A247687 -id:A247687 - cp's OEIS frontend

A279387 Irregular triangle read by rows: suppose the symmetric representation of sigma(n) consists of m = A250068(n) layers of width 1, arranged in increasing order; then T(n,k) (n >= 1, 1 <= k <= m) is the number of subparts in the k-th layer.

1, 1, 2, 1, 2, 1, 1, 2, 1, 3, 2, 2, 1, 1, 2, 2, 3, 1, 1, 2, 1, 2, 2, 1, 1, 4, 2, 2, 1, 1, 3, 2, 4, 1, 1, 2, 1, 3, 2, 1, 4, 2, 3, 1, 1, 2, 2, 2, 4, 1, 1, 2, 1, 3, 2, 2, 3, 3, 2, 2, 1, 1, 3, 3, 4, 2, 2, 1, 3, 4, 1, 1, 4, 2, 2, 1, 1, 2, 2, 2, 5, 1, 1, 4, 1, 3, 2, 2, 4, 3, 1, 2, 1, 1, 1, 2, 2

Offset: 1

Omar E. Pol, Dec 12 2016

The "subparts" of the symmetric representation of sigma(n) are defined to be the regions that arise after the dissection of the symmetric representation of sigma(n) into successive layers of width 1.
The number of layers of width 1 in the symmetric representation of sigma(n) is given in A250068.
The number of subparts in the first layer of the symmetric representation of sigma(n) is equal to A237271(n).
We can find the symmetric representation of sigma(n) as the terraces at the n-th level (starting from the top) of the stepped pyramid described in A245092.
(All above comments are essentially the same as the comments dated Nov 05 2016 at the old version of A275601, which was the same as A001227).
The sum of row n equals the number of subparts in the symmetric representation of sigma(n).
Conjecture:
The number of subparts in the symmetric representation of sigma(n) equals A001227(n), the number of odd divisors of n.
From Hartmut F. W. Hoft, Dec 16 2016: (Start)
Proof:
Each row of the irregular triangle of A262045 can be interpreted as a step function of step sizes 1, 0, and -1. The numbers in row n are the widths of the segments in the parts of the symmetric representation of sigma(n). Each new subpart in a segment (in the left half) of row n starts at the same odd index that represents an odd divisor d of n in the irregular triangle of A237048. Either a subpart ends at an even index e, representing a second odd divisor, which satisfies d * e = oddpart(n), and thus the entire subpart is duplicated in the symmetric portion of the representation, or a subpart runs through the center and continues contiguously into the right half of the symmetric portion of the representation. In other words, the number of subparts in row n equals the number of odd divisors of n, i.e., the conjecture is true. (End)

			Triangle begins (first 18 rows):
1;
1;
2;
1;
2;
1, 1;
2;
1;
3;
2;
2;
1, 1;
2;
2;
3, 1;
1;
2;
1, 2;
...
For n = 12, 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"
.   A237271(12) = 1.                   that contain 23 and 5 cells
.                                      respectively, so the 12th row of
.                                      this triangle is [1, 1], and the
.                                      row sum is A001227(12) = 2,
.                                      equaling the number of odd divisors
.                                      of 12.
.
For n = 15, 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 A237271(15) = 3.       The first layer has three subparts:
.                                      [8, 7, 8]. The second layer has
.                                      only one subpart of size 1, so
.                                      the 15th row of this triangle is
.                                      [3, 1], and the row sum is
.                                      A001227(15) = 4, equaling the
.                                      number of odd divisors of 15.
.
For n = 360, the 359th row of triangle A237593 is [180, 61, 30, 19, 12, 9, 7, 6, 4, 4, 3, 3, 2, 3, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1] and the 360th row of the same triangle is [181, 60, 31, 18, 13, 9, 7, 5, 5, 4, 3, 2, 2, 2, 2, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1], so have that the symmetric representation of sigma(360) = 1170 has only one part, five layers, and six subparts: [(719), (237), (139), (71), (2, 2)], so the 360th row of this triangle is [1, 1, 1, 1, 2], and the row sum is A001227(360) = 6, equaling the number of odd divisors of 360 (the diagram is too large to include).
From _Hartmut F. W. Hoft_, Dec 16 2016: (Start)
45 has 6 subparts of which 2 have symmetric duplicates and 2 span the center. Row length is 18 and "|" indicates the center marker for a row.
1 2 3 4 5 6 7 8 9|9 8 7 6 5 4 3 2 1  : position indices
1 0 1 1 2 1 1 1 2|2 1 1 1 2 1 1 0 1  : row 45 of A262045
1   1 1 1 1 1 1 1|1 1 1 1 1 1 1   1  : layer 1
        1       1|1       1          : layer 2
1 1 1 0 1 1 0 0 1|                   : row 45 of A237048 (odd divisors)
+ - + . + - . . +|                   : change in level ("." no change)
90 has 6 subparts and 3 layers (row length is 24).
1 2 3 4 5 6 7 8..10..12|.14..16..18..20..22..24 : position indices
1 1 2 1 2 2 2 2 3 3 3 2|2 3 3 3 2 2 2 2 1 2 1 1 : row 90 of A262045
1 1 1 1 1 1 1 1 1 1 1 1|1 1 1 1 1 1 1 1 1 1 1 1 : layer 1
    1   1 1 1 1 1 1 1 1|1 1 1 1 1 1 1 1   1     : layer 2
                1 1 1  |  1 1 1                 : layer 3
1 0 1 1 1 0 0 0 1 0 0 1|                        : row 90 of A237048
+ . + - + . . . + . . -|                        : change in level ("." no change)
The process of successive levels provides two "default" dissections of the symmetric representation into subparts from the boundary at n towards the boundary at n-1 or in the reverse direction. (End)
From _Omar E. Pol_, Nov 24 2020: (Start)
For n = 18 we have that the 17th row of triangle A237593 is [9, 4, 2, 1, 1, 1, 1, 2, 4, 9] and the 18th row of the same triangle is [10, 3, 2, 2, 1, 1, 2, 2, 3, 10], so the diagram of the symmetric representation of sigma(18) = 39 is constructed as shown below in Figure 5:
.                                     _                                      _
.                                    | |                                    | |
.                                    | |                                    | |
._                                   | |                                    | |
.                                    | |                                    | |
.                                    | |                                    | |
.                                    | |                                    | |
.                                    | |                                    | |
.                                    | |                                    | |
.                             _ _ _ _| |                             _ _ _ _| |
.                            |    _ _ _|                            |  _ _ _ _|
.                           _|   |                                 _| | |
.                         _|  _ _|                               _|  _|_|
.                     _ _|  _|                               _ _|  _|    2
.                    |     |  39                            |  _ _|
.                    |  _ _|                                | |_ _|
.                    | |                                    | |    2
.   _ _ _ _ _ _ _ _ _| |                   _ _ _ _ _ _ _ _ _| |
.  |_ _ _ _ _ _ _ _ _ _|                  |_ _ _ _ _ _ _ _ _ _|
.                                                              35
.
.   Figure 5. The symmetric               Figure 6. After the dissection
.   representation of sigma(18)           of the symmetric representation
.   has one part of size 39, so           of sigma(18) into layers of
.   A237271(18) = 1.                      width 1 we can see three "subparts".
.                                         The first layer has one subpart of
.                                         size 35. The second layer has
.                                         two subparts of size 2, so
.                                         the 18th row of this triangle is
.                                         [1, 2], and the row sum is
.                                         A001227(18) = 3.
(End)

