A286000 -id:A286000 - cp's OEIS frontend

A237593 Triangle read by rows in which row n lists the elements of the n-th row of A237591 followed by the same elements in reverse order.

1, 1, 2, 2, 2, 1, 1, 2, 3, 1, 1, 3, 3, 2, 2, 3, 4, 1, 1, 1, 1, 4, 4, 2, 1, 1, 2, 4, 5, 2, 1, 1, 2, 5, 5, 2, 2, 2, 2, 5, 6, 2, 1, 1, 1, 1, 2, 6, 6, 3, 1, 1, 1, 1, 3, 6, 7, 2, 2, 1, 1, 2, 2, 7, 7, 3, 2, 1, 1, 2, 3, 7, 8, 3, 1, 2, 2, 1, 3, 8, 8, 3, 2, 1, 1, 1, 1, 2, 3, 8

Offset: 1

Omar E. Pol, Feb 22 2014

Row n is a palindromic composition of 2*n.
T(n,k) is also the length of the k-th segment in a Dyck path on the first quadrant of the square grid, connecting the x-axis with the y-axis, from (n, 0) to (0, n), starting with a segment in vertical direction, see example.
Conjecture 1: the area under the n-th Dyck path equals A024916(n), the sum of all divisors of all positive integers <= n.
If the conjecture is true then the n-th Dyck path represents the boundary segments after the alternating sum of the elements of the n-th row of A236104.
Conjecture 2: two adjacent Dyck paths never cross (checked by hand up to n = 128), hence the total area between the n-th Dyck path and the (n-1)-st Dyck path is equal to sigma(n) = A000203(n), the sum of divisors of n.
The connection between A196020 and A237271 is as follows: A196020 --> A236104 --> A235791 --> A237591 --> this sequence --> A239660 --> A237270 --> A237271.
PARI scripts area(n) and chkcross(n) have been written to check the 2 properties and have been run up to n=10000. - Michel Marcus, Mar 27 2014
Mathematica functions have been written that verified the 2 properties through n=30000. - Hartmut F. W. Hoft, Apr 07 2014
Comments from Franklin T. Adams-Watters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014: (Start)
The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.
You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.
Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n).
Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)
The symmetric representation of sigma, so defined, is row n of A237270. - Peter Munn, Jan 06 2025
It appears that, for the n-th set, the number of cells lying on the first diagonal is equal to A067742(n), the number of middle divisors of n. - Michel Marcus, Jun 21 2014
Checked Michel Marcus's conjecture with two Mathematica functions up to n=100000, for more information see A240542. - Hartmut F. W. Hoft, Jul 17 2014
A003056(n) is also the number of peaks of the Dyck path related to the n-th row of triangle. - Omar E. Pol, Nov 03 2015
The number of peaks of the Dyck path associated to the row A000396(n) of this triangle equals the n-th Mersenne prime A000668(n), hence Mersenne primes are visible in two ways at the pyramid described in A245092. - Omar E. Pol, Dec 19 2016
The limit as n approaches infinity (area under the Dyck path described in the n-th row of triangle divided by n^2) equals Pi^2/12 = zeta(2)/2. (Cf. A072691.) - Omar E. Pol, Dec 18 2021
The connection between the isosceles triangle and the stepped pyramid is due to the fact that this object can also be interpreted as a pop-up card. - Omar E. Pol, Nov 09 2022

