

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 k1 zeros, and the first element of column k is in row k(k+1)/2.


16



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
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OFFSET

1,2


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 kth 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)


LINKS

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


FORMULA

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


EXAMPLE

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].


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 *)


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.
Sequence in context: A299115 A179590 A167504 * A208083 A181515 A183045
Adjacent sequences: A286010 A286011 A286012 * A286014 A286015 A286016


KEYWORD

nonn,tabf


AUTHOR

Omar E. Pol, Apr 30 2017


STATUS

approved