The sum of row n equals A001227(n).
Hence, if n is odd, the sum of row n equals A000005(n).
Row n has length A250068(n).
Column 1 gives A237271.
For more information about "subparts" see A279388 and A279391.
Cf. A000203, A196020, A235791, A236104, A237048, A237270, A237591, A237593, A239657, A243982, A244050, A245092, A249223, A249351, A250070, A262045, A262611, A261699, A262626, A279693, A280850, A280851, A296508.
Cf. A235791, A237048, A237270, A237591, A237593, A247687, A249223, A250070, A264102, A280851, A341969.

(* function a341969[ ] is defined in A341969 *)
a279387[n_] := Module[{widthL=a341969[n], partL, cL, top, ft, sL}, partL=Select[SplitBy[widthL, #==0&], #!={0}&]; cL=Table[0, Max[widthL]]; While[partL!={}, top=Last[partL]; ft=First[top]; sL=Select[SplitBy[top, #==ft&], #!={ft}&];
cL[[ft]]++; partL=Join[Most[partL], sL]]; cL]
Flatten[a279387[74]] (* the first 74 rows of the table; Hartmut F. W. Hoft, Feb 24 2021 *)

Definition edited by Omar E. Pol and N. J. A. Sloane, Nov 25 2020

A249223 Triangle read by rows: row n gives partial alternating sums of row n of A237048.

Original entry on oeis.org

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

Offset: 1

Hartmut F. W. Hoft, Oct 23 2014

All entries in the triangle are nonnegative since the number of 1's in odd-numbered columns of A237048 prior to column j, 1 <= j <= row(n), is at least as large as the number of 1's in even-numbered columns through column j. As a consequence:
(a) The two adjacent symmetric Dyck paths whose legs are defined by adjacent rows of triangle A237593 never cross each other (see also A235791 and A237591) and the rows in this triangle describe the widths between the legs.
(b) Let legs(n) denote the n-th row of triangle A237591, widths(n) the n-th row of this triangle, and c(n) the rightmost entry in the n-th row of this triangle (center of the Dyck path). Then area(n) = 2 * legs(n) . width(n) - c(n), where "." is the inner product, is the area between the two adjacent symmetric Dyck paths.
(c) For certain sequences of integers, it is known that area(n) = sigma(n); see A238443, A245685, A246955, A246956 and A247687.
Right border gives A067742. - Omar E. Pol, Jan 21 2017
For a proof that T(n, k) = |{ d : d|n and k/2 < d <= k }|, for 1 <= k <= row(n), where row(n) is the length of row n, an identity suggested by Peter Munn, see the link. A corollary to it is that the number of divisors of n in the half-open interval (row(n)/2, row(n)] equals the width of the symmetric representation of n at the diagonal: T(n, row(n)) = | { d : d|n and row(n)/2 < d <= row(n) } |. See also the comments and conjectures of Michel Marcus in A067742 and A237593. - Hartmut F. W. Hoft, Jun 24 2024
From Omar E. Pol, Jul 24 2024: (Start)
Conjecture 1: Every column is a periodic sequence.
Conjecture 2: The periods of the columns 1..8 are respectively: 1, 2, 6, 12, 60, 60, 420, 840.
Question 1: Is the period of the column k equal to A003418(k)? (End).
From Omar E. Pol, Jul 26 2024: (Start)
Column 1 gives A000012.
Column 2 gives A000035.
Conjecture 3: Column 3 gives [2, 0] together with A115357, hence column 3 gives 2 together with A171182.
Question 2: Except the first nine terms of A337976, is the column 4 the same as A337976?
Question 3: Except the first 14 terms of A366981, is the column 5 the same as A366981? (End)
From Hartmut F. W. Hoft, Aug 01 2024: (Start)
Conjectures 1 and 2 are true and the answer to question 1 is affirmative.
By definition, each column k in triangle T237048(n, k) of sequence A237048 is a periodic sequence of period k. Since the k-th term in row n of the triangle T(n, k) = Sum_{i=1..k} (-1)^(i+1) * T237048(n, i), with 1 <= k <= A003056(n), each initial subsequence T(n, 1) .. T(n, k) of row n in this triangle is periodic of period lcm(1, .. , k) = A003418(k). This implies that each column k in this sequence has period A003418(k).
Conjecture 3 and Question 2 are true. Since T237048(n, 1) = 1, T237208(n, 2) = 1 if n odd and 0 if n even, T237048(n, 3) = 1 if 3|n and 0 otherwise, and T237048(n, 4) = 1 if 4|(n-2) and 0 otherwise, equations T249223(n, 3) = 1 - (n mod 2) + delta( n mod 3) and T249223(n, 4) = 1 - (n mod 2) + delta( n mod 3) - delta( (n-2) mod 4) hold where delta(k) = 1 if k = 0 and 0 otherwise. With the 3rd column starting at n = A000217(3) = 6, each period starting in a row that is a multiple of 6 is [ 2 0 1 1 1 0 ], and appropriate shifts yield A115357 and A171182. With the 4th column starting at n = A000217(4) = 10, each period starting in a row n with 12|(n+2) is [ 0 0 2 0 0 1 1 0 1 0 1 1 ], and with a shift of 9 yields the apparently periodic A337976(10), A337976(11), ... (End)

