A244050 -id:A244050 - cp's OEIS frontend

A239931 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(4n-3).

1, 3, 3, 5, 3, 5, 7, 7, 9, 9, 11, 5, 5, 11, 13, 5, 13, 15, 15, 17, 7, 7, 17, 19, 19, 21, 21, 23, 32, 23, 25, 7, 25, 27, 27, 29, 11, 11, 29, 31, 31, 33, 9, 9, 33, 35, 13, 13, 35, 37, 37, 39, 18, 39, 41, 15, 9, 15, 41, 43, 11, 11, 43, 45, 45, 47, 17, 17, 47, 49, 49, 51, 51, 53, 43, 43, 53, 55, 55, 57, 57, 59, 21, 22, 21, 59, 61, 11, 61, 63, 15, 15, 63

Offset: 1

Omar E. Pol, Mar 29 2014

Row n is a palindromic composition of sigma(4n-3).
Row n is also the row 4n-3 of A237270.
Row n has length A237271(4n-3).
Row sums give A112610.
Also row n lists the parts of the symmetric representation of sigma in the n-th arm of the first quadrant of the spiral described in A239660, see example.
For the parts of the symmetric representation of sigma(4n-2), see A239932.
For the parts of the symmetric representation of sigma(4n-1), see A239933.
For the parts of the symmetric representation of sigma(4n), see A239934.
We can find the spiral (mentioned above) on the terraces of the pyramid described in A244050. - Omar E. Pol, Dec 06 2016

			The irregular triangle begins:
   1;
   3,  3;
   5,  3,  5;
   7,  7;
   9,  9;
  11,  5,  5, 11;
  13,  5, 13;
  15, 15;
  17,  7,  7, 17;
  19, 19;
  21, 21;
  23, 32, 23;
  25,  7, 25;
  27, 27;
  29, 11, 11, 29;
  31, 31;
  ...
Illustration of initial terms in the first quadrant of the spiral described in A239660:
.
.     _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 15
.    |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
.                                  |
.                                  |
.     _ _ _ _ _ _ _ _ _ _ _ _ _ 13 |
.    |_ _ _ _ _ _ _ _ _ _ _ _ _|   |
.                              |   |_ _ _
.                              |         |
.     _ _ _ _ _ _ _ _ _ _ _ 11 |         |_
.    |_ _ _ _ _ _ _ _ _ _ _|   |_ _ _      |_
.                          |         |_ _ 5  |_
.                          |         |_  |_    |_ _
.     _ _ _ _ _ _ _ _ _ 9  |_ _ _      |_  |       |
.    |_ _ _ _ _ _ _ _ _|   |_ _  |_ 5    |_|_      |
.                      |       |_ _|_ 5      |     |_ _ _ _ _ _ 15
.                      |           | |_      |               | |
.     _ _ _ _ _ _ _ 7  |_ _        |_  |     |_ _ _ _ _ 13   | |
.    |_ _ _ _ _ _ _|       |_        | |             | |     | |
.                  |         |_      |_|_ _ _ _ 11   | |     | |
.                  |_ _        |             | |     | |     | |
.     _ _ _ _ _ 5      |_      |_ _ _ _ 9    | |     | |     | |
.    |_ _ _ _ _|         |           | |     | |     | |     | |
.              |_ _ 3    |_ _ _ 7    | |     | |     | |     | |
.              |_  |         | |     | |     | |     | |     | |
.     _ _ _ 3    |_|_ _ 5    | |     | |     | |     | |     | |
.    |_ _ _|         | |     | |     | |     | |     | |     | |
.          |_ _ 3    | |     | |     | |     | |     | |     | |
.            | |     | |     | |     | |     | |     | |     | |
.     _ 1    | |     | |     | |     | |     | |     | |     | |
.    |_|     |_|     |_|     |_|     |_|     |_|     |_|     |_|
.
For n = 7 we have that 4*7-3 = 25 and the 25th row of A237593 is [13, 5, 3, 1, 2, 1, 1, 2, 1, 3, 5, 13] and the 24th row of A237593 is [13, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 13] therefore between both Dyck paths there are three regions (or parts) of sizes [13, 5, 13], so row 7 is [13, 5, 13].
The sum of divisors of 25 is 1 + 5 + 25 = A000203(25) = 31. On the other hand the sum of the parts of the symmetric representation of sigma(25) is 13 + 5 + 13 = 31, equaling the sum of divisors of 25.

Cf. A000203, A112610, A196020, A236104, A235791, A237270, A237271, A237591, A237593, A239660, A239932-A239934, A244050, A245092, A262626.

A239934 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(4n).

Original entry on oeis.org

7, 15, 28, 31, 42, 60, 56, 63, 91, 90, 42, 42, 124, 49, 49, 120, 168, 127, 63, 63, 195, 70, 70, 186, 224, 180, 84, 84, 252, 217, 210, 280, 248, 105, 105, 360, 112, 112, 255

Offset: 1

Omar E. Pol, Mar 29 2014

Row n is a palindromic composition of sigma(4n).
Row n is also the row 4n of A237270.
Row n has length A237271(4n).
Row sums give A193553.
First differs from A193553 at a(11).
Also row n lists the parts of the symmetric representation of sigma in the n-th arm of the fourth quadrant of the spiral described in A239660, see example.
For the parts of the symmetric representation of sigma(4n-3), see A239931.
For the parts of the symmetric representation of sigma(4n-2), see A239932.
For the parts of the symmetric representation of sigma(4n-1), see A239933.
We can find the spiral (mentioned above) on the terraces of the pyramid described in A244050. - Omar E. Pol, Dec 06 2016

