|
|
A211343
|
|
Triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the positive integers interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.
|
|
27
|
|
|
1, 2, 3, 1, 4, 0, 5, 2, 6, 0, 1, 7, 3, 0, 8, 0, 0, 9, 4, 2, 10, 0, 0, 1, 11, 5, 0, 0, 12, 0, 3, 0, 13, 6, 0, 0, 14, 0, 0, 2, 15, 7, 4, 0, 1, 16, 0, 0, 0, 0, 17, 8, 0, 0, 0, 18, 0, 5, 3, 0, 19, 9, 0, 0, 0, 20, 0, 0, 0, 2, 21, 10, 6, 0, 0, 1, 22, 0, 0, 4, 0, 0, 23, 11, 0, 0, 0, 0, 24, 0, 7, 0, 0, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
The number of positive terms in row n is A001227(n).
If n = 2^j then the only positive integer in row n is T(n,1) = n
If n is an odd prime then the only two positive integers in row n are T(n,1) = n and T(n,2) = (n - 1)/2.
Conjecture 1: T(n,k) is the smallest part of the partition of n into k consecutive parts, if T(n,k) > 0.
Conjecture 2: the last positive integer in the row n is in the column A109814(n). (End)
|
|
LINKS
|
|
|
FORMULA
|
T(n,k) = floor((1 + A196020(n,k))/2).
|
|
EXAMPLE
|
Triangle begins:
1;
2;
3, 1;
4, 0;
5, 2;
6, 0, 1;
7, 3, 0;
8, 0, 0;
9, 4, 2;
10, 0, 0, 1;
11, 5, 0, 0;
12, 0, 3, 0;
13, 6, 0, 0;
14, 0, 0, 2;
15, 7, 4, 0, 1;
16, 0, 0, 0, 0;
17, 8, 0, 0, 0;
18, 0, 5, 3, 0;
19, 9, 0, 0, 0;
20, 0, 0, 0, 2;
21, 10, 6, 0, 0, 1;
22, 0, 0, 4, 0, 0;
23, 11, 0, 0, 0, 0;
24, 0, 7, 0, 0, 0;
25, 12, 0, 0, 3, 0;
26, 0, 0, 5, 0, 0;
27, 13, 8, 0, 0, 2;
28, 0, 0, 0, 0, 0, 1;
...
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]. The smallest parts of these partitions are 15, 7, 4, 1, respectively, so the 15th row of the triangle is [15, 7, 4, 0, 1]. - Omar E. Pol, Apr 30 2017
|
|
MATHEMATICA
|
a196020[n_, k_]:=If[Divisible[n - k(k + 1)/2, k], 2n/k - k, 0]; T[n_, k_]:= Floor[(1 + a196020[n, k])/2]; Table[T[n, k], {n, 28}, {k, Floor[(Sqrt[8n+1]-1)/2]}] // Flatten (* Indranil Ghosh, Apr 30 2017 *)
|
|
PROG
|
(Python)
from sympy import sqrt
import math
def a196020(n, k):return 2*n/k - k if (n - k*(k + 1)/2)%k == 0 else 0
def T(n, k): return int((1 + a196020(n, k))/2)
for n in range(1, 29): print([T(n, k) for k in range(1, int((sqrt(8*n + 1) - 1)/2) + 1)]) # Indranil Ghosh, Apr 30 2017
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|