			Triangle begins:
---------------------------
   n \ k  1  2  3  4  5  6
---------------------------
   1 |    1;
   2 |    1;
   3 |    1, 0;
   4 |    1, 1;
   5 |    1, 0;
   6 |    1, 1, 2;
   7 |    1, 0, 0;
   8 |    1, 1, 1;
   9 |    1, 0, 1;
  10 |    1, 1, 1, 0;
  11 |    1, 0, 0, 0;
  12 |    1, 1, 2, 2;
  13 |    1, 0, 0, 0;
  14 |    1, 1, 1, 0;
  15 |    1, 0, 1, 1, 2;
  16 |    1, 1, 1, 1, 1;
  17 |    1, 0, 0, 0, 0;
  18 |    1, 1, 2, 1, 1;
  19 |    1, 0, 0, 0, 0;
  20 |    1, 1, 1, 1, 2;
  21 |    1, 0, 1, 1, 1, 0;
  22 |    1, 1, 1, 0, 0, 0;
  23 |    1, 0, 0, 0, 0, 0;
  24 |    1, 1, 2, 2, 2, 2;
  ...
The triangle shows that area(n) has width 1 for powers of 2 and that area(p) for primes p consists of only 1 horizontal leg of width 1 (and its symmetric vertical leg in the mirror symmetric duplicate of this triangle).

G. C. Greubel, Table of n, a(n) for the first 150 rows, flattened
Hartmut F. W. Hoft, Divisors d of n in half open interval (k/2, k]

Row n has length A003056(n).
Column k starts in row A000217(k).
Cf. A000012, A000035, A000203, A003418, A067742, A115357, A171182, A237048, A237270, A237271, A237593, A238443, A245685, A246955, A246956, A247687, A337976, A366981.

r := proc(n) floor((sqrt(1+8*n)-1)/2) ; end proc: # R. J. Mathar 2015 A003056
A237048:=proc(n,k) local i; global r;
if n<(k-1)*k/2 or k>r(n) then return(0); fi;
if (k mod 2)=1 and (n mod k)=0 then return(1); fi;
if (k mod 2)=0 and ((n-k/2) mod k) = 0 then return(1); fi;
return(0);
end;
A249223:=proc(n,k) local i; global r,A237048;
if n<(k-1)*k/2 or k>r(n) then return(0); fi;
add( (-1)^(i+1)*A237048(n,i),i=1..k);
end;
for n from 1 to 12 do lprint([seq(A249223(n,k),k=1..r(n))]); od; # N. J. A. Sloane, Jan 15 2021

cd[n_, k_] := If[Divisible[n, k], 1, 0]; row[n_] := Floor[(Sqrt[8 n + 1] - 1)/2]; a237048[n_, k_] := If[OddQ[k], cd[n, k], cd[n - k/2, k]];
a1[n_, k_] := Sum[(-1)^(j + 1)*a237048[n, j], {j, 1, k}];
a2[n_] := Drop[FoldList[Plus, 0, Map[(-1)^(# + 1) &, Range[row[n]]] a237048[n]], 1]; Flatten[Map[a2, Range[24]]] (* data *) (* Corrected by G. C. Greubel, Apr 16 2017 *)

t237048(n,k) = if (k % 2, (n % k) == 0, ((n - k/2) % k) == 0);
kmax(n) = (sqrt(1+8*n)-1)/2;
t(n,k) = sum(j=1, k, (-1)^(j+1)*t237048(n,j));
tabf(nn) = {for (n=1, nn, for (k=1, kmax(n), print1(t(n,k), ", ");); print(););} \\ Michel Marcus, Sep 20 2015

T(n, k) = Sum_{j=1..k} (-1)^(j+1)*A237048(n, j), for n>=1 and 1 <= k <= floor((sqrt(8*n + 1) - 1)/2). - corrected by Hartmut F. W. Hoft, Jan 25 2018

A249351 Triangle read by rows in which row n lists the widths of the symmetric representation of sigma(n).

