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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.

%I A299765 #74 May 26 2025 09:53:39
%S A299765 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,
%T A299765 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,
%U A299765 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
%N 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.
%C A299765 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
%C A299765 For m >= 0, row 2^m consists of just one element (2^m). - _Paolo Xausa_, May 24 2025
%H A299765 Paolo Xausa, <a href="/A299765/b299765.txt">Table of n, a(n) for n = 1..10350</a> (rows 1..500 of triangle, flattened)
%e A299765 Triangle begins:
%e A299765    [1];
%e A299765    [2];
%e A299765    [3], [2, 1];
%e A299765    [4];
%e A299765    [5], [3, 2];
%e A299765    [6], [3, 2, 1];
%e A299765    [7], [4, 3];
%e A299765    [8];
%e A299765    [9], [5, 4], [4, 3, 2];
%e A299765   [10], [4, 3, 2, 1];
%e A299765   [11], [6, 5];
%e A299765   [12], [5, 4, 3];
%e A299765   [13], [7, 6];
%e A299765   [14], [5, 4, 3, 2];
%e A299765   [15], [8, 7], [6, 5, 4], [5, 4, 3, 2, 1];
%e A299765   [16];
%e A299765   [17], [9, 8];
%e A299765   [18], [7, 6, 5], [6, 5, 4, 3];
%e A299765   [19], [10, 9];
%e A299765   [20], [6, 5, 4, 3, 2];
%e A299765   [21], [11, 10], [8, 7, 6], [6, 5, 4, 3, 2, 1];
%e A299765   [22], [7, 6, 5, 4];
%e A299765   [23], [12, 11];
%e A299765   [24], [9, 8, 7];
%e A299765   [25], [13, 12], [7, 6, 5, 4, 3];
%e A299765   [26], [8, 7, 6, 5];
%e A299765   [27], [14, 13], [10, 9, 8], [7, 6, 5, 4, 3, 2];
%e A299765   [28], [7, 6, 5, 4, 3, 2, 1];
%e A299765 ...
%e A299765 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.
%e A299765 .                                                           _
%e A299765 .                                                         _|1|
%e A299765 .                                                       _|2 _|
%e A299765 .                                                     _|3  |2|
%e A299765 .                                                   _|4   _|1|
%e A299765 .                                                 _|5    |3 _|
%e A299765 .                                               _|6     _|2|3|
%e A299765 .                                             _|7      |4  |2|
%e A299765 .                                           _|8       _|3 _|1|
%e A299765 .                                         _|9        |5  |4 _|
%e A299765 .                                       _|10        _|4  |3|4|
%e A299765 .                                     _|11         |6   _|2|3|
%e A299765 .                                   _|12          _|5  |5  |2|
%e A299765 .                                 _|13           |7    |4 _|1|
%e A299765 .                               _|14            _|6   _|3|5 _|
%e A299765 .                             _|15             |8    |6  |4|5|
%e A299765 .                           _|16              _|7    |5  |3|4|
%e A299765 .                         _|17               |9     _|4 _|2|3|
%e A299765 .                       _|18                _|8    |7  |6  |2|
%e A299765 .                     _|19                 |10     |6  |5 _|1|
%e A299765 .                   _|20                  _|9     _|5  |4|6 _|
%e A299765 .                 _|21                   |11     |8   _|3|5|6|
%e A299765 .               _|22                    _|10     |7  |7  |4|5|
%e A299765 .             _|23                     |12      _|6  |6  |3|4|
%e A299765 .           _|24                      _|11     |9    |5 _|2|3|
%e A299765 .         _|25                       |13       |8   _|4|7  |2|
%e A299765 .       _|26                        _|12      _|7  |8  |6 _|1|
%e A299765 .     _|27                         |14       |10   |7  |5|7 _|
%e A299765 .    |28                           |13       |9    |6  |4|6|7|
%e A299765 ...
%e A299765 The diagram is infinite. For more information about the diagram see A286000.
%e A299765 For an amazing connection with sum of divisors function (A000203) see A237593.
%t A299765 intervals[n_]:=Module[{x,y},SolveValues[(x^2-y^2+x+y)/2==n&&0<x<=n&&0<y<=n,{x,y},Integers]];
%t A299765 A299765row[n_]:=Flatten[SortBy[Map[Range[First[#],Last[#],-1]&,intervals[n]],Length]];
%t A299765 nrows=25;Array[A299765row,nrows] (* _Paolo Xausa_, Jun 19 2022 *)
%o A299765 (PARI) iscons(p) = my(v = vector(#p-1, k, p[k+1] - p[k])); v == vector(#p-1, i, 1);
%o A299765 row(n) = my(list = List()); forpart(p=n, if (iscons(p), listput(list, Vecrev(p)));); Vec(list); \\ _Michel Marcus_, May 11 2022
%Y A299765 Row n has length A204217(n).
%Y A299765 Row sums give A245579.
%Y A299765 Right border gives A118235.
%Y A299765 Column 1 gives A000027.
%Y A299765 Records give A000027.
%Y A299765 The number of partitions into consecutive parts in row n is A001227(n).
%Y A299765 For tables of partitions into consecutive parts see A286000 and A286001.
%Y A299765 Cf. A328365 (mirror).
%Y A299765 Cf. A352425 (a subsequence).
%Y A299765 Cf. A000203, A000217, A026792, A235791, A237048, A237591, A237593, A245092, A285914, A286013, A288529, A288772, A288773, A288774.
%K A299765 nonn,tabf,look
%O A299765 1,2
%A A299765 _Omar E. Pol_, Feb 26 2018
%E A299765 Name clarified by _Omar E. Pol_, May 11 2022