A352425 -id:A352425 - cp's OEIS frontend

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

Omar E. Pol, Feb 26 2018

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

For m >= 0, row 2^m consists of just one element (2^m). - Paolo Xausa, May 24 2025

			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.

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

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.

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

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

Name clarified by Omar E. Pol, May 11 2022

A352499 Irregular triangle read by rows: T(n,k) is the sum of all parts of the partition of n into consecutive parts that contains 2*k-1 parts, and the first element of the column k is in row A000384(k).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 6, 7, 0, 8, 0, 9, 9, 10, 0, 11, 0, 12, 12, 13, 0, 14, 0, 15, 15, 15, 16, 0, 0, 17, 0, 0, 18, 18, 0, 19, 0, 0, 20, 0, 20, 21, 21, 0, 22, 0, 0, 23, 0, 0, 24, 24, 0, 25, 0, 25, 26, 0, 0, 27, 27, 0, 28, 0, 0, 28, 29, 0, 0, 0, 30, 30, 30, 0, 31, 0, 0, 0, 32, 0, 0, 0

Offset: 1

Omar E. Pol, Mar 19 2022

This triangle is formed from the odd-indexed columns of the triangle A285891.

			Triangle begins:
   1;
   2;
   3;
   4;
   5;
   6,  6;
   7,  0;
   8,  0;
   9,  9;
  10,  0;
  11,  0;
  12, 12;
  13,  0;
  14,  0;
  15, 15, 15;
  16,  0,  0;
  17,  0,  0;
  18, 18,  0;
  19,  0,  0;
  20,  0, 20;
  21, 21,  0;
  22,  0,  0;
  23,  0,  0;
  24, 24,  0;
  25,  0, 25;
  26,  0,  0;
  27, 27,  0;
  28,  0,  0, 28;
  ...
For n = 21 the partitions of 21 into on odd number of consecutive parts are [21] and [8, 7, 6], so T(21,1) = 1 and T(21,2) = 8 + 7 + 6 = 21. There is no partition of 21 into five consecutive parts so T(21,3) = 0.

Paolo Xausa, Table of n, a(n) for n = 1..10490 (rows 1..800 of triangle, flattened).

Row sums give A352257.
Row n has A351846(n) terms.
The number of nonzero terms in row n equals A082647(n).
Cf. A000384, A237048, A245579, A285891, A299765, A351824, A352425.

A352499[rowmax_]:=Table[Boole[Divisible[n,2k-1]]n,{n,rowmax},{k,Floor[(Sqrt[8n+1]+1)/4]}];A352499[50] (* Paolo Xausa, Apr 12 2023 *)

T(n,k) = n*A351824(n,k).

T(n,k) = n*[(2*k-1)|n], where 1 <= k <= floor((sqrt(8*n+1)+1)/4) and [] is the Iverson bracket. - Paolo Xausa, Apr 12 2023

A352269 Irregular triangle read by rows: T(n,k) is the number of parts in the partition of n into 2*k-1 consecutive parts, 1 <= k <= A351846(n). T(n,k) = 0 if no such partition exists.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 0, 1, 0, 1, 3, 1, 0, 1, 0, 1, 3, 1, 0, 1, 0, 1, 3, 5, 1, 0, 0, 1, 0, 0, 1, 3, 0, 1, 0, 0, 1, 0, 5, 1, 3, 0, 1, 0, 0, 1, 0, 0, 1, 3, 0, 1, 0, 5, 1, 0, 0, 1, 3, 0, 1, 0, 0, 7, 1, 0, 0, 0, 1, 3, 5, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 3, 0, 0, 1, 0, 0, 0, 1, 0, 5, 7, 1, 3, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0

Offset: 1

Omar E. Pol, Mar 18 2022

Column k lists 2*k-1 interleaved with 2*k-2 zeros, and the first element of column k is in row A000384(k).
This triangle is formed from the odd-indexed columns of the triangle A285914.
Conjecture 1: row sums give A341309.
Conjecture 2: A347529, A351819, A347263 have the same indices of zero and nonzero terms as A351824 and this sequence.

			Triangle begins:
-----------------------
   n / k   1  2  3  4
-----------------------
   1 |     1;
   2 |     1;
   3 |     1;
   4 |     1;
   5 |     1;
   6 |     1, 3;
   7 |     1, 0;
   8 |     1, 0;
   9 |     1, 3;
  10 |     1, 0;
  11 |     1, 0;
  12 |     1, 3;
  13 |     1, 0;
  14 |     1, 0;
  15 |     1, 3, 5;
  16 |     1, 0, 0;
  17 |     1, 0, 0;
  18 |     1, 3, 0;
  19 |     1, 0, 0;
  20 |     1, 0, 5;
  21 |     1, 3, 0;
  22 |     1, 0, 0;
  23 |     1, 0, 0;
  24 |     1, 3, 0;
  25 |     1, 0, 5;
  26 |     1, 0, 0;
  27 |     1, 3, 0;
  28 |     1, 0, 0, 7;
  ...

Paolo Xausa, Table of n, a(n) for n = 1..10490 (rows 1..800 of triangle, flattened).

Row lengths give A351846.
The number of nonzeros terms in row n equals A082647(n).
Cf. A000384, A285914, A299765, A341309, A347529, A351819, A347263, A351824, A352425.

A352269[rowmax_]:=Table[If[Divisible[n,2k-1],2k-1,0],{n,rowmax},{k,Floor[(Sqrt[8n+1]+1)/4]}];A352269[50] (* Paolo Xausa, Apr 09 2023 *)

T(n,k) = (2*k-1)*A351824(n,k). [Corrected by Paolo Xausa, Apr 09 2023]

T(n,k) = (2*k-1)*[(2*k-1)|n], where 1 <= k <= floor((sqrt(8*n+1)+1)/4) and [] is the Iverson bracket. - Paolo Xausa, Apr 09 2023

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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.

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A352499 Irregular triangle read by rows: T(n,k) is the sum of all parts of the partition of n into consecutive parts that contains 2*k-1 parts, and the first element of the column k is in row A000384(k).

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A352269 Irregular triangle read by rows: T(n,k) is the number of parts in the partition of n into 2*k-1 consecutive parts, 1 <= k <= A351846(n). T(n,k) = 0 if no such partition exists.

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