OFFSET
1,4
COMMENTS
Conjecture 1: T(n,k) is the number of parts in the partition of n into k consecutive parts, if T(n,k) > 0.
Conjecture 2: row sums give A204217, which should be also the total number of parts in all partitions of n into consecutive parts.
(The conjectures are true. See Joerg Arndt's proof in the Links section.) - Omar E. Pol, Jun 14 2017
From Omar E. Pol, May 05 2020: (Start)
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.
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.
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)
LINKS
Joerg Arndt, Proof of the conjectures of A204217 and A285914, SeqFan Mailing Lists, Jun 03 2017.
FORMULA
T(n,k) = k*A237048(n,k).
EXAMPLE
Triangle begins (rows 1..28):
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, 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, 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;
1, 0, 0, 0, 0, 0, 7;
...
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].
Illustration of initial terms:
Row _
1 _|1|
2 _|1 _|
3 _|1 |2|
4 _|1 _|0|
5 _|1 |2 _|
6 _|1 _|0|3|
7 _|1 |2 |0|
8 _|1 _|0 _|0|
9 _|1 |2 |3 _|
10 _|1 _|0 |0|4|
11 _|1 |2 _|0|0|
12 _|1 _|0 |3 |0|
13 _|1 |2 |0 _|0|
14 _|1 _|0 _|0|4 _|
15 _|1 |2 |3 |0|5|
16 _|1 _|0 |0 |0|0|
17 _|1 |2 _|0 _|0|0|
18 _|1 _|0 |3 |4 |0|
19 _|1 |2 |0 |0 _|0|
20 _|1 _|0 _|0 |0|5 _|
21 _|1 |2 |3 _|0|0|6|
22 _|1 _|0 |0 |4 |0|0|
23 _|1 |2 _|0 |0 |0|0|
24 _|1 _|0 |3 |0 _|0|0|
25 _|1 |2 |0 _|0|5 |0|
26 _|1 _|0 _|0 |4 |0 _|0|
27 _|1 |2 |3 |0 |0|6 _|
28 |1 |0 |0 |0 |0|0|7|
...
Note that the k's are placed exactly below the k-th horizontal line segment of every row.
MATHEMATICA
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 *)
PROG
(Python)
from sympy import sqrt
import math
def a237048(n, k):
return int(n%k == 0) if k%2 else int(((n - k//2)%k) == 0)
def T(n, k): return k*a237048(n, k)
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
(PARI) t(n, k) = if (k % 2, (n % k) == 0, ((n - k/2) % k) == 0); \\ A237048
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
CROSSREFS
Row n has length A003056(n).
Column k starts in row A000217(k).
The number of positive terms in row n is A001227(n), the number of partitions of n into consecutive parts.
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Apr 28 2017
STATUS
approved