Original entry on oeis.org

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

Offset: 1

Omar E. Pol, Oct 26 2014

Here T(n,k) is defined to be the "k-th width" of the symmetric representation of sigma(n), with n>=1 and 1<=k<=2n-1. Explanation: consider the diagram of the symmetric representation of sigma(n) described in A236104, A237593 and other related sequences. Imagine that the diagram for sigma(n) contains 2n-1 equidistant segments which are parallel to the main diagonal [(0,0),(n,n)] of the quadrant. The segments are located on the diagonal of the cells. The distance between two parallel segment is equal to sqrt(2)/2. T(n,k) is the length of the k-th segment divided by sqrt(2). Note that the triangle contains nonnegative terms because for some n the value of some widths is equal to zero. For an illustration of some widths see Hartmut F. W. Hoft's contribution in the Links section of A237270.
Row n has length 2*n-1.
Row sums give A000203.
If n is a power of 2 then all terms of row n are 1's.
If n is an even perfect number then all terms of row n are 1's except the middle term which is 2.
If n is an odd prime then row n lists (n+1)/2 1's, n-2 zeros, (n+1)/2 1's.
The number of blocks of positive terms in row n gives A237271(n).
The sum of the k-th block of positive terms in row n gives A237270(n,k).
It appears that the middle diagonal is also A067742 (which was conjectured by Michel Marcus in the entry A237593 and checked with two Mathematica functions up to n = 100000 by Hartmut F. W. Hoft).
It appears that the trapezoidal numbers (A165513) are also the numbers k > 1 with the property that some of the noncentral widths of the symmetric representation of sigma(k) are not equal to 1. - Omar E. Pol, Mar 04 2023

			Triangle begins:
  1;
  1,1,1;
  1,1,0,1,1;
  1,1,1,1,1,1,1;
  1,1,1,0,0,0,1,1,1;
  1,1,1,1,1,2,1,1,1,1,1;
  1,1,1,1,0,0,0,0,0,1,1,1,1;
  1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
  1,1,1,1,1,0,0,1,1,1,0,0,1,1,1,1,1;
  1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1;
  1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1;
  1,1,1,1,1,1,1,1,1,2,2,2,2,2,1,1,1,1,1,1,1,1,1;
  ...
---------------------------------------------------------------------------
.        Written as an isosceles triangle              Diagram of
.              the sequence begins:               the symmetry of sigma
---------------------------------------------------------------------------
.                                                _ _ _ _ _ _ _ _ _ _ _ _
.                      1;                       |_| | | | | | | | | | | |
.                    1,1,1;                     |_ _|_| | | | | | | | | |
.                  1,1,0,1,1;                   |_ _|  _|_| | | | | | | |
.                1,1,1,1,1,1,1;                 |_ _ _|    _|_| | | | | |
.              1,1,1,0,0,0,1,1,1;               |_ _ _|  _|  _ _|_| | | |
.            1,1,1,1,1,2,1,1,1,1,1;             |_ _ _ _|  _| |  _ _|_| |
.          1,1,1,1,0,0,0,0,0,1,1,1,1;           |_ _ _ _| |_ _|_|    _ _|
.        1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;         |_ _ _ _ _|  _|     |
.      1,1,1,1,1,0,0,1,1,1,0,0,1,1,1,1,1;       |_ _ _ _ _| |      _|
.    1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1;     |_ _ _ _ _ _|  _ _|
.  1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1;   |_ _ _ _ _ _| |
.1,1,1,1,1,1,1,1,1,2,2,2,2,2,1,1,1,1,1,1,1,1,1; |_ _ _ _ _ _ _|
...
From _Omar E. Pol_, Nov 22 2020: (Start)
Also consider the infinite double-staircases diagram defined in A335616.
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 this seq:  [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 A280851    [              8,      7,    1,            8              ]
.
The number of horizontal steps (or 1's) in the successive columns of the above diagram gives the 15th row of this triangle.
For more information about the parts of the symmetric representation of sigma(n) see A237270. For more information about the subparts see A239387, A296508, A280851.
More generally, it appears there is the same correspondence between the original diagram of the symmetric representation of sigma(n) and the "Ziggurat" diagram of n. (End)

Index entries for sequences related to sigma(n)

Cf. A000203, A003056, A067742, A071562, A165513, A196020, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A238443, A239660, A239932-A239934, A240542, A241008, A241010, A245092, A245685, A246955, A246956, A247687, A249223, A250068, A250070, A250071, A262626, A280850, A280851, A296508, A235616, A347186.

(* function segments are defined in A237270 *)
a249351[n_] := Flatten[Map[segments, Range[n]]]
a249351[10] (* Hartmut F. W. Hoft, Jul 20 2022 *)

A341969 Irregular triangle read by rows in which row n lists the sequence of widths, each contiguous sequence of identical widths w in A249223 replaced by a single entry of w, in the symmetric representation of sigma(n).

Original entry on oeis.org

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

Offset: 1

Hartmut F. W. Hoft, Feb 24 2021

This sequence is a companion to A279387 in which each contiguous sequence of identical widths w in A249223 are replaced by a single entry of w. Using the resulting distribution pattern of widths across all parts of the symmetric representation of sigma(n) the subparts at each level are counted in A279387.
The sequence of widths are computed first to the diagonal of the symmetric representation of sigma only for those numbers in set F defined in A341971. Then the reversed list less its first number is appended so that the width at the diagonal is not listed twice. Thus every row contains an odd number of entries and is symmetric about its center entry.
Let 1 <= n, 1 <= d <= s = A001227(n) and 1 <= k <= r = floor((sqrt(8*n + 1) - 1)/2). Let Q(n,d) be row n in the triangle of A341970, R(n,d) be row n in the triangle of A341970 and S(n,d) = R(n,Q(n,d)), then T(n,e) = S(n,e) for 1 <= e <= s and T(n,e) = S(n,2*s - e) for s < e <= 2*s - 1 is row n for this sequence.

