cp's OEIS Frontend

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.

%I A286013 #33 Jul 22 2017 03:03:30
%S A286013 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,
%T A286013 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,
%U A286013 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
%N 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.
%C A286013 Conjecture 1: T(n,k) is the largest part of the partition of n into k consecutive parts, if T(n,k) > 0.
%C A286013 Conjecture 2: row sums give A286015.
%C A286013 Trapezoidal interpretation from _Peter Munn_, Jun 18 2017: (Start)
%C A286013 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).
%C A286013 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.
%C A286013 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).
%C A286013 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.
%C A286013 (End)
%H A286013 Michael De Vlieger, <a href="/A286013/b286013.txt">Table of n, a(n) for n = 1..10944</a> (Rows 1 <= n <= 528, 528 being first row with 32 columns).
%F A286013 For k >= 1, m >= 0, T(A000217(k)+km,k) = k+m. - _Peter Munn_, Jun 19 2017
%e A286013 Triangle begins:
%e A286013 1;
%e A286013 2;
%e A286013 3,   2;
%e A286013 4,   0;
%e A286013 5,   3;
%e A286013 6,   0,  3;
%e A286013 7,   4,  0;
%e A286013 8,   0,  0;
%e A286013 9,   5,  4;
%e A286013 10,  0,  0,  4;
%e A286013 11,  6,  0,  0;
%e A286013 12,  0,  5,  0;
%e A286013 13,  7,  0,  0;
%e A286013 14,  0,  0,  5;
%e A286013 15,  8,  6,  0,  5;
%e A286013 16,  0,  0,  0,  0;
%e A286013 17,  9,  0,  0,  0;
%e A286013 18,  0,  7,  6,  0;
%e A286013 19, 10,  0,  0,  0;
%e A286013 20,  0,  0,  0,  6;
%e A286013 21, 11,  8,  0,  0,  6;
%e A286013 22,  0,  0,  7,  0,  0;
%e A286013 23, 12,  0,  0,  0,  0;
%e A286013 24,  0,  9,  0,  0,  0;
%e A286013 25, 13,  0,  0,  7,  0;
%e A286013 26,  0,  0,  8,  0,  0;
%e A286013 27, 14, 10,  0,  0,  7;
%e A286013 28,  0,  0,  0,  0,  0,  7;
%e A286013 ...
%e A286013 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].
%t A286013 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 *)
%Y A286013 Row n has length A003056(n).
%Y A286013 Column k starts in row A000217(k).
%Y A286013 The number of positive terms in row n is A001227(n), the number of partitions of n into consecutive parts.
%Y A286013 The last positive term in row n is in column A109814(n).
%Y A286013 Cf. A196020, A204217, A211343, A235791, A237048, A237591, A237593, A245579, A270877, A286014, A286015.
%K A286013 nonn,tabf
%O A286013 1,2
%A A286013 _Omar E. Pol_, Apr 30 2017