A286013 Irregular triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the positive integers starting with k, interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

1, 2, 3, 2, 4, 0, 5, 3, 6, 0, 3, 7, 4, 0, 8, 0, 0, 9, 5, 4, 10, 0, 0, 4, 11, 6, 0, 0, 12, 0, 5, 0, 13, 7, 0, 0, 14, 0, 0, 5, 15, 8, 6, 0, 5, 16, 0, 0, 0, 0, 17, 9, 0, 0, 0, 18, 0, 7, 6, 0, 19, 10, 0, 0, 0, 20, 0, 0, 0, 6, 21, 11, 8, 0, 0, 6, 22, 0, 0, 7, 0, 0, 23, 12, 0, 0, 0, 0, 24, 0, 9, 0, 0, 0, 25, 13, 0, 0, 7, 0

Offset: 1

Author

Omar E. Pol, Apr 30 2017

Keywords

• nonn
• tabf

Comments

Conjecture 1: T(n,k) is the largest part of the partition of n into k consecutive parts, if T(n,k) > 0.

Conjecture 2: row sums give A286015.

Trapezoidal interpretation from Peter Munn, Jun 18 2017: (Start)

There is one to one correspondence between nonzero T(n,k) and trapezoidal area patterns of n dots on a triangular grid, if we include the limiting cases of triangular patterns, straight lines (k=1) or a single dot (k=n=1). The corresponding pattern has T(n,k) dots in its longest side, k dots in the two adjacent sides and T(n,k)-k+1 dots in the fourth side (where a count of 1 dot may be understood as signifying that side's absence).

Reason: From the definition, for k >= 1, m >= 0, T(A000217(k)+km,k) = k+m, where A000217(k) = k(k+1)/2, the k-th triangular number. First element of column k is T(A000217(k),k) = k: this matches a triangular pattern of A000217(k) dots with 3 sides of k dots. Looking at this pattern as k rows of 1..k dots, extend each row by m dots to create a trapezoidal pattern of A000217(k)+km dots with a longest side of k+m dots and adjacent sides of k dots: this matches T(A000217(k)+km,k) = k+m. As nonzero elements in column k occur at intervals of k, every nonzero T(n,k) has a match. Every trapezoidal pattern can be produced by extending a triangular pattern as described, so they all have a match.

The truth of conjecture 1 follows, since each nonzero T(n,k) = k+m corresponds to a trapezoidal pattern of n dots having k rows with lengths (1+m)..(k+m).

The A270877 sieve is related to this sequence because it eliminates n if it is the sum of consecutive numbers whose largest term has survived the sifting (which may likewise be seen in terms of a trapezoidal dot pattern and its longest side). So the sieve eliminates n if any lesser numbers in A270877 are in row n of this sequence.

(End)

Examples

			Triangle begins:
1;
2;
3,   2;
4,   0;
5,   3;
6,   0,  3;
7,   4,  0;
8,   0,  0;
9,   5,  4;
10,  0,  0,  4;
11,  6,  0,  0;
12,  0,  5,  0;
13,  7,  0,  0;
14,  0,  0,  5;
15,  8,  6,  0,  5;
16,  0,  0,  0,  0;
17,  9,  0,  0,  0;
18,  0,  7,  6,  0;
19, 10,  0,  0,  0;
20,  0,  0,  0,  6;
21, 11,  8,  0,  0,  6;
22,  0,  0,  7,  0,  0;
23, 12,  0,  0,  0,  0;
24,  0,  9,  0,  0,  0;
25, 13,  0,  0,  7,  0;
26,  0,  0,  8,  0,  0;
27, 14, 10,  0,  0,  7;
28,  0,  0,  0,  0,  0,  7;
...
In accordance with the conjecture, 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 largest parts are 15, 8, 6, 5, respectively, so the 15th row of the triangle is [15, 8, 6, 0, 5].
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10944 (Rows 1 <= n <= 528, 528 being first row with 32 columns).

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.

The last positive term in row n is in column A109814(n).

Cf. A196020, A204217, A211343, A235791, A237048, A237591, A237593, A245579, A270877, A286014, A286015.

Programs

Mathematica

With[{n = 7}, DeleteCases[#, m_ /; m < 0] & /@ Transpose@ Table[Apply[Join @@ {ConstantArray[-1, #2 - 1], Array[(k + #/k) Boole[Mod[#, k] == 0] &, #1 - #2 + 1, 0]} &, # (# + 1)/2 & /@ {n, k}], {k, n}]] // Flatten (* Michael De Vlieger, Jul 21 2017 *)

Formula

For k >= 1, m >= 0, T(A000217(k)+km,k) = k+m. - Peter Munn, Jun 19 2017