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