%I #70 Jun 20 2022 10:13:03
%S 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 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 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 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 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
%H Paolo Xausa, <a href="/A299765/b299765.txt">Table of n, a(n) for n = 1..10350</a> (rows 1..500 of triangle, flattened)
%F T(2^m,1) = 2^m, for m >= 0. - _Paolo Xausa_, Jun 19 2022
%e Triangle begins:
%e [1];
%e [2];
%e [3], [2, 1];
%e [4];
%e [5], [3, 2];
%e [6], [3, 2, 1];
%e [7], [4, 3];
%e [8];
%e [9], [5, 4], [4, 3, 2];
%e [10], [4, 3, 2, 1];
%e [11], [6, 5];
%e [12], [5, 4, 3];
%e [13], [7, 6];
%e [14], [5, 4, 3, 2];
%e [15], [8, 7], [6, 5, 4], [5, 4, 3, 2, 1];
%e [16];
%e [17], [9, 8];
%e [18], [7, 6, 5], [6, 5, 4, 3];
%e [19], [10, 9];
%e [20], [6, 5, 4, 3, 2];
%e [21], [11, 10], [8, 7, 6], [6, 5, 4, 3, 2, 1];
%e [22], [7, 6, 5, 4];
%e [23], [12, 11];
%e [24], [9, 8, 7];
%e [25], [13, 12], [7, 6, 5, 4, 3];
%e [26], [8, 7, 6, 5];
%e [27], [14, 13], [10, 9, 8], [7, 6, 5, 4, 3, 2];
%e [28], [7, 6, 5, 4, 3, 2, 1];
%e ...
%e 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 . _
%e . _|1|
%e . _|2 _|
%e . _|3 |2|
%e . _|4 _|1|
%e . _|5 |3 _|
%e . _|6 _|2|3|
%e . _|7 |4 |2|
%e . _|8 _|3 _|1|
%e . _|9 |5 |4 _|
%e . _|10 _|4 |3|4|
%e . _|11 |6 _|2|3|
%e . _|12 _|5 |5 |2|
%e . _|13 |7 |4 _|1|
%e . _|14 _|6 _|3|5 _|
%e . _|15 |8 |6 |4|5|
%e . _|16 _|7 |5 |3|4|
%e . _|17 |9 _|4 _|2|3|
%e . _|18 _|8 |7 |6 |2|
%e . _|19 |10 |6 |5 _|1|
%e . _|20 _|9 _|5 |4|6 _|
%e . _|21 |11 |8 _|3|5|6|
%e . _|22 _|10 |7 |7 |4|5|
%e . _|23 |12 _|6 |6 |3|4|
%e . _|24 _|11 |9 |5 _|2|3|
%e . _|25 |13 |8 _|4|7 |2|
%e . _|26 _|12 _|7 |8 |6 _|1|
%e . _|27 |14 |10 |7 |5|7 _|
%e . |28 |13 |9 |6 |4|6|7|
%e ...
%e The diagram is infinite. For more information about the diagram see A286000.
%e For an amazing connection with sum of divisors function (A000203) see A237593.
%t intervals[n_]:=Module[{x,y},SolveValues[(x^2-y^2+x+y)/2==n&&0<x<=n&&0<y<=n,{x,y},Integers]];
%t A299765row[n_]:=Flatten[SortBy[Map[Range[First[#],Last[#],-1]&,intervals[n]],Length]];
%t nrows=25;Array[A299765row,nrows] (* _Paolo Xausa_, Jun 19 2022 *)
%o (PARI) iscons(p) = my(v = vector(#p-1, k, p[k+1] - p[k])); v == vector(#p-1, i, 1);
%o row(n) = my(list = List()); forpart(p=n, if (iscons(p), listput(list, Vecrev(p)));); Vec(list); \\ _Michel Marcus_, May 11 2022
%Y Row n has length A204217(n).
%Y Row sums give A245579.
%Y Right border gives A118235.
%Y Column 1 gives A000027.
%Y Records give A000027.
%Y The number of partitions into consecutive parts in row n is A001227(n).
%Y For tables of partitions into consecutive parts see A286000 and A286001.
%Y Cf. A328365 (mirror).
%Y Cf. A352425 (a subsequence).
%Y Cf. A000203, A000217, A026792, A235791, A237048, A237591, A237593, A245092, A285914, A286013, A288529, A288772, A288773, A288774.
%K nonn,tabf,look
%O 1,2
%A _Omar E. Pol_, Feb 26 2018
%E Name clarified by _Omar E. Pol_, May 11 2022