			The irregular triangle begins:
    7;
   15;
   28;
   31;
   42;
   60;
   56;
   63;
   91;
   90;
   42, 42;
  124;
   49, 49;
  120;
  168;
  ...
Illustration of initial terms in the fourth quadrant of the spiral described in A239660:
.
.           7       15      28      31      42      60      56      63
.           _       _       _       _       _       _       _       _
.          | |     | |     | |     | |     | |     | |     | |     | |
.         _| |     | |     | |     | |     | |     | |     | |     | |
.     _ _|  _|     | |     | |     | |     | |     | |     | |     | |
.    |_ _ _|    _ _| |     | |     | |     | |     | |     | |     | |
.             _|  _ _|     | |     | |     | |     | |     | |     | |
.            |  _|    _ _ _| |     | |     | |     | |     | |     | |
.     _ _ _ _| |    _|    _ _|     | |     | |     | |     | |     | |
.    |_ _ _ _ _|  _|     |    _ _ _| |     | |     | |     | |     | |
.                |      _|   |  _ _ _|     | |     | |     | |     | |
.                |  _ _|    _| |    _ _ _ _| |     | |     | |     | |
.     _ _ _ _ _ _| |      _|  _|   |  _ _ _ _|     | |     | |     | |
.    |_ _ _ _ _ _ _|  _ _|  _|  _ _| |    _ _ _ _ _| |     | |     | |
.                    |  _ _|  _|    _|   |    _ _ _ _|     | |     | |
.                    | |     |     |  _ _|   |    _ _ _ _ _| |     | |
.     _ _ _ _ _ _ _ _| |  _ _|  _ _|_|       |   |  _ _ _ _ _|     | |
.    |_ _ _ _ _ _ _ _ _| |  _ _|  _|      _ _|   | |    _ _ _ _ _ _| |
.                        | |     |      _|    _ _| |   |  _ _ _ _ _ _|
.                        | |  _ _|    _|  _ _|  _ _|   | |
.     _ _ _ _ _ _ _ _ _ _| | |       |   |    _|    _ _| |
.    |_ _ _ _ _ _ _ _ _ _ _| |  _ _ _|  _|  _|     |  _ _|
.                            | |       |  _|      _| |
.                            | |  _ _ _| |      _|  _|
.     _ _ _ _ _ _ _ _ _ _ _ _| | |  _ _ _|  _ _|  _|
.    |_ _ _ _ _ _ _ _ _ _ _ _ _| | |       |  _ _|
.                                | |  _ _ _| |
.                                | | |  _ _ _|
.     _ _ _ _ _ _ _ _ _ _ _ _ _ _| | | |
.    |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _| | |
.                                    | |
.                                    | |
.     _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _| |
.    |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
.
For n = 7 we have that 4*7 = 28 and the 28th row of A237593 is [15, 5, 3, 2, 1, 1, 1, 1, 1, 1, 2, 3, 5, 15] and the 27th row of A237593 is [14, 5, 3, 2, 1, 2, 2, 1, 2, 3, 5, 14] therefore between both Dyck paths there are only one region (or part) of size 56, so row 7 is 56.
The sum of divisors of 28 is 1 + 2 + 4 + 7 + 14 + 28 = A000203(28) = 56. On the other hand the sum of the parts of the symmetric representation of sigma(28) is 56, equaling the sum of divisors of 28.
For n = 11 we have that 4*11 = 44 and the 44th row of A237593 is [23, 8, 4, 3, 2, 1, 1, 2, 2, 1, 1, 2, 3, 4, 8, 23] and the 43rd row of A237593 is [22, 8, 4, 3, 2, 1, 2, 1, 1, 2, 1, 2, 3, 4, 8, 23] therefore between both Dyck paths there are two regions (or parts) of sizes [42, 42], so row 11 is [42, 42].
The sum of divisors of 44 is 1 + 2 + 4 + 11 + 22 + 44 = A000203(44) = 84. On the other hand the sum of the parts of the symmetric representation of sigma(44) is 42 + 42 = 84, equaling the sum of divisors of 44.

Cf. A000203, A193553, A196020, A236104, A235791, A237048, A237270, A237271, A237591, A237593, A239660, A239931, A239932, A239933, A244050, A245092, A262626.

A296508 Irregular triangle read by rows: T(n,k) is the size of the subpart that is adjacent to the k-th peak of the largest Dyck path of the symmetric representation of sigma(n), or T(n,k) = 0 if the mentioned subpart is already associated to a previous peak or if there is no subpart adjacent to the k-th peak, with n >= 1, k >= 1.

Original entry on oeis.org

1, 3, 2, 2, 7, 0, 3, 3, 11, 1, 0, 4, 0, 4, 15, 0, 0, 5, 3, 5, 9, 0, 9, 0, 6, 0, 0, 6, 23, 5, 0, 0, 7, 0, 0, 7, 12, 0, 12, 0, 8, 7, 1, 0, 8, 31, 0, 0, 0, 0, 9, 0, 0, 0, 9, 35, 2, 0, 2, 0, 10, 0, 0, 0, 10, 39, 0, 3, 0, 0, 11, 5, 0, 5, 0, 11, 18, 0, 0, 0, 18, 0, 12, 0, 0, 0, 0, 12, 47, 13, 0, 0, 0, 0, 13, 0, 5, 0, 0, 13

Offset: 1

Omar E. Pol, Feb 10 2018

Conjecture: row n is formed by the odd-indexed terms of the n-th row of triangle A280850 together with the even-indexed terms of the same row but listed in reverse order. Examples: the 15th row of A280850 is [8, 8, 7, 0, 1] so the 15th row of this triangle is [8, 7, 1, 0, 8]. The 75th row of A280850 is [38, 38, 21, 0, 3, 3, 0, 0, 0, 21, 0] so the 75h row of this triangle is [38, 21, 3, 0, 0, 0, 21, 0, 3, 0, 38].
For the definition of "subparts" see A279387.
For more information about the mentioned Dyck paths see A237593.
T(n,k) could be called the "charge" of the k-th peak of the largest Dyck path of the symmetric representation of sigma(n).
The number of zeros in row n is A238005(n). - Omar E. Pol, Sep 11 2021