			The irregular triangle for A279387 and this sequence:
  row  A279387  A341969
  1    1        1
  2    1        1
  3    2        1  0  1
  4    1        1
  5    2        1  0  1
  6    1  1     1  2  1
  7    2        1  0  1
  8    1        1
  9    3        1  0  1  0  1
  10   2        1  0  1
  11   2        1  0  1
  12   1  1     1  2  1
  13   2        1  0  1
  14   2        1  0  1
  15   3  1     1  0  1  2  1  0  1
  16   1        1
  17   2        1  0  1
  18   1  2     1  2  1  2  1
  19   2        1  0  1
  20   1  1     1  2  1
  21   4        1  0  1  0  1  0  1
  ..   ..       ..
  30   1  3     1  2  1  2  1  2  1
  ..   ..       ..
  45   3  3     1  0  1  2  1  2  1  2  1  0  1
  ..   ..       ..
a(17)..a(21) = { 1, 0, 1, 0, 1 } is row 9; the symmetric representation of sigma(9) consists of 3 parts of width 1 - see A247687.
a(37)..a(43) = { 1, 0, 1, 2, 1, 0, 1} is row 15; the symmetric representation of sigma(15) consists of 2 outer parts of width 1 and a central part of width 2 only at the diagonal - see A338488.
a(59)..a(65) = { 1, 0, 1, 0, 1, 0, 1 } is row 21; the symmetric representation of sigma(21) consists of 4 parts of width 1, and 21 is the smallest such number - see A264102.
a(234)..a(240) = { 1, 2, 3, 2, 3, 2, 1 } is row 60; the symmetric representation of sigma(60) consists of 1 part of maximum width 3 which occurs in two subparts, and 60 is the smallest number with width 3 - see A250070.

Cf. A235791, A237048, A237270, A237591, A237593, A247687, A249223, A249351, A250070, A264102, A279387, A280851, A338488, A341970, A341971.

(* function widthL[ ] is defined in A341971 *)
a341969[n_] := Module[{wL=widthL[n]}, Join[wL, Rest[Reverse[wL]]]]
Flatten[Table[a341969[n], {n, 28}]] (* the first 28 rows of the table *)

a(2*A060831(n-1) - (n-1) + e) = T(n,e), 1 <= n, 1 <= e <= 2*A001227(n) - 1.

A250068 Maximum width of any region in the symmetric representation of sigma(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2

Offset: 1

Hartmut F. W. Hoft, Nov 11 2014

nonn

Since the width of the single region of the symmetric representation of sigma( 2^ceiling((p-1)*(log_2 3) - 1) * 3^(p-1) ), for prime number p, at the diagonal equals p, this sequence contains an increasing subsequence (see A250071).

a(n) is also the number of layers of width 1 in the symmetric representation of sigma(n). For more information see A001227. - Omar E. Pol, Dec 13 2016

			a(6) = 2 since the sequence of widths at each unit step in the symmetric representation of sigma(6) = 12 is 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1. For visual examples see A237270, A237593 and sequences referenced in these.

N. J. A. Sloane, Table of n, a(n) for n = 1..10000

Cf. A000203, A001227, A237048, A237270, A237271, A237591, A237593, A241008, A241010, A246955, A247687, A249223, A249351 (widths), A279387, A279388, A279391.

See A250070, A250071, A340506 for records.

(* function a2[ ] is defined in A249223 *)
a250068[n_]:=Max[a2[n]]
a250068[{m_,n_}]:=Map[a250068,Range[m,n]]
a250068[{1,100}](* data *)

t237048(n,k) = if (k % 2, (n % k) == 0, ((n - k/2) % k) == 0);
kmax(n) = (sqrt(1+8*n)-1)/2;
t249223(n,k) = sum(j=1, k, (-1)^(j+1)*t237048(n,j));
a(n) = my(wm = t249223(n, 1)); for (k=2, kmax(n), wm = max(wm, t249223(n, k))); wm; \\ Michel Marcus, Sep 20 2015

a(n) = max_{k=1..floor((sqrt(8*n+1) - 1)/2)} (Sum_{j=1..k}(-1)^(j+1)*A237048(n, j)), for n >= 1.

A250070 Smallest number k such that the symmetric representation of sigma(k) has at least one part of width n.

Original entry on oeis.org

1, 6, 60, 120, 360, 840, 3360, 2520, 5040, 10080, 15120, 32760, 27720, 50400, 98280, 83160, 110880, 138600, 221760, 277200, 332640, 360360, 554400, 960960, 831600, 942480, 720720, 2217600, 1965600, 1441440, 3160080, 2827440, 2162160, 2882880, 3603600, 5765760, 5654880, 4324320, 9979200

Offset: 1

Hartmut F. W. Hoft, Nov 11 2014

nonn

The 26 entries starting with a(2) = 6 are products of powers of consecutive primes starting with 2, except for a(12) = 32760 and a(15) = 98280 (which are missing 11), and a(26) = 942480 (which is missing 13).
a(n) is the smallest number k such that the symmetric representation of sigma(k) has n layers. For more information see A279387. - Omar E. Pol, Dec 16 2016
Row 1 of A253258. - Omar E. Pol, Apr 15 2018
From Hartmut F. W. Hoft, Jun 10 2024: (Start)
All terms a(n) <= 1.75*10^7 have a symmetric representation of sigma that consists of a single part and they are abundant for n > 2. Numbers a(1) = 1, a(2) = 6, and a(4) = 120 are unimodal while numbers a(6) = 840, a(14) = 50400, a(18) = 138600, a(24) = 960960, a(26) = 942480, a(32) = 2827440, a(44) = 8648640 have a single extent of maximum width, but are not unimodal.
Conjecture: The symmetric representation of sigma for every term consists of a single part and it is unimodal only for a(1), a(2), and a(4).
As a consequence, this sequence would be a subsequence of A174973, and all a(n), n > 2, would be abundant. (End)

			a(3) = 60 since the symmetric representation of sigma(60) = 168 consists of a single region of whose successive widths are 41 1's, 9 2's, 6 3's, 7 2's, 6 3's, 9 2's, and 41 1's.
a(6) = 840 has a single extent of 12 units of width 6 centered around point (583,583) on the diagonal, but is not unimodal. - _Hartmut F. W. Hoft_, Jun 10 2024

Hartmut F. W. Hoft, Table of n, a(n) for n = 1..48

Indices of records in A250068.
Cf. A000203, A237270, A237271, A237593, A241008, A241010, A246955, A247687, A249223, A250071, A253258, A279387.
Cf. A174973.