			Triangle begins:
   n
   1 |  1, 1;
   2 |  2, 2;
   3 |  2, 1, 1, 2;
   4 |  3, 1, 1, 3;
   5 |  3, 2, 2, 3;
   6 |  4, 1, 1, 1, 1, 4;
   7 |  4, 2, 1, 1, 2, 4;
   8 |  5, 2, 1, 1, 2, 5;
   9 |  5, 2, 2, 2, 2, 5;
  10 |  6, 2, 1, 1, 1, 1, 2, 6;
  11 |  6, 3, 1, 1, 1, 1, 3, 6;
  12 |  7, 2, 2, 1, 1, 2, 2, 7;
  13 |  7, 3, 2, 1, 1, 2, 3, 7;
  14 |  8, 3, 1, 2, 2, 1, 3, 8;
  15 |  8, 3, 2, 1, 1, 1, 1, 2, 3, 8;
  16 |  9, 3, 2, 1, 1, 1, 1, 2, 3, 9;
  17 |  9, 4, 2, 1, 1, 1, 1, 2, 4, 9;
  18 | 10, 3, 2, 2, 1, 1, 2, 2, 3, 10;
  19 | 10, 4, 2, 2, 1, 1, 2, 2, 4, 10;
  20 | 11, 4, 2, 1, 2, 2, 1, 2, 4, 11;
  21 | 11, 4, 3, 1, 1, 1, 1, 1, 1, 3, 4, 11;
  22 | 12, 4, 2, 2, 1, 1, 1, 1, 2, 2, 4, 12;
  23 | 12, 5, 2, 2, 1, 1, 1, 1, 2, 2, 5, 12;
  24 | 13, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 13;
  ...
Illustration of rows 8 and 9 interpreted as Dyck paths in the first quadrant and the illustration of the symmetric representation of sigma(9) = 5 + 3 + 5 = 13, see below:
.
y                       y
.                       .
.                       ._ _ _ _ _                _ _ _ _ _ 5
._ _ _ _ _              .         |              |_ _ _ _ _|
.         |             .         |_ _                     |_ _ 3
.         |_            .             |                    |_  |
.           |_ _        .             |_ _                   |_|_ _ 5
.               |       .                 |                      | |
.   Area = 56   |       .    Area = 69    |          Area = 13   | |
.               |       .                 |                      | |
.               |       .                 |                      | |
. . . . . . . . | . x   . . . . . . . . . | . x                  |_|
.
.    Fig. 1                    Fig. 2                  Fig. 3
.
Figure 1. For n = 8 the 8th row of triangle is [5, 2, 1, 1, 2, 5] and the area under the symmetric Dyck path is equal to A024916(8) = 56.
Figure 2. For n = 9 the 9th row of triangle is [5, 2, 2, 2, 2, 5] and the area under the symmetric Dyck path is equal to A024916(9) = 69.
Figure 3. The symmetric representation of sigma(9): between both symmetric Dyck paths there are three regions (or parts) of sizes [5, 3, 5].
The sum of divisors of 9 is 1 + 3 + 9 = A000203(9) = 13. On the other hand the difference between the areas under the Dyck paths equals the sum of the parts of the symmetric representation of sigma(9) = 69 - 56 = 5 + 3 + 5 = 13, equaling the sum of divisors of 9.
.
Illustration of initial terms as Dyck paths in the first quadrant:
(row n = 1..28)
.  _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
  |_ _ _ _ _ _ _ _ _ _ _ _ _ _  |
  |_ _ _ _ _ _ _ _ _ _ _ _ _ _| |
  |_ _ _ _ _ _ _ _ _ _ _ _ _  | |
  |_ _ _ _ _ _ _ _ _ _ _ _ _| | |
  |_ _ _ _ _ _ _ _ _ _ _ _  | | |_ _ _
  |_ _ _ _ _ _ _ _ _ _ _ _| | |_ _ _  |
  |_ _ _ _ _ _ _ _ _ _ _  | | |_ _  | |_
  |_ _ _ _ _ _ _ _ _ _ _| | |_ _ _| |_  |_
  |_ _ _ _ _ _ _ _ _ _  | |       |_ _|   |_
  |_ _ _ _ _ _ _ _ _ _| | |_ _    |_  |_ _  |_ _
  |_ _ _ _ _ _ _ _ _  | |_ _ _|     |_  | |_ _  |
  |_ _ _ _ _ _ _ _ _| | |_ _  |_      |_|_ _  | |
  |_ _ _ _ _ _ _ _  | |_ _  |_ _|_        | | | |_ _ _ _ _
  |_ _ _ _ _ _ _ _| |     |     | |_ _    | |_|_ _ _ _ _  |
  |_ _ _ _ _ _ _  | |_ _  |_    |_  | |   |_ _ _ _ _  | | |
  |_ _ _ _ _ _ _| |_ _  |_  |_ _  | | |_ _ _ _ _  | | | | |
  |_ _ _ _ _ _  | |_  |_  |_    | |_|_ _ _ _  | | | | | | |
  |_ _ _ _ _ _| |_ _|   |_  |   |_ _ _ _  | | | | | | | | |
  |_ _ _ _ _  |     |_ _  | |_ _ _ _  | | | | | | | | | | |
  |_ _ _ _ _| |_      | |_|_ _ _  | | | | | | | | | | | | |
  |_ _ _ _  |_ _|_    |_ _ _  | | | | | | | | | | | | | | |
  |_ _ _ _| |_  | |_ _ _  | | | | | | | | | | | | | | | | |
  |_ _ _  |_  |_|_ _  | | | | | | | | | | | | | | | | | | |
  |_ _ _|   |_ _  | | | | | | | | | | | | | | | | | | | | |
  |_ _  |_ _  | | | | | | | | | | | | | | | | | | | | | | |
  |_ _|_  | | | | | | | | | | | | | | | | | | | | | | | | |
  |_  | | | | | | | | | | | | | | | | | | | | | | | | | | |
  |_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|
.
n: 1 2 3 4 5 6 7 8 9 10..12..14..16..18..20..22..24..26..28
.
It appears that the total area (also the total number of cells) in the first n set of symmetric regions of the diagram is equal to A024916(n), the sum of all divisors of all positive integers <= n.
It appears that the total area (also the total number of cells) in the n-th set of symmetric regions of the diagram is equal to sigma(n) = A000203(n) (checked by hand up n = 128).
From _Omar E. Pol_, Aug 18 2015: (Start)
The above diagram is also the top view of the stepped pyramid described in A245092 and it is also the top view of the staircase described in A244580, in both cases the figure represents the first 28 levels of the structure. Note that the diagram contains (and arises from) a hidden pattern which is shown below.
.
Illustration of initial terms as an isosceles triangle:
Row                                 _ _
1                                 _|1|1|_
2                               _|2 _|_ 2|_
3                             _|2  |1|1|  2|_
4                           _|3   _|1|1|_   3|_
5                         _|3    |2 _|_ 2|    3|_
6                       _|4     _|1|1|1|1|_     4|_
7                     _|4      |2  |1|1|  2|      4|_
8                   _|5       _|2 _|1|1|_ 2|_       5|_
9                 _|5        |2  |2 _|_ 2|  2|        5|_
10              _|6         _|2  |1|1|1|1|  2|_         6|_
11            _|6          |3   _|1|1|1|1|_   3|          6|_
12          _|7           _|2  |2  |1|1|  2|  2|_           7|_
13        _|7            |3    |2 _|1|1|_ 2|    3|            7|_
14      _|8             _|3   _|1|2 _|_ 2|1|_   3|_             8|_
15    _|8              |3    |2  |1|1|1|1|  2|    3|              8|_
16   |9                |3    |2  |1|1|1|1|  2|    3|                9|
...
This diagram is the simpler representation of the sequence.
The number of horizontal line segments in the n-th level in each side of the diagram equals A001227(n), the number of odd divisors of n.
The number of horizontal line segments in the left side of the diagram plus the number of the horizontal line segment in the right side equals A054844(n).
The total number of vertical line segments in the n-th level of the diagram equals A131507(n).
Note that this symmetric pattern also emerges from the front view of the stepped pyramid described in A245092, which is related to sigma A000203, the sum-of-divisors function, and other related sequences. The diagram represents the first 16 levels of the pyramid. (End)