			Triangle begins (rows 1..28):
   1;
   3;
   2,  2;
   7,  0;
   3,  3;
  11,  1,  0;
   4,  0,  4;
  15,  0,  0;
   5,  3,  5;
   9,  0,  9,  0;
   6,  0,  0,  6;
  23,  5,  0,  0;
   7,  0,  0,  7;
  12,  0, 12,  0;
   8,  7,  1,  0,  8;
  31,  0,  0,  0,  0;
   9,  0,  0,  0,  9;
  35,  2,  0,  2,  0;
  10,  0,  0,  0, 10;
  39,  0,  3,  0,  0;
  11,  5,  0,  5,  0, 11;
  18,  0,  0,  0, 18,  0;
  12,  0,  0,  0,  0, 12;
  47, 13,  0,  0,  0,  0;
  13,  0,  5,  0,  0, 13;
  21,  0,  0,  0  21,  0;
  14,  6,  0,  6,  0, 14;
  55,  0,  0,  1,  0,  0,  0;
  ...
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) is constructed in the third quadrant as shown below in Figure 1:
.    _                                  _
.   | |                                | |
.   | |                                | |
.   | |                                | |
. 8 | |                                | |
.   | |                                | |
.   | |                                | |
.   | |                                | |
.   |_|_ _ _                           |_|_ _ _
.         | |_ _                      8      | |_ _
.         |_    |                            |_ _  |
.           |_  |_                          7  |_| |_
.          8  |_ _|                           1  |_ _|
.                 |                             0    |
.                 |_ _ _ _ _ _ _ _                   |_ _ _ _ _ _ _ _
.                 |_ _ _ _ _ _ _ _|                  |_ _ _ _ _ _ _ _|
.                         8                         8
.
.   Figure 1. The symmetric            Figure 2. 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        so the 15th row of this triangle is
.   triangle A237270 is [8, 8, 8].     [8, 7, 1, 0, 8]. See also below.
.
Illustration of first 50 terms (rows 1..16 of triangle) in an irregular spiral which can be find in the top view of the pyramid described in A244050:
.
.               12 _ _ _ _ _ _ _ _
.                 |  _ _ _ _ _ _ _|_ _ _ _ _ _ _ 7
.                 | |             |_ _ _ _ _ _ _|
.              0 _| |                           |
.               |_ _|9 _ _ _ _ _ _              |_ _ 0
.         12 _ _|     |  _ _ _ _ _|_ _ _ _ _ 5      |_ 0
.    0 _ _ _| |    0 _| |         |_ _ _ _ _|         |
.     |  _ _ _|  9 _|_ _|                   |_ _ 3    |_ _ _ 7
.     | |    0 _ _| |   11 _ _ _ _          |_  |         | |
.     | |     |  _ _|  1 _|  _ _ _|_ _ _ 3    |_|_ _ 5    | |
.     | |     | |    0 _|_| |     |_ _ _|         | |     | |
.     | |     | |     |  _ _|           |_ _ 3    | |     | |
.     | |     | |     | |    3 _ _        | |     | |     | |
.     | |     | |     | |     |  _|_ 1    | |     | |     | |
.    _|_|    _|_|    _|_|    _|_| |_|    _|_|    _|_|    _|_|    _
.   | |     | |     | |     | |         | |     | |     | |     | |
.   | |     | |     | |     |_|_ _     _| |     | |     | |     | |
.   | |     | |     | |    2  |_ _|_ _|  _|     | |     | |     | |
.   | |     | |     |_|_     2    |_ _ _|  0 _ _| |     | |     | |
.   | |     | |    4    |_               7 _|  _ _|0    | |     | |
.   | |     |_|_ _     0  |_ _ _ _        |  _|    _ _ _| |     | |
.   | |    6      |_      |_ _ _ _|_ _ _ _| |  0 _|  _ _ _|0    | |
.   |_|_ _ _     0  |_   4        |_ _ _ _ _|  _|  _| |    _ _ _| |
.  8      | |_ _   0  |                     15|  _|  _|   |  _ _ _|
.         |_ _  |     |_ _ _ _ _ _            | |_ _|  0 _| |      0
.        7  |_| |_    |_ _ _ _ _ _|_ _ _ _ _ _| |    5 _|  _|
.          1  |_ _|  6            |_ _ _ _ _ _ _|  _ _|  _|  0
.            0    |                             23|  _ _|  0
.                 |_ _ _ _ _ _ _ _                | |    0
.                 |_ _ _ _ _ _ _ _|_ _ _ _ _ _ _ _| |
.                8                |_ _ _ _ _ _ _ _ _|
.                                                    31
.
The diagram contains 30 subparts equaling A060831(16), the total number of partitions of all positive integers <= 16 into consecutive parts.
For the construction of the spiral see A239660.
From _Omar E. Pol_, Nov 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 n A280850 and the conjecture 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 this seq:  [              8,      7,    1,    0,      8              ]
The 15th row
of A280851:   [              8,      7,    1,            8              ]
.
(End)

Index entries for sequences related to sigma(n)

Row sums give A000203.
Row n has length A003056(n).
Column k starts in row A000217(k).
Nonzero terms give A280851.
The number of nonzero terms in row n is A001227(n).
The triangle with n rows contain A060831(n) nonzero terms.
Cf. A024916, A196020, A235791, A236104, A237048, A237270, A237591, A237593, A238005, A239657, A239660, A239931-A239934, A240542, A244050, A245092, A250068, A250070, A261699, A262626, A279387, A279388, A279391, A280850.

A175254 a(n) = Sum_{k<=n} A000203(k)*(n-k+1), where A000203(m) is the sum of divisors of m.

Original entry on oeis.org

1, 5, 13, 28, 49, 82, 123, 179, 248, 335, 434, 561, 702, 867, 1056, 1276, 1514, 1791, 2088, 2427, 2798, 3205, 3636, 4127, 4649, 5213, 5817, 6477, 7167, 7929, 8723, 9580, 10485, 11444, 12451, 13549, 14685, 15881, 17133, 18475, 19859, 21339, 22863, 24471, 26157

Offset: 1

Jaroslav Krizek, Mar 14 2010

Partial sums of A024916. - Omar E. Pol, Jul 03 2014
a(n) is also the volume of the stepped pyramid with n levels described in A245092. - Omar E. Pol, Aug 12 2015
Also the alternating row sums of A262612. - Omar E. Pol, Nov 23 2015
From Omar E. Pol, Jan 20 2021: (Start)
Convolution of A000203 and A000027.
Convolution of A340793 and the nonzero terms of A000217.
Antidiagonal sums of A319073.
Row sums of A274824. (End)
Row sums of A345272. - Omar E. Pol, Jun 14 2021
Also the alternating row sums of A353690. - Omar E. Pol, Jun 05 2022

			For n = 4: a(4) = sigma(1)*4 + sigma(2)*3 + sigma(3)*2 + sigma(4)*1 = 1*4 + 3*3 + 4*2 + 7*1 = 28.