(* function a2[ ] is defined in A249223 *)
a250070[{j_, k_}, b_] := Module[{i, max, acc={{1, 1}}}, For[i=j, i<=k, i++, max={Max[a2[i]], i}; If[max[[1]]>b && !MemberQ[Transpose[acc][[1]], max[[1]]], AppendTo[acc,max]]]; acc]
(* returns (argument,result) data pairs since sequence is non-monotonic *)
Sort[a250070[{1, 1000000}, 1]] (* computed in steps *)

a(n) = min(k such that A250068(k) = n), n >= 1.

a(28)-a(48) from Hartmut F. W. Hoft, Jun 10 2024

A246955 Numbers j for which the symmetric representation of sigma(j) has two parts, each of width one.

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, 79, 82, 83, 86, 89, 92, 94, 97, 101, 103, 106, 107, 109, 113, 116, 118, 122, 124, 127, 131, 134, 136, 137, 139, 142, 146, 148, 149, 151, 152, 157, 158, 163, 164, 166, 167, 172, 173, 178, 179, 181, 184, 188, 191, 193, 194, 197, 199

Offset: 1

Hartmut F. W. Hoft, Sep 08 2014

nonn

The sequence is the intersection of A239929 (sigma(j) has two parts) and of A241008 (sigma(j) has an even number of parts of width one).
The numbers in the sequence are precisely those defined by the formula for the triangle, see the link. The symmetric representation of sigma(j) has two parts, each part having width one, precisely when j = 2^(k - 1) * p where 2^k <= row(j) < p, p is prime and row(j) = floor((sqrt(8*j + 1) - 1)/2). Therefore, the sequence can be written naturally as a triangle as shown in the Example section.
The symmetric representation of sigma(j) = 2*j - 2 consists of two regions of width 1 that meet on the diagonal precisely when j = 2^(2^m - 1)*(2^(2^m) + 1) where 2^(2^m) + 1 is a Fermat prime (see A019434). This subsequence of numbers j is 3, 10, 136, 32896, 2147516416, ...[?]... (A191363).
The k-th column of the triangle starts in the row whose initial entry is the first prime larger than 2^(k+1) (that sequence of primes is A014210, except for 2).
Observation: at least the first 82 terms coincide with the numbers j with no middle divisors whose largest divisor <= sqrt(j) is a power of 2, or in other words, coincide with the intersection of A071561 and A365406. - Omar E. Pol, Oct 11 2023

			We show portions of the first eight columns, 0 <= k <= 7, of the triangle.
0    1    2     3     4     5     6     7
3
5    10
7    14
11   22   44
13   26   52
17   34   68    136
19   38   76    152
23   46   92    184
29   58   116   232
31   62   124   248
37   74   148   296   592
41   82   164   328   656
43   86   172   344   688
47   94   188   376   752
53   106  212   424   848
59   118  236   472   944
61   122  244   488   976
67   134  268   536   1072  2144
71   142  284   568   1136  2272
.    .    .     .     .     .
.    .    .     .     .     .
127  254  508   1016  2032  4064
131  262  524   1048  2096  4192  8384
137  274  548   1096  2192  4384  8768
.    .    .     .     .     .     .
.    .    .     .     .     .     .
251  502  1004  2008  4016  8032  16064
257  514  1028  2056  4112  8224  16448  32896
263  526  1052  2104  4208  8416  16832  33664
Since 2^(2^4) + 1 = 65537 is the 6543rd prime, column k = 15 starts with 2^15*(2^(2^16) + 1) = 2147516416 in row 6542 with 65537 in column k = 0.
For an image of the symmetric representations of sigma(m) for all values m <= 137 in the triangle see the link.
The first column is the sequence of odd primes, see A065091.
The second column is the sequence of twice the primes starting with 10, see A001747.
The third column is the sequence of four times the primes starting with 44, see A001749.
For related references also see A033676 (largest divisor of n less than or equal to sqrt(n)).

Hartmut F. W. Hoft, Equivalence proof of sequence and triangle
Hartmut F. W. Hoft, Image for sigma(n) values

Cf. A000203, A033676, A071561, A163280, A237270, A237271, A237593, A241008, A241010, A247687, A250068, A250070, A250071, A365406.

(* functions path[] and a237270[ ] are defined in A237270 *)
atmostOneDiagonalsQ[n_]:=SubsetQ[{0, 1}, Union[Flatten[Drop[Drop[path[n], 1], - 1] - path[n - 1], 1]]]
(* data *)
Select[Range[200], Length[a237270[#]]==2 && atmostOneDiagonalsQ[#]&]
(* function for computing triangle in the Example section through row 55 *)
TableForm[Table[2^k Prime[n], {n, 2, 56}, {k, 0, Floor[Log[2, Prime[n]] - 1]}], TableDepth->2]

Formula for the triangle of numbers associated with the sequence:

P(n, k) = 2^k * prime(n) where n >= 2, 0 <= k <= floor(log_2(prime(n)) - 1).

A250071 Smallest number k such that the symmetric representation of sigma(k) has maximum width n for those k whose representation has nondecreasing width up to the diagonal.

