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

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

Author

Omar E. Pol, Oct 26 2014

Keywords

• nonn
• tabf

Comments

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

Examples

			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)
		

Links

• Index entries for sequences related to sigma(n)

Crossrefs

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.

Programs

Mathematica

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