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 2209 terms from Indranil Ghosh)

Cf. A000203, A000217, A006218, A024916, A072481, A237593, A244050, A245092, A262612, A274824, A319073, A340793, A345272, A353690.

Cf. A143128, A353908.

b:= proc(n) option remember; `if`(n<1, [0$2],
      (p-> p+[numtheory[sigma](n), p[1]])(b(n-1)))
    end:
a:= n-> b(n+1)[2]:
seq(a(n), n=1..45);  # Alois P. Heinz, Oct 07 2021

Table[Sum[DivisorSigma[1, k] (n - k + 1), {k, n}], {n, 45}] (* Michael De Vlieger, Nov 24 2015 *)

a(n) = sum(x=1, n, sigma(x)*(n-x+1)) \\ Michel Marcus, Mar 18 2013

from math import isqrt
def A175254(n): return (((s:=isqrt(n))**2*(s+1)*((s+1)*(2*s+1)-6*(n+1))>>1) + sum((q:=n//k)*(-k*(q+1)*(3*k+2*q+1)+3*(n+1)*(2*k+q+1)) for k in range(1,s+1)))//6 # Chai Wah Wu, Oct 21 2023

Conjecture: a(n) = Sum_{k=0..n} A006218(n-k). - R. J. Mathar, Oct 17 2012
a(n) = A000330(n) - A072481(n). - Omar E. Pol, Aug 12 2015
a(n) ~ Pi^2*n^3/36. - Vaclav Kotesovec, Sep 25 2016
G.f.: (1/(1 - x)^2)*Sum_{k>=1} k*x^k/(1 - x^k). - Ilya Gutkovskiy, Jan 03 2017
a(n) = Sum_{k=1..n} Sum_{i=1..k} k - (k mod i). - Wesley Ivan Hurt, Sep 13 2017
a(n) = A244050(n)/4. - Omar E. Pol, Jan 22 2021
a(n) = (n+1)*A024916(n) - A143128(n). - Vaclav Kotesovec, May 11 2022

Corrected by Jaroslav Krizek, Mar 17 2010

More terms from Michel Marcus, Mar 18 2013

A280850 Irregular triangle read by rows in which row n is constructed with an algorithm using the n-th row of triangle A196020 (see Comments for precise definition).

Original entry on oeis.org

1, 3, 2, 2, 7, 0, 3, 3, 11, 0, 1, 4, 4, 0, 15, 0, 0, 5, 5, 3, 9, 0, 0, 9, 6, 6, 0, 0, 23, 0, 5, 0, 7, 7, 0, 0, 12, 0, 0, 12, 8, 8, 7, 0, 1, 31, 0, 0, 0, 0, 9, 9, 0, 0, 0, 35, 0, 2, 2, 0, 10, 10, 0, 0, 0, 39, 0, 0, 0, 3, 11, 11, 5, 0, 0, 5, 18, 0, 0, 18, 0, 0, 12, 12, 0, 0, 0, 0, 47, 0, 13, 0, 0, 0

Offset: 1

Omar E. Pol, Jan 09 2017

For the construction of the n-th row of this triangle start with a copy of the n-th row of the triangle A196020.
Then replace each element of the m-th pair of positive integers (x, y) with the value (x - y)/2, where "y" is the m-th even-indexed term of the row, and "x" is its previous nearest odd-indexed term not used in another pair in the same row, if such a pair exist. Otherwise T(n,k) = A196020(n,k). (See example).
Observation 1: at least for the first 28 rows of the triangle the nonzero terms in the n-th row are also the subparts of the symmetric representation of sigma(n), assuming the ordering of the subparts in the same row does not matter.
Question 1: are always the nonzero terms of the n-th row the same as all the subparts of the symmetric representation of sigma(n)? If not, what is the index of the row in which appears the first counterexample?
Note that the "subparts" are the regions that arise after the dissection of the symmetric representation of sigma(n) into successive layers of width 1.
For more information about "subparts" see A279387 and A237593.
About the question 1, it appears that the n-th row of the triangle A280851 and the n-th row of this triangle contain the same nonzero numbers, though in different order; checked through n = 250000. - Hartmut F. W. Hoft, Jan 31 2018
From Omar E. Pol, Feb 02 2018: (Start)
Observation 2: at least for the first 28 rows of the triangle we have that in the n-th row the odd-indexed terms, from left to right, together with the even-indexed terms, from right to left, form a finite sequence in which the nonzero terms are the same as the n-th row of triangle A280851, which lists the subparts of the symmetric representation of sigma(n).
Question 2: Are always the same for all rows? If not, what is the index of the row in which appears the first counterexample? (End)
Conjecture: the odd-indexed terms of the n-th row together with the even-indexed terms of the same row but listed in reverse order give the n-th row of triangle A296508 (this is the same conjecture from A296508). - Omar E. Pol, Apr 20 2018