Original entry on oeis.org

1, 6, 72, 120, 5184, 1440, 373248, 6720, 28800, 103680, 1934917632, 80640, 278628139008, 7464960, 2073600, 483840, 1444408272617472, 1612800, 103997395628457984, 5806080, 298598400, 77396705280, 539122498937926189056, 7096320, 1658880000, 5572562780160, 90316800, 418037760, 402452788967166148425547776, 116121600

Offset: 1

Hartmut F. W. Hoft, Nov 11 2014

nonn

The symmetric representation of sigma(k) has nondecreasing width to the diagonal precisely when all odd divisors counted in the k-th row of A237048 occur at odd indices. If we write k = 2^m * q with m >= 0 and q odd, this property is equivalent to q < 2^(m+1).
The values for a(11), a(13), a(17) and a(19) were computed directly using the formula k = 2^m * 3^(p-1) where p is one of the four primes and m the smallest exponent so that 3^(p-1) < 2^(m+1). Each of these numbers has a symmetric representation of nondecreasing width ending in a prime number width, and they are the first such numbers since the number of divisors of an odd number is a prime precisely when the number is a power of a prime.
The other numbers listed whose symmetric representations of sigma(k) have nondecreasing width are smaller than 7500000. The only additional numbers k <= 100000000 are a(24) = 7096320, a(27) = 90316800 and a(32) = 85155840.
See A340506 for another way to look at this data. - N. J. A. Sloane, Jan 23 2021

			a(6) = 1440 = 2^5 * 3^2 * 5 has 6 odd divisors. It is the smallest number of the form 2^m * q with m > 0, q odd and such that q < 2^(m+1).

Cf. A000203, A237048, A237270, A237271, A237593, A241008, A241010, A246955, A247687, A249223, A250068, A250070.
Cf. also A340506.
Cf. A162247.

(* function a2[ ] is defined in A249223 *)
smallQ[n_] := Module[{x=2^IntegerExponent[n,2]}, n/x<2x]
ndWidth[{m_,n_}] := Select[Range[m, n], smallQ]
a250071[x_List] := Module[{i, max, acc={{1, 1}}}, For[i=1, i<=Length[x], i++, max={Max[a2[x[[i]]]], x[[i]]}; If[!MemberQ[Transpose[acc][[1]], max[[1]]], AppendTo[acc, max]]]; acc]
(* returns (argument,result) data pairs since sequence is non-monotonic *)
Sort[a250071[ndWidth[{1,100000000}]]] (* computed in steps *)
(* alternate implementation using function f[ ] by T. D. Noe in A162247 *)
sF[n_] := Min[Map[Apply[Times, Prime[Range[2, Length[#]+1]]^#]&, Map[Reverse[#-1]&, f[n]]]]
f1U[n_] := Module[{s=sF[n], k}, k=Floor[Log[2, s]]; 2^k s]
a250071[n_] := Map[f1U, Range[n]]
a250071[30] (* Hartmut F. W. Hoft, Nov 27 2024 *)

a(n) = min(2^m * q, m >= 0 & q odd & sigma_0(q) = n & q < 2^(m+1)) where sigma_0 is the number of divisors.

a(p) = 2^ceiling((p-1)*(log_2(3)) - 1) * 3^(p-1) for primes p.

a(21)-a(30) from Hartmut F. W. Hoft, Nov 27 2024

A241010 Numbers n with the property that the number of parts in the symmetric representation of sigma(n) is odd, and that all parts have width 1.

Original entry on oeis.org

1, 2, 4, 8, 9, 16, 25, 32, 49, 50, 64, 81, 98, 121, 128, 169, 242, 256, 289, 338, 361, 484, 512, 529, 578, 625, 676, 722, 729, 841, 961, 1024, 1058, 1156, 1250, 1369, 1444, 1681, 1682, 1849, 1922, 2048, 2116, 2209, 2312, 2401, 2738, 2809, 2888, 3025, 3249, 3362, 3364, 3481, 3698, 3721, 3844

Offset: 1

Hartmut F. W. Hoft, Aug 07 2014

nonn

The first eight entries in A071562 but not in this sequence are 6, 12, 15, 18, 20, 24, 28 & 30.
The first eight entries in A238443 but not in this sequence are 6, 12, 18, 20, 24, 28, 30 & 36.
The union of A241008 and of this sequence equals A174905 (for a proof see link in A174905).
Let n = 2^m * product(p_i^e_i, i=1,...,k) = 2^m * q with m >= 0, k >= 0, 2 < p_1, ...< p_k primes and e_i >= 1, for all 1 <= i <= k. For each number n in this sequence all e_i are even, and for any two odd divisors f < g of n, 2^(m+1) * f < g. The sum of the areas of the regions r(n, z) equals sigma(n). For a proof of the characterization and the formula see the theorem in the link below.
Numbers 3025 = 5^2 * 11^2 and 510050 = 2^1 * 5^2 * 101^2 are the smallest odd and even numbers, respectively, in the sequence with two distinct odd prime factors.
Among the 706 numbers in the sequence less than 1000000 (see link to the table) there are 143 that have two different odd prime factors, but none with three. All numbers with three different odd prime factors are larger than 500000000. Number 4450891225 = 5^2 * 11^2 * 1213^2 is in the sequence, but may not be the smallest one with three different odd prime factors. Note that 1213 is the first prime that extends the list of divisors of 3025 while preserving the property for numbers in this sequence.
The subsequence of numbers n = 2^(k-1) * p^2 satisfying the constraints above is A247687.
n = 3^(2*h) = 9^h = A001019(h), h>=0, is the smallest number for which the symmetric representation of sigma(n) has 2*h+1 regions of width one, for example for h = 1, 2, 3 and 5, but not for h = 4 in which case 3025 = 5^2 * 11^2 < 3^8 = 6561 is the smallest (see A318843). [Comment corrected by Hartmut F. W. Hoft, Sep 04 2018]
Computations using this characterization are more efficient than those of Dyck paths for the symmetric representations of sigma(n), e.g., the Mathematica code below.