Robert Price, Table of n, a(n) for n = 1..15008 (rows n = 1..412, flattened)
Michel Marcus, A colored version of the symmetric representation of sigma(n), multipage, n = 1..85
Omar E. Pol, An infinite stepped pyramid
Omar E. Pol, Illustration of initial terms of A000203 in the pyramid
Omar E. Pol, Illustration of initial terms of A001065 in the pyramid
Omar E. Pol, Illustration of initial terms of A048050 in the pyramid
Omar E. Pol, Illustration of initial terms of A067742 in the pyramid
Omar E. Pol, Illustration of initial terms of A224613, (black spiders)
Omar E. Pol, Illustration of initial terms as an isosceles triangle (rows: 1..28)
Omar E. Pol, The symmetric representation of sigma(n), n = 1..64, (version with six colors)
Index entries for sequences related to sigma(n)

Row n has length 2*A003056(n).
Row sums give A005843, n >= 1.
Column k starts in row A008805(k-1).
Column 1 = right border = A008619, n >= 1.
Bisections are in A259176, A259177.
For further information see A262626.
Cf. A000203, A000217, A001065, A001227, A024916, A048050, A054844, A067742, A072691, A131507, A196020, A221529, A224613, A235791, A236104, A237048, A237270, A237271, A237590, A237591, A239660, A239931-A239934, A244050, A244580, A245092, A249351, A261350, A261699, A262611, A262612, A279387, A280850, A280851, A286000, A286001, A296508, A335616, A340035.

row[n_]:=Floor[(Sqrt[8n+1]-1)/2]
s[n_,k_]:=Ceiling[(n+1)/k-(k+1)/2]-Ceiling[(n+1)/(k+1)-(k+2)/2]
f[n_,k_]:=If[k<=row[n],s[n,k],s[n,2 row[n]+1-k]]
TableForm[Table[f[n,k],{n,1,50},{k,1,2 row[n]}]] (* Hartmut F. W. Hoft, Apr 08 2014 *)

row(n) = {my(orow = row237591(n)); vector(2*#orow, i, if (i <= #orow, orow[i], orow[2*#orow-i+1]));}
area(n) = {my(rown = row(n)); surf = 0; h = n; odd = 1; for (i=1, #row, if (odd, surf += h*rown[i], h -= rown[i];); odd = !odd;); surf;}
heights(v, n) = {vh = vector(n); ivh = 1; h = n; odd = 1; for (i=1, #v, if (odd, for (j=1, v[i], vh[ivh] = h; ivh++), h -= v[i];); odd = !odd;); vh;}
isabove(hb, ha) = {for (i=1, #hb, if (hb[i] < ha[i], return (0));); return (1);}
chkcross(nn) = {hga = concat(heights(row(1), 1), 0); for (n=2, nn, hgb = heights(row(n), n); if (! isabove(hgb, hga), print("pb cross at n=", n)); hga = concat(hgb, 0););} \\ Michel Marcus, Mar 27 2014

from sympy import sqrt
import math
def row(n): return int(math.floor((sqrt(8*n + 1) - 1)/2))
def s(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2)) - int(math.ceil((n + 1)/(k + 1) - (k + 2)/2))
def T(n, k): return s(n, k) if k<=row(n) else s(n, 2*row(n) + 1 - k)
for n in range(1, 11): print([T(n, k) for k in range(1, 2*row(n) + 1)]) # Indranil Ghosh, Apr 21 2017

Let j(n)= floor((sqrt(8n+1)-1)/2) then T(n,k) = A237591(n,k), if k <= j(n); otherwise T(n,k) = A237591(n,2*j(n)+1-k). - Hartmut F. W. Hoft, Apr 07 2014 (corrected by Omar E. Pol, May 31 2015)

A minor edit to the definition. - N. J. A. Sloane, Jul 31 2025

A237591 Irregular triangle read by rows: T(n,k) is the difference between the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the total number of partitions of all positive integers <= n into exactly k+1 consecutive parts (n>=1, 1<=k<=A003056(n)).

Original entry on oeis.org

1, 2, 2, 1, 3, 1, 3, 2, 4, 1, 1, 4, 2, 1, 5, 2, 1, 5, 2, 2, 6, 2, 1, 1, 6, 3, 1, 1, 7, 2, 2, 1, 7, 3, 2, 1, 8, 3, 1, 2, 8, 3, 2, 1, 1, 9, 3, 2, 1, 1, 9, 4, 2, 1, 1, 10, 3, 2, 2, 1, 10, 4, 2, 2, 1, 11, 4, 2, 1, 2, 11, 4, 3, 1, 1, 1, 12, 4, 2, 2, 1, 1, 12, 5, 2, 2, 1, 1, 13, 4, 3, 2, 1, 1, 13, 5, 3, 1, 2, 1, 14, 5, 2, 2, 2, 1