			Triangle begins (rows 1..28):
   1;
   3;
   2,  2;
   7,  0;
   3,  3;
  11,  0,  1;
   4,  4,  0;
  15,  0,  0;
   5,  5,  3;
   9,  0,  0,  9;
   6,  6,  0,  0;
  23,  0,  5,  0;
   7,  7,  0,  0;
  12,  0,  0, 12;
   8,  8,  7,  0,  1;
  31,  0,  0,  0,  0;
   9,  9,  0,  0,  0;
  35,  0,  2,  2,  0;
  10, 10,  0,  0,  0;
  39,  0,  0,  0,  3;
  11, 11,  5,  0,  0,  5;
  18,  0,  0, 18,  0,  0;
  12, 12,  0,  0,  0,  0;
  47,  0, 13,  0,  0,  0;
  13, 13,  0,  0,  5,  0;
  21,  0,  0, 21,  0,  0;
  14, 14,  6,  0,  0,  6;
  55,  0,  0,  0,  0,  0,  1;
  ...
An example of the algorithm.
For n = 75, the construction of the 75th row of this triangle is as shown below:
.
75th row of A196020:             [149,  73, 47, 0, 25, 19, 0, 0, 0,  5, 0]
.
Odd-indexed terms:                149       47     25      0     0      0
Even-indexed terms:                     73      0      19     0      5
.
First even-indexed nonzero term:        73
First pair:                       149   73
.                                   *----*
Difference: 149 - 73 =                76
76/2 = 38                           *----*
New first pair:                    38   38
.
Second even-indexed nonzero term:                      19
Second pair:                                       25  19
.                                                   *---*
Difference: 25 - 19 =                                 6
6/2 = 3                                             *---*
New second pair:                                    3   3
.
Third even-indexed nonzero term:                                     5
Third pair:                                  47                      5
.                                             *----------------------*
Difference: 47 - 5 =                                     42
42/2 = 21                                     *----------------------*
New third pair:                              21                     21
.
So the 75th row
of this triangle is                [38,  38, 21, 0, 3,  3, 0, 0, 0, 21, 0]
.
On the other hand, the 75th row of A237593 is [38, 13, 7, 4, 3, 3, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 3, 3, 4, 7, 13, 38], and the 74th row of the same triangle is [38, 13, 6, 5, 3, 2, 2, 1, 2, 1, 1, 1, 1, 2, 1, 2, 2, 3, 5, 6, 13, 38], therefore between both symmetric Dyck paths (described in A237593 and A279387) there are six subparts: [38, 38, 21, 21, 3, 3]. (The diagram of the symmetric representation of sigma(75) is too large to include.) At least in this case the nonzero terms of the 75th row of the triangle coincide with the subparts of the symmetric representation of sigma(75). The ordering of the elements does not matter.
Continuing with the original example, in the 75th row of this triangle we have that the odd-indexed terms, from left to right, together with the even-indexed terms, from right to left, form the finite sequence [38, 21, 3, 0, 0, 0, 21, 0, 3, 0, 38] which is the 75th row of a triangle. At least in this case the nonzero terms coincide with the 75th row of triangle A280851: [38, 21, 3, 21, 3, 38], which lists the six subparts of the symmetric representation of sigma(75) in order of appearance from left to right. - _Omar E. Pol_, Feb 02 2018
In accordance with the conjecture from the Comments section, the finite sequence [38, 21, 3, 0, 0, 0, 21, 0, 3, 0, 38] mentioned above should be the 75th row of triangle A296508. - _Omar E. Pol_, Apr 20 2018

Index entries for sequences related to sigma(n)

Row sums give A000203.
The number of positive terms in row n is A001227(n).
Row n has length A003056(n).
Column k starts in row A000217(k).
Cf. A196020, A235791, A236104, A237048, A237270, A237591, A237593, A239657, A239660, A244050, A245092, A250068, A250070, A261699, A262626, A279387, A279388, A279391, A280851, A296508.

(* functions row[], line[] and their support are defined in A196020 *)
(* maintain a stack of odd indices with nonzero entries for matching *)
a280850[n_] := Module[{a=line[n], r=row[n], stack={1}, i, j, b}, For[i=2, i<=r, i++, If[a[[i]]!=0, If[OddQ[i], AppendTo[stack, i], j=Last[stack]; b=(a[[j]]-a[[i]])/2; a[[i]]=b; a[[j]]=b; stack=Drop[stack, -1]]]]; a]
Flatten[Map[a280850,Range[24]]] (* data *)
TableForm[Map[a280850, Range[28]], TableDepth->2] (* triangle in Example *)
(* Hartmut F. W. Hoft, Jan 31 2018 *)

Name edited by Omar E. Pol, Nov 11 2018

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

A239050 a(n) = 4*sigma(n).

Original entry on oeis.org

4, 12, 16, 28, 24, 48, 32, 60, 52, 72, 48, 112, 56, 96, 96, 124, 72, 156, 80, 168, 128, 144, 96, 240, 124, 168, 160, 224, 120, 288, 128, 252, 192, 216, 192, 364, 152, 240, 224, 360, 168, 384, 176, 336, 312, 288, 192, 496, 228, 372, 288, 392, 216, 480, 288, 480, 320, 360, 240, 672, 248, 384, 416, 508

Offset: 1

Omar E. Pol, Mar 09 2014

4 times the sum of divisors of n.

a(n) is also the total number of horizontal cells in the terraces of the n-th level of an irregular stepped pyramid (starting from the top) where the structure of every three-dimensional quadrant arises after the 90-degree zig-zag folding of every row of the diagram of the isosceles triangle A237593. The top of the pyramid is a square formed by four cells (see links and examples). - Omar E. Pol, Jul 04 2016

			For n = 4 the sum of divisors of 4 is 1 + 2 + 4 = 7, so a(4) = 4*7 = 28.
For n = 5 the sum of divisors of 5 is 1 + 5 = 6, so a(5) = 4*6 = 24.
.
Illustration of initial terms:                                    _ _ _ _ _ _
.                                           _ _ _ _ _ _          |_|_|_|_|_|_|
.                           _ _ _ _       _|_|_|_|_|_|_|_     _ _|           |_ _
.             _ _ _ _     _|_|_|_|_|_    |_|_|       |_|_|   |_|               |_|
.     _ _    |_|_|_|_|   |_|       |_|   |_|           |_|   |_|               |_|
.    |_|_|   |_|   |_|   |_|       |_|   |_|           |_|   |_|               |_|
.    |_|_|   |_|_ _|_|   |_|       |_|   |_|           |_|   |_|               |_|
.            |_|_|_|_|   |_|_ _ _ _|_|   |_|_         _|_|   |_|               |_|
.                          |_|_|_|_|     |_|_|_ _ _ _|_|_|   |_|_             _|_|
.                                          |_|_|_|_|_|_|         |_ _ _ _ _ _|
.                                                                |_|_|_|_|_|_|
.
n:     1          2             3                4                     5
S(n):  1          3             4                7                     6
a(n):  4         12            16               28                    24
.
For n = 1..5, the figure n represents the reflection in the four quadrants of the symmetric representation of S(n) = sigma(n) = A000203(n). For more information see A237270 and A237593.
The diagram also represents the top view of the first four terraces of the stepped pyramid described in Comments section. - _Omar E. Pol_, Jul 04 2016

