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%I #38 Jun 20 2022 10:13:38
%S 1,2,1,2,3,4,2,3,5,1,2,3,6,3,4,7,8,2,3,4,4,5,9,1,2,3,4,10,5,6,11,3,4,
%T 5,12,6,7,13,2,3,4,5,14,1,2,3,4,5,4,5,6,7,8,15,16,8,9,17,3,4,5,6,5,6,
%U 7,18,9,10,19,2,3,4,5,6,20,1,2,3,4,5,6,6,7,8,10,11,21,4,5,6,7,22,11,12,23,7,8,9,24
%N Irregular triangle read by rows, T(n,k), n >= 1, k >= 1, in which row n lists in reverse order the partitions of n into consecutive parts.
%H Paolo Xausa, <a href="/A328365/b328365.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 [1, 2], [3];
%e [4];
%e [2, 3], [5];
%e [1, 2, 3], [6];
%e [3, 4], [7];
%e [8];
%e [2, 3, 4], [4, 5], [9];
%e [1, 2, 3, 4], [10];
%e [5, 6], [11];
%e [3, 4, 5], [12];
%e [6, 7], [13];
%e [2, 3, 4, 5], [14];
%e [1, 2, 3, 4, 5], [4, 5, 6], [7, 8], [15];
%e [16];
%e [8, 9], [17];
%e [3, 4, 5, 6], [5, 6, 7], [18];
%e [9, 10], [19];
%e [2, 3, 4, 5, 6], [20];
%e [1, 2, 3, 4, 5, 6], [6, 7, 8], [10, 11], [21];
%e [4, 5, 6, 7], [22];
%e [11, 12], [23];
%e [7, 8, 9], [24];
%e [3, 4, 5, 6, 7], [12, 13], [25];
%e [5, 6, 7, 8], [26];
%e [2, 3, 4, 5, 6, 7], [8, 9, 10], [13, 14], [27];
%e [1, 2, 3, 4, 5, 6, 7], [28];
%e ...
%e For n = 9 there are three partitions of 9 into consecutive parts, they are [9], [5, 4], [4, 3, 2], so the 9th row of triangle is [2, 3, 4], [4, 5], [9].
%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 [1, 2, 3, 4, 5], [4, 5, 6], [7, 8], [15], equaling the 15th row of the above triangle.
%e Row _
%e 1 |1|_
%e 2 |_ 2|_
%e 3 |1| 3|_
%e 4 |2|_ 4|_
%e 5 |_ 2| 5|_
%e 6 |1|3|_ 6|_
%e 7 |2| 3| 7|_
%e 8 |3|_ 4|_ 8|_
%e 9 |_ 2| 4| 9|_
%e 10 |1|3| 5|_ 10|_
%e 11 |2|4|_ 5| 11|_
%e 12 |3| 3| 6|_ 12|_
%e 13 |4|_ 4| 6| 13|_
%e 14 |_ 2|5|_ 7|_ 14|_
%e 15 |1|3| 4| 7| 15|_
%e 16 |2|4| 5| 8|_ 16|_
%e 17 |3|5|_ 6|_ 8| 17|_
%e 18 |4| 3| 5| 9|_ 18|_
%e 19 |5|_ 4| 6| 9| 19|_
%e 20 |_ 2|5| 7|_ 10|_ 20|_
%e 21 |1|3|6|_ 6| 10| 21|_
%e 22 |2|4| 4| 7| 11|_ 22|_
%e 23 |3|5| 5| 8|_ 11| 23|_
%e 24 |4|6|_ 6| 7| 12|_ 24|_
%e 25 |5| 3|7|_ 8| 12| 25|_
%e 26 |6|_ 4| 5| 9|_ 13|_ 26|_
%e 27 |_ 2|5| 6| 8| 13| 27|_
%e 28 |1|3|6| 7| 9| 14| 28|
%e ...
%e The diagram is infinite. For more information about the diagram see A286001.
%e For an amazing connection with sum of divisors function (A000203) see A237593.
%t Table[With[{h = Floor[n/2] - Boole[EvenQ@ n]},Append[Array[Which[Total@ # == n, #, Total@ Most@ # == n, Most[#], True, Nothing] &@ NestWhile[Append[#, #[[-1]] + 1] &, {#}, Total@ # <= n &, 1, h - # + 1] &, h], {n}]], {n, 24}] // Flatten (* _Michael De Vlieger_, Oct 22 2019 *)
%Y Mirror of A299765.
%Y Row n has length A204217(n).
%Y Row sums give A245579.
%Y Column 1 gives A118235.
%Y Right border gives A000027.
%Y Records give A000027.
%Y Where records occur gives A285899.
%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. A000203, A026792, A235791, A237048, A237591, A237593, A245092, A285914, A286013, A288529, A288772, A288773, A288774, A328361, A328362.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Oct 22 2019
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