Offset: 1

Omar E. Pol, Feb 22 2014

The original name was: Triangle read by rows: T(n,k) = A235791(n,k) - A235791(n,k+1), assuming that the virtual right border of triangle A235791 is A000004.
T(n,k) is also the length of the k-th segment in a zig-zag path on the first quadrant of the square grid, connecting the point (n, 0) with the point (m, m), starting with a segment in vertical direction, where m <= n.
Conjecture: the area of the polygon defined by the x-axis, this zig-zag path and the diagonal [(0, 0), (m, m)], is equal to A024916(n)/2, one half of the sum of all divisors of all positive integers <= n. Therefore the reflected polygon, which is adjacent to the y-axis, with the zig-zag path connecting the point (0, n) with the point (m, m), has the same property. And so on for each octant in the four quadrants.
For the representation of A024916 and A000203 we use two octants, for example: the first octant and the second octant, or the 6th octant and the 7th octant, etc., see A237593.
At least up to n = 128, two zig-zag paths never cross (checked by hand).
The finite sequence formed by the n-th row of triangle together with its mirror row gives the n-th row of triangle A237593.
The connection between A196020 and A237271 is as follows: A196020 --> A236104 --> A235791 --> this sequence --> A237593 --> A239660 --> A237270 --> A237271.
Comments from Franklin T. Adams-Watters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014. (Start)
The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.
You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.
Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n).
Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)
From Hartmut F. W. Hoft, Apr 07 2014: (Start)
The row sum is A235791(n,1) - A235791(n,floor((sqrt(8n+1)-1)/2)+1) = n - 0.
Mathematica function has been written to check the conjecture as well as non-crossing zig-zag paths (Dyck paths rotated by 90 degrees) up through n=30000 (same applies to A237593). (End)
The n-th zig-zag path ending at the point (m, m), where m = A240542(n). - Omar E. Pol, Apr 16 2014
From Omar E. Pol, Aug 23 2015: (Start)
n is an odd prime if and only if T(n,2) = 1 + T(n-1,2) and T(n,k) = T(n-1,k) for the rest of the values of k.
The elements of the n-th row of triangle together with the elements of the n-th row of triangle A261350 give the n-th row of triangle A237593.
T(n,k) is also the area (or the number of cells) of the k-th vertical side at the n-th level (starting from the top) in the left hand part of the front view of the stepped pyramid described in A245092, see Example section.
(End)
From Omar E. Pol, Nov 19 2015: (Start)
T(n,k) is also the number of cells between the k-th and the (k+1)st line segments (from left to right) in the n-th row of the diagram as shown in Example section.
Note that the number of horizontal line segments in the n-th row of the diagram equals A001227(n), the number of odd divisors of n. (End)
Conjecture: the values f(n,k) in the n-th row of the triangle are either 1 or 2 for all k with ceiling((sqrt(4*n+1)-1)/2) <= k <= floor((sqrt(8*n+1)-1)/2) = r(n), the length of the n-th row, though the lower bound need not be minimal; tested through 2500000. See also A285356. - Hartmut F. W. Hoft, Apr 17 2017
Conjecture: T(n,k) is the difference between the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the total number of partitions of all positive integers <= n into exactly k+1 consecutive parts. - Omar E. Pol, Apr 30 2017
From Omar E. Pol, Aug 31 2021: (Start)
It appears that T(n,2)/T(n,1) converges to 1/3.
It appears that T(n,3)/T(n,2) converges to 1/2.
It appears that T(n,4)/T(n,3) converges to 3/5.
It appears that T(n,5)/T(n,4) converges to 2/3. (End)
In other words: T(n,k) is the length of the k-th line segment of the largest Dyck path of the symmetric representation of sigma(n). - Omar E. Pol, Sep 08 2021

			Triangle begins:
   1;
   2;
   2, 1;
   3, 1;
   3, 2;
   4, 1, 1;
   4, 2, 1;
   5, 2, 1;
   5, 2, 2;
   6, 2, 1, 1;
   6, 3, 1, 1;
   7, 2, 2, 1;
   7, 3, 2, 1;
   8, 3, 1, 2;
   8, 3, 2, 1, 1;
   9, 3, 2, 1, 1;
   9, 4, 2, 1, 1;
  10, 3, 2, 2, 1;
  10, 4, 2, 2, 1;
  11, 4, 2, 1, 2;
  11, 4, 3, 1, 1, 1;
  12, 4, 2, 2, 1, 1;
  12, 5, 2, 2, 1, 1;
  13, 4, 3, 2, 1, 1;
  13, 5, 3, 1, 2, 1;
  14, 5, 2, 2, 2, 1;
  14, 5, 3, 2, 1, 2;
  15, 5, 3, 2, 1, 1, 1;
  ...
For n = 10 the 10th row of triangle A235791 is [10, 4, 2, 1] so row 10 is [6, 2, 1, 1].
From _Omar E. Pol_, Aug 23 2015: (Start)
Illustration of initial terms:
  Row                                                         _
   1                                                        _|1|
   2                                                      _|2 _|
   3                                                    _|2  |1|
   4                                                  _|3   _|1|
   5                                                _|3    |2 _|
   6                                              _|4     _|1|1|
   7                                            _|4      |2  |1|
   8                                          _|5       _|2 _|1|
   9                                        _|5        |2  |2 _|
  10                                      _|6         _|2  |1|1|
  11                                    _|6          |3   _|1|1|
  12                                  _|7           _|2  |2  |1|
  13                                _|7            |3    |2 _|1|
  14                              _|8             _|3   _|1|2 _|
  15                            _|8              |3    |2  |1|1|
  16                          _|9               _|3    |2  |1|1|
  17                        _|9                |4     _|2 _|1|1|
  18                      _|10                _|3    |2  |2  |1|
  19                    _|10                 |4      |2  |2 _|1|
  20                  _|11                  _|4     _|2  |1|2 _|
  21                _|11                   |4      |3   _|1|1|1|
  22              _|12                    _|4      |2  |2  |1|1|
  23            _|12                     |5       _|2  |2  |1|1|
  24          _|13                      _|4      |3    |2 _|1|1|
  25        _|13                       |5        |3   _|1|2  |1|
  26      _|14                        _|5       _|2  |2  |2 _|1|
  27    _|14                         |5        |3    |2  |1|2 _|
  28   |15                           |5        |3    |2  |1|1|1|
  ...
Also the diagram represents the left part of the front view of the pyramid described in A245092. For the other half front view see A261350. For more information about the pyramid and the symmetric representation of sigma see A237593. (End)
From _Omar E. Pol_, Sep 08 2021: (Start)
For n = 12 the symmetric representation of sigma(12) in the fourth quadrant is as shown below:
.                           _
                           | |
                           | |
                           | |
                           | |
                           | |
                      _ _ _| |
                    _|    _ _|
                  _|     |
                 |      _|
                 |  _ _|1
      _ _ _ _ _ _| |  2
     |_ _ _ _ _ _ _|2
            7
.
The lengths of the successive line segments from the first vertex to the central vertex of the largest Dyck path are [7, 2, 2, 1] respectively, the same as the 12th row of triangle. (End)

