

A285914


Irregular triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists k's interleaved with k1 zeros, and the first element of column k is in row k(k+1)/2.


27



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
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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 k1 zeros, and the first element of column k is in the row that is the kth (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

Table of n, a(n) for n=1..112.
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 _03
7 _1 2 0
8 _1 _0 _0
9 _1 2 3 _
10 _1 _0 04
11 _1 2 _00
12 _1 _0 3 0
13 _1 2 0 _0
14 _1 _0 _04 _
15 _1 2 3 05
16 _1 _0 0 00
17 _1 2 _0 _00
18 _1 _0 3 4 0
19 _1 2 0 0 _0
20 _1 _0 _0 05 _
21 _1 2 3 _006
22 _1 _0 0 4 00
23 _1 2 _0 0 00
24 _1 _0 3 0 _00
25 _1 2 0 _05 0
26 _1 _0 _0 4 0 _0
27 _1 2 3 0 06 _
28 1 0 0 0 007
...
Note that the k's are placed exactly below the kth horizontal line segment of every row.
The above structure is related to the triangle A237591, also to the lefthand part of the triangle A237593, and also to the lefthand part of the front view of the pyramid described in A245092.


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.
Cf. A000203, A066839, A196020, A204217, A235791, A236104, A237048, A237591, A237593, A245092, A261699, A285898, A262626, A330889.
Triangles of the same family are A127093, this sequence, A330466, A330888.
Sequence in context: A104276 A216191 A268834 * A096875 A194514 A324667
Adjacent sequences: A285911 A285912 A285913 * A285915 A285916 A285917


KEYWORD

nonn,tabf


AUTHOR

Omar E. Pol, Apr 28 2017


STATUS

approved



