A299765 Irregular triangle read by rows, T(n,k), n >= 1, k >= 1, in which row n lists the partitions of n into consecutive parts, with the partitions ordered by increasing number of parts.

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

Offset

1,2

Comments

In the triangle the first partition with m parts appears as the last partition in row A000217(m), m >= 1. - Omar E. Pol, Mar 23 2022

Links

Paolo Xausa, Table of n, a(n) for n = 1..10350 (rows 1..500 of triangle, flattened)

Formula

T(2^m,1) = 2^m, for m >= 0. - Paolo Xausa, Jun 19 2022

Example

Triangle begins:
   [1];
   [2];
   [3], [2, 1];
   [4];
   [5], [3, 2];
   [6], [3, 2, 1];
   [7], [4, 3];
   [8];
   [9], [5, 4], [4, 3, 2];
  [10], [4, 3, 2, 1];
  [11], [6, 5];
  [12], [5, 4, 3];
  [13], [7, 6];
  [14], [5, 4, 3, 2];
  [15], [8, 7], [6, 5, 4], [5, 4, 3, 2, 1];
  [16];
  [17], [9, 8];
  [18], [7, 6, 5], [6, 5, 4, 3];
  [19], [10, 9];
  [20], [6, 5, 4, 3, 2];
  [21], [11, 10], [8, 7, 6], [6, 5, 4, 3, 2, 1];
  [22], [7, 6, 5, 4];
  [23], [12, 11];
  [24], [9, 8, 7];
  [25], [13, 12], [7, 6, 5, 4, 3];
  [26], [8, 7, 6, 5];
  [27], [14, 13], [10, 9, 8], [7, 6, 5, 4, 3, 2];
  [28], [7, 6, 5, 4, 3, 2, 1];
...
Note that in the below diagram the number of horizontal line segments in the n-th row equals A001227(n), the number of partitions of n into consecutive parts, so we can find the partitions of n into consecutive parts as follows: consider the vertical blocks of numbers that start exactly in the n-th row of the diagram, for example: for n = 15 consider the vertical blocks of numbers that start exactly in the 15th row. They are [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1], equaling the 15th row of the above triangle.
.                                                           _
.                                                         _|1|
.                                                       _|2 _|
.                                                     _|3  |2|
.                                                   _|4   _|1|
.                                                 _|5    |3 _|
.                                               _|6     _|2|3|
.                                             _|7      |4  |2|
.                                           _|8       _|3 _|1|
.                                         _|9        |5  |4 _|
.                                       _|10        _|4  |3|4|
.                                     _|11         |6   _|2|3|
.                                   _|12          _|5  |5  |2|
.                                 _|13           |7    |4 _|1|
.                               _|14            _|6   _|3|5 _|
.                             _|15             |8    |6  |4|5|
.                           _|16              _|7    |5  |3|4|
.                         _|17               |9     _|4 _|2|3|
.                       _|18                _|8    |7  |6  |2|
.                     _|19                 |10     |6  |5 _|1|
.                   _|20                  _|9     _|5  |4|6 _|
.                 _|21                   |11     |8   _|3|5|6|
.               _|22                    _|10     |7  |7  |4|5|
.             _|23                     |12      _|6  |6  |3|4|
.           _|24                      _|11     |9    |5 _|2|3|
.         _|25                       |13       |8   _|4|7  |2|
.       _|26                        _|12      _|7  |8  |6 _|1|
.     _|27                         |14       |10   |7  |5|7 _|
.    |28                           |13       |9    |6  |4|6|7|
...
The diagram is infinite. For more information about the diagram see A286000.
For an amazing connection with sum of divisors function (A000203) see A237593.

Mathematica

intervals[n_]:=Module[{x, y}, SolveValues[(x^2-y^2+x+y)/2==n&&0<x<=n&&0<y<=n, {x, y}, Integers]];
A299765row[n_]:=Flatten[SortBy[Map[Range[First[#], Last[#], -1]&, intervals[n]], Length]];
nrows=25; Array[A299765row, nrows] (* Paolo Xausa, Jun 19 2022 *)

Prog

(PARI) iscons(p) = my(v = vector(#p-1, k, p[k+1] - p[k])); v == vector(#p-1, i, 1);
row(n) = my(list = List()); forpart(p=n, if (iscons(p), listput(list, Vecrev(p))); ); Vec(list); \\ Michel Marcus, May 11 2022

Crossrefs

Row n has length A204217(n).

Row sums give A245579.

Right border gives A118235.

Column 1 gives A000027.

Records give A000027.

The number of partitions into consecutive parts in row n is A001227(n).

For tables of partitions into consecutive parts see A286000 and A286001.

Cf. A328365 (mirror).

Cf. A352425 (a subsequence).

Cf. A000203, A000217, A026792, A235791, A237048, A237591, A237593, A245092, A285914, A286013, A288529, A288772, A288773, A288774.

Sequence in context: A210500 A336126 A336153 * A104411 A216084 A055101

Adjacent sequences: A299762 A299763 A299764 * A299766 A299767 A299768

Keyword

nonn,tabf,look

Author

Omar E. Pol, Feb 26 2018

Extensions

Name clarified by Omar E. Pol, May 11 2022

Status

approved