			This irregular triangle presents in each column those elements of the sequence that have the same factor of a power of 2.
  row/col      2^0    2^1   2^2   2^3    2^4    2^5  ...
   2^k:          1      2     4     8     16     32  ...
   3^2:          9
   5^2:         25     50
   7^2:         49     98
   3^4:         81
  11^2:        121    242   484
  13^2:        169    338   676
  17^2:        289    578  1156  2312
  19^2:        361    722  1444  2888
  23^2:        529   1058  2116  4232
   5^4:        625   1250
   3^6:        729
  29^2:        841   1682  3364  6728
  31^2:        961   1922  3844  7688
  37^2:       1369   2738  5476 10952 21904
  41^2:       1681   3362  6724 13448 26896
  43^2:       1849   3698  7396 14792 29584
  47^2:       2209   4418  8836 17672 35344
   7^4:       2401   4802
  53^2:       2809   5618 11236 22472 44944
  5^2*11^2:   3025
  3^2*19^2:   3249
  59^2:       3481   6962 13924 27848 55696
  61^2:       3721   7442 14884 29768 59536
  67^2:       4489   8978 17956 35912 71824 143648
  3^2*23^2:   4761
  71^2:       5041
  ...
  5^2*101^2:225025 510050
  ...
Number 3025 = 5^2 * 11^2 is in the sequence since its divisors are 1, 5, 11, 25, 55, 121, 275, 605 and 3025. Number 6050 = 2^1 * 5^2 * 11^2 is not in the sequence since 2^2 * 5 > 11 while 5 < 11.
Number 510050 = 2^1 * 5^2 * 101^2 is in the sequence since its 9 odd divisors 1, 5, 25, 101, 505, 2525, 10201, 51005 and 225025 are separated by factors larger than 2^2. The areas of its 9 regions are 382539, 76515, 15339, 3939, 1515, 3939, 15339, 76515 and 382539. However, 2^2 * 5^2 * 101^2 is not in the sequence.
The first row is A000079.
The rows, except the first, are indexed by products of even powers of the odd primes satisfying the property, sorted in increasing order.
The first column is a subsequence of A244579.
A row labeled p^(2*h), h>=1 and p>=3 with p = A000040(n), has A098388(n) entries.
Starting with the second column, dividing the entries of a column by 2 creates a proper subsequence of the prior column.
See A259417 for references to other sequences of even powers of odd primes that are subsequences of column 1.
The first entry greater than 16 in column labeled 2^4 is 21904 since 37 is the first prime larger than 2^5. The rightmost entry in the row labeled 19^2 is 2888 in the column labeled 2^3 since 2^4 < 19 < 2^5.

Hartmut F. W. Hoft, Table of n, a(n) for n = 1..563 (values less than 1000000)
Hartmut F. W. Hoft, Proof of characterization theorem
Hartmut F. W. Hoft, Table of n, a(n) for n = 1...706(values less than 1000000)
Hartmut F. W. Hoft, Illustration of the symmetric representation of sigma(a(n)), for n=1..15
Hartmut F. W. Hoft, Proof of property for n and formula for regions of sigma(n)

Cf. A000203, A174905, A236104, A237270 (symmetric representation of sigma(n)), A237271, A237593, A238443, A241008, A071562, A246955, A247687, A250068, A250070, A250071.

Cf. A318843

(* path[n] and a237270[n] are defined in A237270 *)
atmostOneDiagonalsQ[n_] := SubsetQ[{0, 1}, Union[Flatten[Drop[Drop[path[n], 1], -1] - path[n-1], 1]]]
Select[Range[1000], atmostOneDiagonalsQ[#] && OddQ[Length[a237270[#]]]&] (* data *)
(* more efficient code based on numeric characterization *)
divisorPairsQ[m_, q_] := Module[{d = Divisors[q]}, Select[2^(m + 1)*Most[d] - Rest[d], # >= 0 &] == {}]
a241010AltQ[n_] := Module[{m, q, p, e}, m=IntegerExponent[n, 2]; q=n/2^m; {p, e} = Transpose[FactorInteger[q]]; q==1||(Select[e, EvenQ]==e && divisorPairsQ[m, q])]
a241010Alt[m_,n_] := Select[Range[m, n], a241010AltQ]
a241010Alt[1,4000] (* data *)

Formula for the z-th region in the symmetric representation of n = 2^m * q in this sequence, 1 <= z <= sigma_0(q) and q odd: r(n, z) = 1/2 * (2^(m+1) - 1) * (d_z + d_(2*x+2-z)) where 1 = d_1 < ... < d_(2*x+1) = q are the odd divisors of n.

More terms and further edited by Hartmut F. W. Hoft, Jun 26 2015 and Jul 02 2015 and corrected Oct 11 2015

cp's OEIS Frontend

A338510 a(n) is smallest number in column n >= 0 of the triangle in A247687.

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A279387 Irregular triangle read by rows: suppose the symmetric representation of sigma(n) consists of m = A250068(n) layers of width 1, arranged in increasing order; then T(n,k) (n >= 1, 1 <= k <= m) is the number of subparts in the k-th layer.

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A249223 Triangle read by rows: row n gives partial alternating sums of row n of A237048.

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A249351 Triangle read by rows in which row n lists the widths of the symmetric representation of sigma(n).

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A341969 Irregular triangle read by rows in which row n lists the sequence of widths, each contiguous sequence of identical widths w in A249223 replaced by a single entry of w, in the symmetric representation of sigma(n).

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A250068 Maximum width of any region in the symmetric representation of sigma(n).

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A250070 Smallest number k such that the symmetric representation of sigma(k) has at least one part of width n.

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A246955 Numbers j for which the symmetric representation of sigma(j) has two parts, each of width one.

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