G. C. Greubel, Table of n, a(n) for the first 150 rows

Row n has length A003056(n) hence column k starts in row A000217(k).
Row sums give A000027.
Column 1 is A008619, n >= 1.
Right border gives A042974.
Cf. A000203, A001227, A024916, A196020, A235791, A236104, A237048, A237270, A237271, A237593, A239660, A239931-A239934, A240542, A244580, A245092, A249351, A259176, A259177, A261350, A261699, A262626, A285356, A286000, A286001.

row[n_]:= Floor[(Sqrt[8*n+1] -1)/2];  f[n_,k_]:= Ceiling[(n+1)/k-(k+1)/2] - Ceiling[(n+1)/(k+1)-(k+2)/2];
Table[f[n,k],{n,1,50},{k,1,row[n]}]//Flatten
(* Hartmut F. W. Hoft, Apr 08 2014 *)

row235791(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i);
row(n) = {my(orow = concat(row235791(n), 0)); vector(#orow -1, i, orow[i] - orow[i+1]);} \\ Michel Marcus, Mar 27 2014

from sympy import sqrt
import math
def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2)) - int(math.ceil((n + 1)/(k + 1) - (k + 2)/2))
for n in range(1, 29): print([T(n, k) for k in range(1, int((sqrt(8*n + 1) - 1)/2) + 1)]) # Indranil Ghosh, Apr 30 2017

T(n,k) = ceiling((n+1)/k - (k+1)/2) - ceiling((n+1)/(k+1) - (k+2)/2), for 1 <= n and 1 <= k <= floor((sqrt(8n+1)-1)/2). - Hartmut F. W. Hoft, Apr 07 2014

3 more rows added by Omar E. Pol, Aug 23 2015

New name from a comment dated Apr 30 2017. - Omar E. Pol, Jun 18 2023

A236104 Triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k copies of the positive squares in nondecreasing order, and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

1, 4, 9, 1, 16, 1, 25, 4, 36, 4, 1, 49, 9, 1, 64, 9, 1, 81, 16, 4, 100, 16, 4, 1, 121, 25, 4, 1, 144, 25, 9, 1, 169, 36, 9, 1, 196, 36, 9, 4, 225, 49, 16, 4, 1, 256, 49, 16, 4, 1, 289, 64, 16, 4, 1, 324, 64, 25, 9, 1, 361, 81, 25, 9, 1, 400, 81, 25, 9, 4

Offset: 1

Omar E. Pol, Jan 23 2014

These are the squares of the entries of the triangle in A235791: T(n,k) = (A235791(n,k))^2.
Row n has length A003056(n) hence the first element of column k is in row A000217(k).
Columns 1-3 (including the initial zeros) are A000290, A008794, A211547.
Also column k lists the partial sums of the k-th column of triangle A196020 which gives an identity for sigma.
Since all the elements of this sequence are squares, we can draw an illustration of the alternating sum of row n step by step, and a symmetric diagram for A000203, A024916, A004125; see example.
For more information about the diagram see A237593.

			Triangle begins:
    1;
    4;
    9,   1;
   16,   1;
   25,   4;
   36,   4,   1;
   49,   9,   1;
   64,   9,   1;
   81,  16,   4;
  100,  16,   4,   1;
  121,  25,   4,   1;
  144,  25,   9,   1;
  169,  36,   9,   1;
  196,  36,   9,   4;
  225,  49,  16,   4,   1;
  256,  49,  16,   4,   1;
  289,  64,  16,   4,   1;
  324,  64,  25,   9,   1;
  361,  81,  25,   9,   1;
  400,  81,  25,   9,   4;
  441, 100,  36,   9,   4,   1;
  484, 100,  36,  16,   4,   1;
  529, 121,  36,  16,   4,   1;
  576, 121,  49,  16,   4,   1;
  ...
For n = 6 the sum of all divisors of all positive integers <= 6 is [1] + [1+2] + [1+3] + [1+2+4] + [1+5] + [1+2+3+6] = 1 + 3 + 4 + 7 + 6 + 12 = 33. On the other hand the 6th row of triangle is 36, 4, 1, therefore the alternating row sum is 36 - 4 + 1 = 33, equaling the sum of all divisors of all positive integers <= 6.
Illustration of the alternating sum of the 6th row as the area of a polygon (or the number of cells), step by step, in the fourth quadrant:
.     _ _ _ _ _ _       _ _ _ _ _ _       _ _ _ _ _ _
.    |           |     |           |     |           |
.    |           |     |           |     |           |
.    |           |     |           |     |           |
.    |           |     |        _ _|     |          _|
.    |           |     |       |         |        _|
.    |_ _ _ _ _ _|     |_ _ _ _|         |_ _ _ _|
.
.          36           36 - 4 = 32     36 - 4 + 1 = 33
.
Then using this method we can draw a symmetric diagram for A000203, A024916, A004125, as shown below:
--------------------------------------------------
n     A000203  A024916            Diagram
--------------------------------------------------
.                         _ _ _ _ _ _ _ _ _ _ _ _
1        1        1      |_| | | | | | | | | | | |
2        3        4      |_ _|_| | | | | | | | | |
3        4        8      |_ _|  _|_| | | | | | | |
4        7       15      |_ _ _|    _|_| | | | | |
5        6       21      |_ _ _|  _|  _ _|_| | | |
6       12       33      |_ _ _ _|  _| |  _ _|_| |
7        8       41      |_ _ _ _| |_ _|_|    _ _|
8       15       56      |_ _ _ _ _|  _|     |* *
9       13       69      |_ _ _ _ _| |      _|* *
10      18       87      |_ _ _ _ _ _|  _ _|* * *
11      12       99      |_ _ _ _ _ _| |* * * * *
12      28      127      |_ _ _ _ _ _ _|* * * * *
.
The total number of cells in the first n set of symmetric regions of the diagram equals A024916(n). It appears that the total number of cells in the n-th set of symmetric regions of the diagram equals sigma(n) = A000203(n). Example: for n = 12 the 12th row of triangle is 144, 25, 9, 1, hence the alternating sums is 144 - 25 + 9 - 1 = 127. On the other hand we have that A000290(12) - A004125(12) = 144 - 17 = A024916(12) = 127, equaling the total number of cells in the diagram after 12 stages. The number of cells in the 12th set of symmetric regions of the diagram is sigma(12) = A000203(12) = 28. Note that in this case there is only one region. Finally, the number of *'s is A004125(12) = 17.
Note that the diagram is also the top view of the stepped pyramid described in A245092. - _Omar E. Pol_, Feb 12 2018