Antti Karttunen, Table of n, a(n) for n = 1..10000
Omar E. Pol, Diagram of the triangle before the 90-degree-zig-zag folding (rows: 1..28)
Omar E. Pol, Folding the first eight rows of triangle
Index entries for sequences related to sigma(n)

Alternating row sums of A239662.
Partial sums give A243980.
k times sigma(n), k=1..6: A000203, A074400, A272027, this sequence, A274535, A274536.
k times sigma(n), k = 1..10: A000203, A074400, A272027, this sequence, A274535, A274536, A319527, A319528, A325299, A326122.
Cf. A008438, A017113, A062731, A112610, A144613, A193553, A196020, A235791, A236104, A237270, A237593, A239052, A239053, A239660, A239662, A244050, A262626.

[4*SumOfDivisors(n): n in [1..70]]; // Vincenzo Librandi, Jul 30 2019

with(numtheory): seq(4*sigma(n), n=1..64); # Omar E. Pol, Jul 04 2016

Array[4 DivisorSigma[1, #] &, 64] (* Michael De Vlieger, Nov 16 2017 *)

a(n) = 4 * sigma(n); \\ Omar E. Pol, Jul 04 2016

a(n) = 4*A000203(n) = 2*A074400(n).
a(n) = A000203(n) + A272027(n). - Omar E. Pol, Jul 04 2016
Dirichlet g.f.: 4*zeta(s-1)*zeta(s). - Ilya Gutkovskiy, Jul 04 2016
Conjecture: a(n) = sigma(3*n) = A144613(n) iff n is not a multiple of 3. - Omar E. Pol, Oct 02 2018
The conjecture above is correct. Write n = 3^e*m, gcd(3, m) = 1, then sigma(3*n) = sigma(3^(e+1))*sigma(m) = ((3^(e+2) - 1)/2)*sigma(m) = ((3^(e+2) - 1)/(3^(e+1) - 1))*sigma(3^e*m), and (3^(e+2) - 1)/(3^(e+1) - 1) = 4 if and only if e = 0. - Jianing Song, Feb 03 2019

A239929 Numbers n with the property that the symmetric representation of sigma(n) has two parts.

Original entry on oeis.org

3, 5, 7, 10, 11, 13, 14, 17, 19, 22, 23, 26, 29, 31, 34, 37, 38, 41, 43, 44, 46, 47, 52, 53, 58, 59, 61, 62, 67, 68, 71, 73, 74, 76, 78, 79, 82, 83, 86, 89, 92, 94, 97, 101, 102, 103, 106, 107, 109, 113, 114, 116, 118, 122, 124, 127, 131, 134, 136, 137, 138

Offset: 1

Omar E. Pol, Apr 06 2014

nonn

All odd primes are in the sequence because the parts of the symmetric representation of sigma(prime(i)) are [m, m], where m = (1 + prime(i))/2, for i >= 2.
There are no odd composite numbers in this sequence.
First differs from A173708 at a(13).
Since sigma(p*q) >= 1 + p + q + p*q for odd p and q, the symmetric representation of sigma(p*q) has more parts than the two extremal ones of size (p*q + 1)/2; therefore, the above comments are true. - Hartmut F. W. Hoft, Jul 16 2014
From Hartmut F. W. Hoft, Sep 16 2015: (Start)
The following two statements are equivalent:
  (1) The symmetric representation of sigma(n) has two parts, and
  (2) n = q * p where q is in A174973, p is prime, and 2 * q < p.
For a proof see the link and also the link in A071561.
This characterization allows for much faster computation of numbers in the sequence - function a239929F[] in the Mathematica section - than computations based on Dyck paths. The function a239929Stalk[] gives rise to the associated irregular triangle whose columns are indexed by A174973 and whose rows are indexed by A065091, the odd primes. (End)
From Hartmut F. W. Hoft, Dec 06 2016: (Start)
For the respective columns of the irregular triangle with fixed m: k = 2^m * p, m >= 1, 2^(m+1) < p and p prime:
(a) each number k is representable as the sum of 2^(m+1) but no fewer consecutive positive integers [since 2^(m+1) < p].
(b) each number k has 2^m as largest divisor <= sqrt(k) [since 2^m < sqrt(k) < p].
(c) each number k is of the form 2^m * p with p prime [by definition].
m = 1: (a) A100484 even semiprimes (except 4 and 6)
       (b) A161344 (except 4, 6 and 8)
       (c) A001747 (except 2, 4 and 6)
m = 2: (a) A270298
       (b) A161424 (except 16, 20, 24, 28 and 32)
       (c) A001749 (except 8, 12, 20 and 28)
m = 3: (a) A270301
       (b) A162528 (except 64, 72, 80, 88, 96, 104, 112 and 128)
       (c) sequence not in OEIS
b(i,j) = A174973(j) * {1,5) mod 6 * A174973(j), for all i,j >= 1; see A091999 for j=2. (End)

			From _Hartmut F. W. Hoft_, Sep 16 2015: (Start)
a(23) = 52 = 2^2 * 13 = q * p with q = 4 in A174973 and 8 < 13 = p.
a(59) = 136 = 2^3 * 17 = q * p with q = 8 in A174973 and 16 < 17 = p.
The first six columns of the irregular triangle through prime 37:
   1    2    4    6    8   12 ...
  -------------------------------
   3
   5   10
   7   14
  11   22   44
  13   26   52   78
  17   34   68  102  136
  19   38   76  114  152
  23   46   92  138  184
  29   58  116  174  232  348
  31   62  124  186  248  372
  37   74  148  222  296  444
  ...
(End)

