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A285914 Irregular triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists k's interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2. 24

%I

%S 1,1,1,2,1,0,1,2,1,0,3,1,2,0,1,0,0,1,2,3,1,0,0,4,1,2,0,0,1,0,3,0,1,2,

%T 0,0,1,0,0,4,1,2,3,0,5,1,0,0,0,0,1,2,0,0,0,1,0,3,4,0,1,2,0,0,0,1,0,0,

%U 0,5,1,2,3,0,0,6,1,0,0,4,0,0,1,2,0,0,0,0,1,0,3,0,0,0,1,2,0,0,5,0,1,0,0,4,0,0,1,2,3,0,0,6

%N Irregular triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists k's interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

%C Conjecture 1: T(n,k) is the number of parts in the partition of n into k consecutive parts, if T(n,k) > 0.

%C Conjecture 2: row sums give A204217, which should be also the total number of parts in all partitions of n into consecutive parts.

%C (The conjectures are true. See _Joerg Arndt_'s proof in the Links section.) - _Omar E. Pol_, Jun 14 2017

%C From _Omar E. Pol_, May 05 2020: (Start)

%C Theorem: Let T(n,k) be an irregular triangle read by rows in which column k lists k's interleaved with k-1 zeros, and the first element of column k is in the row that is the k-th (m+2)-gonal number, with n >= 1, k >= 1, m >= 0. T(n,k) is also the number of parts in the partition of n into k consecutive parts that differ by m, including n as a valid partition. Hence the sum of row n gives the total number of parts in all partitions of n into consecutive parts that differ by m.

%C About the above theorem, this is the case for m = 1. For m = 0 see the triangle A127093, in which row sums give A000203. For m = 2 see the triangle A330466, in which row sums give A066839 (conjectured). For m = 3 see the triangle A330888, in which row sums give A330889.

%C Note that there are infinitely many triangles of this kind, with m >= 0. Also, every triangle can be represented with a diagram of overlapping curves, in which every column of triangle is represented by a periodic curve. (End)

%H Joerg Arndt, <a href="http://list.seqfan.eu/pipermail/seqfan/2017-June/017633.html">Proof of the conjectures of A204217 and A285914</a>, SeqFan Mailing Lists, Jun 03 2017.

%F T(n,k) = k*A237048(n,k).

%e Triangle begins (rows 1..28):

%e 1;

%e 1;

%e 1, 2;

%e 1, 0;

%e 1, 2;

%e 1, 0, 3;

%e 1, 2, 0;

%e 1, 0, 0;

%e 1, 2, 3;

%e 1, 0, 0, 4;

%e 1, 2, 0, 0;

%e 1, 0, 3, 0;

%e 1, 2, 0, 0;

%e 1, 0, 0, 4;

%e 1, 2, 3, 0, 5;

%e 1, 0, 0, 0, 0;

%e 1, 2, 0, 0, 0;

%e 1, 0, 3, 4, 0;

%e 1, 2, 0, 0, 0;

%e 1, 0, 0, 0, 5;

%e 1, 2, 3, 0, 0, 6;

%e 1, 0, 0, 4, 0, 0;

%e 1, 2, 0, 0, 0, 0;

%e 1, 0, 3, 0, 0, 0;

%e 1, 2, 0, 0, 5, 0;

%e 1, 0, 0, 4, 0, 0;

%e 1, 2, 3, 0, 0, 6;

%e 1, 0, 0, 0, 0, 0, 7;

%e ...

%e In accordance with the conjectures, for n = 15 there are four partitions of 15 into consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1]. These partitions are formed by 1, 2, 3 and 5 consecutive parts respectively, so the 15th row of the triangle is [1, 2, 3, 0, 5].

%e Illustration of initial terms:

%e Row _

%e 1 _|1|

%e 2 _|1 _|

%e 3 _|1 |2|

%e 4 _|1 _|0|

%e 5 _|1 |2 _|

%e 6 _|1 _|0|3|

%e 7 _|1 |2 |0|

%e 8 _|1 _|0 _|0|

%e 9 _|1 |2 |3 _|

%e 10 _|1 _|0 |0|4|

%e 11 _|1 |2 _|0|0|

%e 12 _|1 _|0 |3 |0|

%e 13 _|1 |2 |0 _|0|

%e 14 _|1 _|0 _|0|4 _|

%e 15 _|1 |2 |3 |0|5|

%e 16 _|1 _|0 |0 |0|0|

%e 17 _|1 |2 _|0 _|0|0|

%e 18 _|1 _|0 |3 |4 |0|

%e 19 _|1 |2 |0 |0 _|0|

%e 20 _|1 _|0 _|0 |0|5 _|

%e 21 _|1 |2 |3 _|0|0|6|

%e 22 _|1 _|0 |0 |4 |0|0|

%e 23 _|1 |2 _|0 |0 |0|0|

%e 24 _|1 _|0 |3 |0 _|0|0|

%e 25 _|1 |2 |0 _|0|5 |0|

%e 26 _|1 _|0 _|0 |4 |0 _|0|

%e 27 _|1 |2 |3 |0 |0|6 _|

%e 28 |1 |0 |0 |0 |0|0|7|

%e ...

%e Note that the k's are placed exactly below the k-th horizontal line segment of every row.

%e The above structure is related to the triangle A237591, also to the left-hand part of the triangle A237593, and also to the left-hand part of the front view of the pyramid described in A245092.

%t With[{nn = 6}, Table[Boole[If[EvenQ@ k, Mod[(n - k/2), k] == 0, Mod[n, k] == 0]] k, {n, nn (nn + 3)/2}, {k, Floor[((Sqrt[8 n + 1] - 1)/2)]}]] // Flatten (* _Michael De Vlieger_, Jun 15 2017, after Python by _Indranil Ghosh_ *)

%o (Python)

%o from sympy import sqrt

%o import math

%o def a237048(n, k):

%o return int(n%k == 0) if k%2 else int(((n - k//2)%k) == 0)

%o def T(n, k): return k*a237048(n, k)

%o for n in range(1, 29): print([T(n, k) for k in range(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)]) # _Indranil Ghosh_, Apr 30 2017

%o (PARI) t(n, k) = if (k % 2, (n % k) == 0, ((n - k/2) % k) == 0); \\ A237048

%o tabf(nn) = {for (n=1, nn, for (k=1, floor((sqrt(1+8*n)-1)/2), print1(k*t(n, k), ", "); ); print(); ); } \\ _Michel Marcus_, Nov 04 2019

%Y Row n has length A003056(n).

%Y Column k starts in row A000217(k).

%Y The number of positive terms in row n is A001227(n), the number of partitions of n into consecutive parts.

%Y Cf. A000203, A066839, A196020, A204217, A235791, A236104, A237048, A237591, A237593, A245092, A261699, A285898, A262626, A330889.

%Y Triangles of the same family are A127093, this sequence, A330466, A330888.

%K nonn,tabf

%O 1,4

%A _Omar E. Pol_, Apr 28 2017

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