Michael De Vlieger, Table of n, a(n) for n = 1..10075 (rows 1 <= n <= 500).

Cf. A000203, A000217, A000290, A001227, A003056, A008794, A024916, A004125, A196020, A211343, A228813, A231345, A231347, A235791, A235794, A235799, A236106, A236112, A236540, A237270, A237591, A237593, A239660, A244050, A245092, A262626, A286000.

Table[Ceiling[(n + 1)/k - (k + 1)/2]^2, {n, 20}, {k, Floor[(Sqrt[8 n + 1] - 1)/2]}] // Flatten (* Michael De Vlieger, Feb 10 2018, after Hartmut F. W. Hoft at A235791 *)

from sympy import sqrt
import math
def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2))
for n in range(1, 21): print([T(n, k)**2 for k in range(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)]) # Indranil Ghosh, Apr 25 2017

Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A024916(n). [Although this was stated as a fact, as far as I can tell, no proof was known. However, Don Reble has recently found a proof, which will be added here soon. - N. J. A. Sloane, Nov 23 2020]

A000203(n) = Sum_{k=1..A003056(n)} (-1)^(k-1) * (T(n,k) - T(n-1,k)), assuming that T(k*(k+1)/2-1,k) = 0. - Omar E. Pol, Oct 10 2018

A235791 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k copies of every positive integer in nondecreasing order, and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

1, 2, 3, 1, 4, 1, 5, 2, 6, 2, 1, 7, 3, 1, 8, 3, 1, 9, 4, 2, 10, 4, 2, 1, 11, 5, 2, 1, 12, 5, 3, 1, 13, 6, 3, 1, 14, 6, 3, 2, 15, 7, 4, 2, 1, 16, 7, 4, 2, 1, 17, 8, 4, 2, 1, 18, 8, 5, 3, 1, 19, 9, 5, 3, 1, 20, 9, 5, 3, 2, 21, 10, 6, 3, 2, 1, 22, 10, 6, 4, 2, 1, 23, 11, 6, 4, 2, 1, 24, 11, 7, 4, 2, 1

Offset: 1

Omar E. Pol, Jan 23 2014

The alternating sum of the squares of the elements of the n-th row equals the sum of all divisors of all positive integers <= n, i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*(T(n,k))^2 = A024916(n).
Row n has length A003056(n) hence the first element of column k is in row A000217(k).
For more information see A236104.
The sum of row n gives A060831(n), the sum of the number of odd divisors of all positive integers <= n. - Omar E. Pol, Mar 01 2014. [An equivalent assertion is that the sum of row n of A237048 is the number of odd divisors of n, and this was proved by Hartmut F. W. Hoft in a comment in A237048. - N. J. A. Sloane, Dec 07 2020]
Comments from Franklin T. Adams-Watters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014: (Start)
The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.
You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.
Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n).
Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)
From Hartmut F. W. Hoft, Apr 07 2014: (Start)
Mathematica function has been written to check the first property up to n = 20000.
T(n,(sqrt(8n+1)-1)/2+1) = 0 for all n >= 1, which is useful for formulas for A237591 and A237593. (End)
Alternating row sums give A240542. - Omar E. Pol, Apr 16 2014
Conjecture: T(n,k) is also the total number of partitions of all positive integers <= n into exactly k consecutive parts, i.e., the partial column sum of A285898, or in accordance with the triangles of the same family: the partial column sum of A237048. - Omar E. Pol, Apr 28 2017, Nov 24 2020
The above conjecture is true. The proof will be added soon (it uses the generating function for the columns). - N. J. A. Sloane, Nov 24 2020
T(n,k) is also the total length of all line segments between the k-th vertex and the central vertex of the largest Dyck path of the symmetric representation of sigma(n). In other words: T(n,k) is the sum of the last (A003056(n)-k+1) terms of the n-th row of A237591. - Omar E. Pol, Sep 07 2021
T(n,k) is also the Manhattan distance between the k-th vertex and the central vertex of the Dyck path described in the n-th row of the triangle A237593. - Omar E. Pol, Jan 11 2023

