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.

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

Offset: 1

Author

Omar E. Pol, Feb 05 2013

Keywords

• nonn
• tabf
• look

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.

From Omar E. Pol, Apr 30 2017: (Start)

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)

Examples

			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
		

Links

• Robert Price, Table of n, a(n) for n = 1..28864 (rows n = 1..1000, flattened)

Crossrefs

Columns 1-3: A000027, A027656, A175676.

Column k starts in row A000217(k).

Row n has length A003056(n).

Cf. A000203, A001227, A196020, A212119, A235791, A236104, A237048, A237591, A237593, A286001.

Programs

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 *)

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

Formula

T(n,k) = floor((1 + A196020(n,k))/2).

T(n,k) = A237048(n,k)*A286001(n,k). - Omar E. Pol, Aug 13 2018