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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.
27

%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