			Triangle begins:
   1;
   2;
   3,  1;
   4,  1;
   5,  2;
   6,  2,  1;
   7,  3,  1;
   8,  3,  1;
   9,  4,  2;
  10,  4,  2,  1;
  11,  5,  2,  1;
  12,  5,  3,  1;
  13,  6,  3,  1;
  14,  6,  3,  2;
  15,  7,  4,  2,  1;
  16,  7,  4,  2,  1;
  17,  8,  4,  2,  1;
  18,  8,  5,  3,  1;
  19,  9,  5,  3,  1;
  20,  9,  5,  3,  2;
  21, 10,  6,  3,  2,  1;
  22, 10,  6,  4,  2,  1;
  23, 11,  6,  4,  2,  1;
  24, 11,  7,  4,  2,  1;
  25, 12,  7,  4,  3,  1;
  26, 12,  7,  5,  3,  1;
  27, 13,  8,  5,  3,  2;
  28, 13,  8,  5,  3,  2,  1;
  ...
For n = 10 the 10th row of triangle is 10, 4, 2, 1, so we have that 10^2 - 4^2 + 2^2 - 1^2 = 100 - 16 + 4 - 1 = 87, the same as A024916(10) = 87, the sum of all divisors of all positive integers <= 10.
From _Omar E. Pol_, Nov 19 2015: (Start)
Illustration of initial terms in the third quadrant:
.                                                            y
Row                                                         _|
1                                                         _|1|
2                                                       _|2 _|
3                                                     _|3  |1|
4                                                   _|4   _|1|
5                                                 _|5    |2 _|
6                                               _|6     _|2|1|
7                                             _|7      |3  |1|
8                                           _|8       _|3 _|1|
9                                         _|9        |4  |2 _|
10                                      _|10        _|4  |2|1|
11                                    _|11         |5   _|2|1|
12                                  _|12          _|5  |3  |1|
13                                _|13           |6    |3 _|1|
14                              _|14            _|6   _|3|2 _|
15                            _|15             |7    |4  |2|1|
16                          _|16              _|7    |4  |2|1|
17                        _|17               |8     _|4 _|2|1|
18                      _|18                _|8    |5  |3  |1|
19                    _|19                 |9      |5  |3 _|1|
20                  _|20                  _|9     _|5  |3|2 _|
21                _|21                   |10     |6   _|3|2|1|
22              _|22                    _|10     |6  |4  |2|1|
23            _|23                     |11      _|6  |4  |2|1|
24          _|24                      _|11     |7    |4 _|2|1|
25        _|25                       |12       |7   _|4|3  |1|
26      _|26                        _|12      _|7  |5  |3 _|1|
27    _|27                         |13       |8    |5  |3|2 _|
28   |28                           |13       |8    |5  |3|2|1|
...
T(n,k) is also the number of cells between the k-th vertical line segment (from left to right) and the y-axis in the n-th row of the structure.
Note that the number of horizontal line segments in the n-th row of the structure equals A001227(n), the number of odd divisors of n.
Also the diagram represents the left part of the front view of the pyramid described in A245092. (End)
For more information about the diagram see A286001. - _Omar E. Pol_, Dec 19 2020
From _Omar E. Pol_, Sep 08 2021: (Start)
For n = 12 the symmetric representation of sigma(12) in the fourth quadrant is as shown below:
                            _
                           | |
                           | |
                           | |
                           | |
                           | |
                      _ _ _| |
                    _|    _ _|
                  _|     |
                 |      _|
                 |  _ _|
      _ _ _ _ _ _| |3   1
     |_ _ _ _ _ _ _|
    12              5
.
For n = 12 and k = 1 the total length of all line segments between the first vertex and the central vertex of the largest Dyck path is equal to 12, so T(12,1) = 12.
For n = 12 and k = 2 the total length of all line segments between the second vertex and the central vertex of the largest Dyck path is equal to 5, so T(12,2) = 5.
For n = 12 and k = 3 the total length of all line segments between the third vertex and the central vertex of the largest Dyck path is equal to 3, so T(12,3) = 3.
For n = 12 and k = 4 the total length of all line segments between the fourth vertex and the central vertex of the largest Dyck path is equal to 1, so T(12,4) = 1.
Hence the 12th row of triangle is [12, 5, 3, 1]. (End)

G. C. Greubel, Table of n, a(n) for the first 150 rows, flattened

Columns 1..3: A000027, A008619, A008620.
Operations on rows: A003056 (number of terms), A237591 (differences between terms), A060831 (sums), A339577 (products), A240542 (alternating sums), A236104 (squares), A339576 (sums of squares), A024916 (alternating sums of squares), A237048 (differences between rows), A042974 (right border).
Cf. A000203, A000217, A001227, A196020, A211343, A228813, A231345, A231347, A235794, A236106, A236112, A237270, A237271, A237593, A239660, A245092, A261699, A262626, A286000, A286001, A280850, A280851, A296508, A335616.

row[n_] := Floor[(Sqrt[8*n + 1] - 1)/2]; f[n_, k_] := Ceiling[(n + 1)/k - (k + 1)/2]; Table[f[n, k], {n, 1, 150}, {k, 1, row[n]}] // Flatten (* Hartmut F. W. Hoft, Apr 07 2014 *)

row(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i); \\ Michel Marcus, Mar 27 2014

from sympy import sqrt
import math
def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2))
for n in range(1, 21): print([T(n, k) for k in range(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)]) # Indranil Ghosh, Apr 25 2017

T(n,k) = ceiling((n+1)/k - (k+1)/2) for 1 <= n, 1 <= k <= floor((sqrt(8n+1)-1)/2) = A003056(n). - Hartmut F. W. Hoft, Apr 07 2014
G.f. for column k (k >= 1): x^(k*(k+1)/2)/( (1-x)*(1-x^k) ). - N. J. A. Sloane, Nov 24 2020
T(n,k) = Sum_{j=1..n} A237048(j,k). - Omar E. Pol, May 18 2017
T(n,k) = sqrt(A236104(n,k)). - Omar E. Pol, Feb 14 2018
Sigma(n) = Sum_{k=1..A003056(n)} (-1)^(k-1) * (T(n,k)^2 - T(n-1,k)^2), assuming that T(k*(k+1)/2-1,k) = 0. - Omar E. Pol, Oct 10 2018
a(s(n,k)) = T(n,k), n >= 1, 1 <= k <= r = floor((sqrt(8*n + 1) - 1)/2), where s(n,k) = r*n - r*(r+1)*(r+2)/6 + k translates position (row n, column k) in the triangle of this sequence to its position in the sequence. - Hartmut F. W. Hoft, Feb 24 2021

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.