Hartmut F. W. Hoft, Proof of Characterization Theorem

Column 2 of A240062.
Cf. A000203, A006254, A065091, A071561, A174973, A196020, A236104, A235791, A237591, A237593, A237270, A237271, A238443, A239660, A239663, A239665, A239931-A239934, A244050, A245062, A262626.
Cf. A000040, A001747, A001749, A091999, A100484, A161344, A161424, A162528, A270298, A270301. - Hartmut F. W. Hoft, Dec 06 2016

isA174973 := proc(n)
    option remember;
    local k,dvs;
    dvs := sort(convert(numtheory[divisors](n),list)) ;
    for k from 2 to nops(dvs) do
        if op(k,dvs) > 2*op(k-1,dvs) then
            return false;
        end if;
    end do:
    true ;
end proc:
A174973 := proc(n)
    if n = 1 then
        1;
    else
        for a from procname(n-1)+1 do
            if isA174973(a) then
                return a;
            end if;
        end do:
    end if;
end proc:
isA239929 := proc(n)
    local i,p,j,a73;
    for i from 1 do
        p := ithprime(i+1) ;
        if p > n then
            return false;
        end if;
        for j from 1 do
            a73 := A174973(j) ;
            if a73 > n then
                break;
            end if;
            if p > 2*a73 and n = p*a73 then
                return true;
            end if;
        end do:
    end do:
end proc:
for n from 1 to 200 do
    if isA239929(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Oct 04 2018

(* sequence of numbers k for m <= k <= n having exactly two parts *)
(* Function a237270[] is defined in A237270 *)
a239929[m_, n_]:=Select[Range[m, n], Length[a237270[#]]==2&]
a239929[1, 260] (* data *)
(* Hartmut F. W. Hoft, Jul 07 2014 *)
(* test for membership in A174973 *)
a174973Q[n_]:=Module[{d=Divisors[n]}, Select[Rest[d] - 2 Most[d], #>0&]=={}]
a174973[n_]:=Select[Range[n], a174973Q]
(* compute numbers satisfying the condition *)
a239929Stalk[start_, bound_]:=Module[{p=NextPrime[2 start], list={}}, While[start p<=bound, AppendTo[list, start p]; p=NextPrime[p]]; list]
a239929F[n_]:=Sort[Flatten[Map[a239929Stalk[#, n]&, a174973[n]]]]
a239929F[138] (* data *)(* Hartmut F. W. Hoft, Sep 16 2015 *)

Entries b(i, j) in the irregular triangle with rows indexed by i>=1 and columns indexed by j>=1 (alternate indexing of the example):

b(i,j) = A000040(i+1) * A174973(j) where A000040(i+1) > 2 * A174973(j). - Hartmut F. W. Hoft, Dec 06 2016

Extended beyond a(56) by Michel Marcus, Apr 07 2014

A239932 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(4n-2).

Original entry on oeis.org

3, 12, 9, 9, 12, 12, 39, 18, 18, 21, 21, 72, 27, 27, 30, 30, 96, 36, 36, 39, 15, 39, 120, 45, 45, 48, 48, 144, 54, 36, 54, 57, 57, 84, 84, 63, 63, 66, 66, 234, 72, 72, 75, 21, 75, 108, 108, 81, 81, 84, 48, 84, 120, 120, 90, 90, 93, 93, 312

Offset: 1

Omar E. Pol, Mar 29 2014

Row n is a palindromic composition of sigma(4n-2).
Row n is also the row 4n-2 of A237270.
Row n has length A237271(4n-2).
Row sums give A239052.
Also row n lists the parts of the symmetric representation of sigma in the n-th arm of the second quadrant of the spiral described in A239660, see example.
For the parts of the symmetric representation of sigma(4n-3), see A239931.
For the parts of the symmetric representation of sigma(4n-1), see A239933.
For the parts of the symmetric representation of sigma(4n), see A239934.
We can find the spiral (mentioned above) on the terraces of the pyramid described in A244050. - Omar E. Pol, Dec 06 2016

			The irregular triangle begins:
3;
12;
9, 9;
12, 12;
39;
18, 18;
21, 21;
72;
27, 27;
30, 30;
96;
36, 36;
39, 15, 39;
120;
45, 45;
48, 48;
...
Illustration of initial terms in the second quadrant of the spiral described in A239660:
.                                 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
.                                |  _ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
.                                | |
.                                | |
.                                | |  _ _ _ _ _ _ _ _ _ _ _ _ _ _
.                           _ _ _| | |  _ _ _ _ _ _ _ _ _ _ _ _ _|
.                          |       | | |
.                       _ _|  _ _ _| | |
.                  72 _|     |       | |  _ _ _ _ _ _ _ _ _ _ _ _
.                   _|      _| 21 _ _| | |  _ _ _ _ _ _ _ _ _ _ _|
.                  |      _|     |_ _ _| | |
.               _ _|    _|    _ _|       | |
.              |    _ _|    _|     18 _ _| |  _ _ _ _ _ _ _ _ _ _
.              |   |       |         |_ _ _| |  _ _ _ _ _ _ _ _ _|
.     _ _ _ _ _|   | 21 _ _|        _|       | |
.    |  _ _ _ _ _ _|   | |        _|      _ _| |
.    | |      _ _ _ _ _| | 18 _ _|       |     |  _ _ _ _ _ _ _ _
.    | |     |  _ _ _ _ _|   | |     39 _|  _ _| |  _ _ _ _ _ _ _|
.    | |     | |      _ _ _ _| |    _ _|  _|     | |
.    | |     | |     |  _ _ _ _|   |    _|   12 _| |
.    | |     | |     | |      _ _ _|   |       |_ _|  _ _ _ _ _ _
.    | |     | |     | |     |  _ _ _ _| 12 _ _|     |  _ _ _ _ _|
.    | |     | |     | |     | |      _ _ _| |    9 _| |
.    | |     | |     | |     | |     |  _ _ _|  9 _|_ _|
.    | |     | |     | |     | |     | |      _ _| |      _ _ _ _
.    | |     | |     | |     | |     | |     |  _ _| 12 _|  _ _ _|
.    | |     | |     | |     | |     | |     | |      _|   |
.    | |     | |     | |     | |     | |     | |     |  _ _|
.    | |     | |     | |     | |     | |     | |     | |    3 _ _
.    | |     | |     | |     | |     | |     | |     | |     |  _|
.    |_|     |_|     |_|     |_|     |_|     |_|     |_|     |_|
.
For n = 7 we have that 4*7-2 = 26 and the 26th row of A237593 is [14, 5, 2, 2, 2, 1, 1, 2, 2, 2, 5, 14] and the 25th row of A237593 is [13, 5, 3, 1, 2, 1, 1, 2, 1, 3, 5, 13] therefore between both Dyck paths there are two regions (or parts) of sizes [21, 21], so row 7 is [21, 21].
The sum of divisors of 26 is 1 + 2 + 13 + 26 = A000203(26) = 42. On the other hand the sum of the parts of the symmetric representation of sigma(26) is 21 + 21 = 42, equaling the sum of divisors of 26.

