Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #54 Mar 12 2021 07:51:47
%S 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,
%T 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,
%U 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
%N 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.
%C The number of positive terms in row n is A001227(n).
%C If n = 2^j then the only positive integer in row n is T(n,1) = n
%C 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.
%C From _Omar E. Pol_, Apr 30 2017: (Start)
%C Conjecture 1: T(n,k) is the smallest part of the partition of n into k consecutive parts, if T(n,k) > 0.
%C Conjecture 2: the last positive integer in the row n is in the column A109814(n). (End)
%H Robert Price, <a href="/A211343/b211343.txt">Table of n, a(n) for n = 1..28864</a> (rows n = 1..1000, flattened)
%F T(n,k) = floor((1 + A196020(n,k))/2).
%F T(n,k) = A237048(n,k)*A286001(n,k). - _Omar E. Pol_, Aug 13 2018
%e Triangle begins:
%e 1;
%e 2;
%e 3, 1;
%e 4, 0;
%e 5, 2;
%e 6, 0, 1;
%e 7, 3, 0;
%e 8, 0, 0;
%e 9, 4, 2;
%e 10, 0, 0, 1;
%e 11, 5, 0, 0;
%e 12, 0, 3, 0;
%e 13, 6, 0, 0;
%e 14, 0, 0, 2;
%e 15, 7, 4, 0, 1;
%e 16, 0, 0, 0, 0;
%e 17, 8, 0, 0, 0;
%e 18, 0, 5, 3, 0;
%e 19, 9, 0, 0, 0;
%e 20, 0, 0, 0, 2;
%e 21, 10, 6, 0, 0, 1;
%e 22, 0, 0, 4, 0, 0;
%e 23, 11, 0, 0, 0, 0;
%e 24, 0, 7, 0, 0, 0;
%e 25, 12, 0, 0, 3, 0;
%e 26, 0, 0, 5, 0, 0;
%e 27, 13, 8, 0, 0, 2;
%e 28, 0, 0, 0, 0, 0, 1;
%e ...
%e 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
%t 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 *)
%o (Python)
%o from sympy import sqrt
%o import math
%o def a196020(n, k):return 2*n/k - k if (n - k*(k + 1)/2)%k == 0 else 0
%o def T(n, k): return int((1 + a196020(n, k))/2)
%o 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
%Y Columns 1-3: A000027, A027656, A175676.
%Y Column k starts in row A000217(k).
%Y Row n has length A003056(n).
%Y Cf. A000203, A001227, A196020, A212119, A235791, A236104, A237048, A237591, A237593, A286001.
%K nonn,tabf,look
%O 1,2
%A _Omar E. Pol_, Feb 05 2013