Original entry on oeis.org

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

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

Original entry on oeis.org

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

Offset: 1

Omar E. Pol, Sep 26 2015

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

			Irregular triangle begins:
  1;
  1, 1;
  3;
  2, 2;
  2, 2;
  2, 1, 1, 2;
  7;
  3, 1, 1, 3;
  3, 3;
  3, 2, 2, 3;
  12;
  4, 1, 1, 1, 1, 4;
  4, 4;
  4, 2, 1, 1, 2, 4;
  15;
  5, 2, 1, 1, 2, 5;
  5, 3, 5;
  5, 2, 2, 2, 2, 5;
  9, 9;
  6, 2, 1, 1, 1, 1, 2, 6;
  6, 6;
  6, 3, 1, 1, 1, 1, 3, 6;
  28;
  7, 2, 2, 1, 1, 2, 2, 7;
  7, 7;
  7, 3, 2, 1, 1, 2, 3, 7;
  12, 12;
  8, 3, 1, 2, 2, 1, 3, 8;
  8, 8, 8;
  8, 3, 2, 1, 1, 1, 1, 2, 3, 8;
  31;
  9, 3, 2, 1, 1, 1, 1, 2, 3, 9;
  ...
Illustration of the odd-indexed rows of triangle as the diagram of the symmetric representation of sigma which is also the top view of the stepped pyramid:
.
   n  A000203    A237270    _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
   1     1   =      1      |_| | | | | | | | | | | | | | | |
   2     3   =      3      |_ _|_| | | | | | | | | | | | | |
   3     4   =    2 + 2    |_ _|  _|_| | | | | | | | | | | |
   4     7   =      7      |_ _ _|    _|_| | | | | | | | | |
   5     6   =    3 + 3    |_ _ _|  _|  _ _|_| | | | | | | |
   6    12   =     12      |_ _ _ _|  _| |  _ _|_| | | | | |
   7     8   =    4 + 4    |_ _ _ _| |_ _|_|    _ _|_| | | |
   8    15   =     15      |_ _ _ _ _|  _|     |  _ _ _|_| |
   9    13   =  5 + 3 + 5  |_ _ _ _ _| |      _|_| |  _ _ _|
  10    18   =    9 + 9    |_ _ _ _ _ _|  _ _|    _| |
  11    12   =    6 + 6    |_ _ _ _ _ _| |  _|  _|  _|
  12    28   =     28      |_ _ _ _ _ _ _| |_ _|  _|
  13    14   =    7 + 7    |_ _ _ _ _ _ _| |  _ _|
  14    24   =   12 + 12   |_ _ _ _ _ _ _ _| |
  15    24   =  8 + 8 + 8  |_ _ _ _ _ _ _ _| |
  16    31   =     31      |_ _ _ _ _ _ _ _ _|
  ...
The above diagram arises from a simpler diagram as shown below.
Illustration of the even-indexed rows of triangle as the diagram of the deployed front view of the corner of the stepped pyramid:
.
.                                 A237593
Level                               _ _
1                                 _|1|1|_
2                               _|2 _|_ 2|_
3                             _|2  |1|1|  2|_
4                           _|3   _|1|1|_   3|_
5                         _|3    |2 _|_ 2|    3|_
6                       _|4     _|1|1|1|1|_     4|_
7                     _|4      |2  |1|1|  2|      4|_
8                   _|5       _|2 _|1|1|_ 2|_       5|_
9                 _|5        |2  |2 _|_ 2|  2|        5|_
10              _|6         _|2  |1|1|1|1|  2|_         6|_
11            _|6          |3   _|1|1|1|1|_   3|          6|_
12          _|7           _|2  |2  |1|1|  2|  2|_           7|_
13        _|7            |3    |2 _|1|1|_ 2|    3|            7|_
14      _|8             _|3   _|1|2 _|_ 2|1|_   3|_             8|_
15    _|8              |3    |2  |1|1|1|1|  2|    3|              8|_
16   |9                |3    |2  |1|1|1|1|  2|    3|                9|
...
The number of horizontal line segments in the n-th level in each side of the diagram equals A001227(n), the number of odd divisors of n.
The number of horizontal line segments in the left side of the diagram plus the number of the horizontal line segment in the right side equals A054844(n).
The total number of vertical line segments in the n-th level of the diagram equals A131507(n).
The diagram represents the first 16 levels of the pyramid.
The diagram of the isosceles triangle and the diagram of the top view of the pyramid shows the connection between the partitions into consecutive parts and the sum of divisors function (see also A286000 and A286001). - _Omar E. Pol_, Aug 28 2018
The connection between the isosceles triangle and the stepped pyramid is due to the fact that this object can also be interpreted as a pop-up card. - _Omar E. Pol_, Nov 09 2022

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

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

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A288529 a(n) is the minimum number of rows from the table described in A286000 that are required to represent the partitions of n into consecutive parts.

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A288772 a(n) is the minimum number of rows from the table described in A286000 that are required to represent the partitions of all positive integers <= n into consecutive parts.

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A288773 a(n) is the largest of all positive integers whose partitions into consecutive parts can be totally represented in the first n rows of the table described in A286000.

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A288774 a(n) is the largest positive integers whose partitions into consecutive parts can be totally represented in the first n rows of the table described in A286000.

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A237593 Triangle read by rows in which row n lists the elements of the n-th row of A237591 followed by the same elements in reverse order.

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A237591 Irregular triangle read by rows: T(n,k) is the difference between the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the total number of partitions of all positive integers <= n into exactly k+1 consecutive parts (n>=1, 1<=k<=A003056(n)).

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A236104 Triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k copies of the positive squares in nondecreasing order, and the first element of column k is in row k(k+1)/2.

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A235791 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k copies of every positive integer in nondecreasing order, and the first element of column k is in row k(k+1)/2.

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