is equal to 4. The number of cells (and the total area) of the structure after n-th stage is equal to 4*A024916(n) = A243980(n), hence the number of cells (and the total area) of the n-th set of symmetric regions added at n-th stage is equal to 4*A000203(n) = A239050(n).

Examples

			Illustration of the structure after 16 stages (Contains 960 toothpicks):
.
.                 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
.                |  _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _  |
.                | |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _| |
.             _ _| |  _ _ _ _ _ _ _ _ _ _ _ _ _ _  | |_ _
.           _|  _ _| |_ _ _ _ _ _ _ _ _ _ _ _ _ _| |_ _  |_
.         _|  _|  _| |  _ _ _ _ _ _ _ _ _ _ _ _  | |_  |_  |_
.        |  _|   |_ _| |_ _ _ _ _ _ _ _ _ _ _ _| |_ _|   |_  |
.   _ _ _| |  _ _|     |  _ _ _ _ _ _ _ _ _ _  |     |_ _  | |_ _ _
.  |  _ _ _|_| |      _| |_ _ _ _ _ _ _ _ _ _| |_      | |_|_ _ _  |
.  | | |  _ _ _|    _|_ _|  _ _ _ _ _ _ _ _  |_ _|_    |_ _ _  | | |
.  | | | | |  _ _ _| |  _| |_ _ _ _ _ _ _ _| |_  | |_ _ _  | | | | |
.  | | | | | | |  _ _|_|  _|  _ _ _ _ _ _  |_  |_|_ _  | | | | | | |
.  | | | | | | | | |  _ _|   |_ _ _ _ _ _|   |_ _  | | | | | | | | |
.  | | | | | | | | | | |  _ _|  _ _ _ _  |_ _  | | | | | | | | | | |
.  | | | | | | | | | | | | |  _|_ _ _ _|_  | | | | | | | | | | | | |
.  | | | | | | | | | | | | | | |  _ _  | | | | | | | | | | | | | | |
.  | | | | | | | | | | | | | | | |   | | | | | | | | | | | | | | | |
.  | | | | | | | | | | | | | | | |_ _| | | | | | | | | | | | | | | |
.  | | | | | | | | | | | | | |_|_ _ _ _|_| | | | | | | | | | | | | |
.  | | | | | | | | | | | |_|_  |_ _ _ _|  _|_| | | | | | | | | | | |
.  | | | | | | | | | |_|_    |_ _ _ _ _ _|    _|_| | | | | | | | | |
.  | | | | | | | |_|_ _  |_  |_ _ _ _ _ _|  _|  _ _|_| | | | | | | |
.  | | | | | |_|_ _  | |_  |_ _ _ _ _ _ _ _|  _| |  _ _|_| | | | | |
.  | | | |_|_ _    |_|_ _| |_ _ _ _ _ _ _ _| |_ _|_|    _ _|_| | | |
.  | |_|_ _ _  |     |_  |_ _ _ _ _ _ _ _ _ _|  _|     |  _ _ _|_| |
.  |_ _ _  | |_|_      | |_ _ _ _ _ _ _ _ _ _| |      _|_| |  _ _ _|
.        | |_    |_ _  |_ _ _ _ _ _ _ _ _ _ _ _|  _ _|    _| |
.        |_  |_  |_  | |_ _ _ _ _ _ _ _ _ _ _ _| |  _|  _|  _|
.          |_  |_ _| |_ _ _ _ _ _ _ _ _ _ _ _ _ _| |_ _|  _|
.            |_ _  | |_ _ _ _ _ _ _ _ _ _ _ _ _ _| |  _ _|
.                | |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _| |
.                | |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _| |
.                |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
.
		

Crossrefs

Formula

a(n) = 4*A244362(n) = 8*A244360(n).

Extensions

a(8) corrected and more terms from Omar E. Pol, Oct 18 2014

A244970 Total number of regions after n-th stage in the diagram of the symmetric representation of sigma on the four quadrants.

Original entry on oeis.org

1, 2, 6, 7, 11, 12, 16, 17, 25, 29, 33, 34, 38, 42, 50, 51, 55, 56, 60, 61, 73, 77, 81, 82, 90, 94, 106, 107, 111, 112, 116, 117, 129, 133, 141, 142, 146, 150, 162, 163, 167, 168, 172, 176, 184, 188, 192, 193, 201, 209, 221, 225, 229, 230, 242, 243, 255, 259, 263, 264
Offset: 1

Views

Author

Omar E. Pol, Jul 08 2014

Keywords

Comments

Partial sums of A244971.
If we use toothpicks of length 1/2, so the area of the central square is equal to 1. The total area of the structure after n-th stage is equal to A024916(n), the sum of all divisors of all positive integers <= n, hence the total area of the n-th set of symmetric regions added at n-th stage is equal to sigma(n) = A000203(n), the sum of divisors of n.
If we use toothpicks of length 1, so the number of cells (and the area) of the central square is equal to 4. The number of cells (and the total area) of the structure after n-th stage is equal to 4*A024916(n) = A243980(n), hence the number of cells (and the total area) of the n-th set of symmetric regions added at n-th stage is equal to 4*A000203(n) = A239050(n).
a(n) is also the total number of terraces of the stepped pyramid with n levels described in A244050. - Omar E. Pol, Apr 20 2016

Examples

			Illustration of the structure after 15 stages (contains 50 regions):
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The diagram is also the top view of the stepped pyramid with 15 levels described in A244050. - _Omar E. Pol_, Apr 20 2016
		

Crossrefs

A262618 Number of parts in the asymmetric representation of sigma(n) in an octant.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 3, 1, 1, 1, 1, 1, 2, 1
Offset: 1

Views

Author

Omar E. Pol, Nov 05 2015

Keywords

Comments

The diagram of the asymmetric representation of sigma in an octant has been obtained according to the following way: A196020 --> A236104 --> A235791 --> A237591.
Consider that the hypotenuse of the first triangle of the diagram has length 2, so the area of the triangle is equal to 1 and the sum of the areas of all parts added at n-th stage equals sigma(n), the sum of the divisors of n.
a(n) is also the number of terraces at n-th level (starting from the top) in an octant of the step pyramids described in A245092 and A244050.

Crossrefs

Formula

a(n) = (A237271(n) + 1)/2, if A237271(n) is odd.
a(n) = A237271(n)/2, if A237271(n) is even.

A262619 Number of parts in the symmetric representation of sigma(n) in two successive octants of two quadrants.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 3, 3, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 5, 1, 3, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 5, 1, 1, 1, 3, 1, 3, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 3, 5, 1, 1, 1, 1, 1, 3, 1
Offset: 1

Views

Author

Omar E. Pol, Nov 06 2015

Keywords

Comments

In the diagram of the top view of the pyramid described in A244050 consider a 90-degree sector on two successive octants of two quadrants. The area of the top triangle is equal to 1 and the sum of the areas of all parts (or regions) added at n-th stage equals sigma(n), the sum of the divisors of n.
a(n) is also the number of terraces at n-th level (starting from the top) in the mentioned sector of the pyramid.
For more information see A237593 and A237270.

Crossrefs

Formula

a(n) = A237271(n), if A237271(n) is odd.
a(n) = A237271(n) - 1, if A237271(n) is even.
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