Cf. A000203, A196020, A236104, A235791, A237270, A237271, A237591, A237593, A239052, A239660, A239931, A239933, A239934, A244050, A245092, A262626.

A239933 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(4n-1).

Original entry on oeis.org

2, 2, 4, 4, 6, 6, 8, 8, 8, 10, 10, 12, 12, 14, 6, 6, 14, 16, 16, 18, 12, 18, 20, 8, 8, 20, 22, 22, 24, 24, 26, 10, 10, 26, 28, 8, 8, 28, 30, 30, 32, 12, 16, 12, 32, 34, 34, 36, 36, 38, 24, 24, 38, 40, 40, 42, 42, 44, 16, 16, 44, 46, 20, 46, 48, 12, 12, 48, 50, 18, 20, 18, 50, 52, 52, 54, 54, 56, 20, 20, 56, 58, 14, 14, 58, 60, 12, 12, 60, 62, 22, 22, 62, 64, 64

Offset: 1

Omar E. Pol, Mar 29 2014

Row n is a palindromic composition of sigma(4n-1).
Row n is also the row 4n-1 of A237270.
Row n has length A237271(4n-1).
Row sums give A239053.
Also row n lists the parts of the symmetric representation of sigma in the n-th arm of the third quadrant of the spiral described in A239660, see example.
For the parts of the symmetric representation of sigma(4n-3), see A239931.
For the parts of the symmetric representation of sigma(4n-2), see A239932.
For the parts of the symmetric representation of sigma(4n), see A239934.
We can find the spiral (mentioned above) on the terraces of the pyramid described in A244050. - Omar E. Pol, Dec 06 2016

			The irregular triangle begins:
2, 2;
4, 4;
6, 6;
8, 8, 8;
10, 10;
12, 12;
14, 6, 6, 14;
16, 16;
18, 12, 18;
20, 8, 8, 20;
22, 22;
24, 24;
26, 10, 10, 26;
28, 8, 8, 28;
30, 30;
32, 12, 16, 12, 32;
...
Illustration of initial terms in the third quadrant of the spiral described in A239660:
.     _       _       _       _       _       _       _       _
.    | |     | |     | |     | |     | |     | |     | |     | |
.    | |     | |     | |     | |     | |     | |     | |     |_|_ _
.    | |     | |     | |     | |     | |     | |     | |    2  |_ _|
.    | |     | |     | |     | |     | |     | |     |_|_     2
.    | |     | |     | |     | |     | |     | |    4    |_
.    | |     | |     | |     | |     | |     |_|_ _        |_ _ _ _
.    | |     | |     | |     | |     | |    6      |_      |_ _ _ _|
.    | |     | |     | |     | |     |_|_ _ _        |_   4
.    | |     | |     | |     | |    8      | |_ _      |
.    | |     | |     | |     |_|_ _ _      |_    |     |_ _ _ _ _ _
.    | |     | |     | |   10        |       |_  |_    |_ _ _ _ _ _|
.    | |     | |     |_|_ _ _ _      |_ _   8  |_ _|  6
.    | |     | |   12          |         |_        |
.    | |     |_|_ _ _ _ _      |_ _        |       |_ _ _ _ _ _ _ _
.    | |   14          | |         |_      |_ _    |_ _ _ _ _ _ _ _|
.    |_|_ _ _ _ _      | |_ _        |_        |  8
.  16            |     |_ _  |         |       |
.                |         |_|_        |_ _    |_ _ _ _ _ _ _ _ _ _
.                |_ _     6    |_ _        |   |_ _ _ _ _ _ _ _ _ _|
.                    |         |_  |       | 10
.                    |_       6  | |_ _    |
.                      |_        |_ _ _|   |_ _ _ _ _ _ _ _ _ _ _ _
.                        |_ _          |   |_ _ _ _ _ _ _ _ _ _ _ _|
.                            |         | 12
.                            |_ _ _    |
.                                  |   |_ _ _ _ _ _ _ _ _ _ _ _ _ _
.                                  |   |_ _ _ _ _ _ _ _ _ _ _ _ _ _|
.                                  | 14
.                                  |
.                                  |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
.                                  |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
.                                16
.
For n = 7 we have that 4*7-1 = 27 and the 27th row of A237593 is [14, 5, 3, 2, 1, 2, 2, 1, 2, 3, 5, 14] and the 26th row of A237593 is [14, 5, 2, 2, 2, 1, 1, 2, 2, 2, 5, 14] therefore between both Dyck paths there are four regions (or parts) of sizes [14, 6, 6, 14], so row 7 is [14, 6, 6, 14].
The sum of divisors of 27 is 1 + 3 + 9 + 27 = A000203(27) = 40. On the other hand the sum of the parts of the symmetric representation of sigma(27) is 14 + 6 + 6 + 14 = 40, equaling the sum of divisors of 27.

Cf. A000203, A196020, A236104, A235791, A237270, A237271, A237591, A237593, A239053, A239660, A239931, A239932, A239934, A244050, A245092, A262626.

cp's OEIS Frontend

A239931 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(4n-3).

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A239934 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(4n).

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A175254 a(n) = Sum_{k<=n} A000203(k)*(n-k+1), where A000203(m) is the sum of divisors of m.

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A280850 Irregular triangle read by rows in which row n is constructed with an algorithm using the n-th row of triangle A196020 (see Comments for precise definition).

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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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A239050 a(n) = 4*sigma(n).

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A239929 Numbers n with the property that the symmetric representation of sigma(n) has two parts.

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