A351824 - OEIS

%I #93 Apr 09 2023 02:32:03

%S 1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,1,0,0,1,0,0,1,1,

%T 0,1,0,0,1,0,1,1,1,0,1,0,0,1,0,0,1,1,0,1,0,1,1,0,0,1,1,0,1,0,0,1,1,0,

%U 0,0,1,1,1,0,1,0,0,0,1,0,0,0,1,1,0,0,1,0,0,0,1,0,1,1,1,1,0,0,1,0,0,0,1,0,0,0

%N Irregular triangle read by rows: T(n,k) is the number of partitions of n into 2*k-1 consecutive parts, n >= 1, k >= 1. Column k lists 1's interleaved with 2*k-2 zeros, and the first element of column k is in row A000384(k).

%C Conjecture 1: T(n,k) is the number of subparts, in an octant of the symmetric representation of sigma(n), that arises from the (2*k-1)-th double-staircase of the double-staircases diagram of n described in A335616.

%C Conjecture 2: Indices of 1's coincide with indices of nonzero terms in A347263, A347529, A351819.

%C For the above conjectures see also the "ziggurat" diagram described in A347186.

%C This triangle is formed by the odd-indexed columns of the triangle A237048.

%C Terms can be 0 or 1.

%H Paolo Xausa, <a href="/A351824/b351824.txt">Table of n, a(n) for n = 1..10490</a> (rows 1..800 of triangle, flattened).

%F T(n,k) = A352499(n,k)/n. - _Omar E. Pol_, Mar 24 2022

%F T(n,k) = [(2*k-1)|n], where 1 <= k <= floor((sqrt(8*n+1)+1)/4) and [] is the Iverson bracket. - _Paolo Xausa_, Apr 01 2023

%e Triangle begins:

%e -----------------------

%e    n / k   1  2  3  4

%e -----------------------

%e    1 |     1;

%e    2 |     1;

%e    3 |     1;

%e    4 |     1;

%e    5 |     1;

%e    6 |     1, 1;

%e    7 |     1, 0;

%e    8 |     1, 0;

%e    9 |     1, 1;

%e   10 |     1, 0;

%e   11 |     1, 0;

%e   12 |     1, 1;

%e   13 |     1, 0;

%e   14 |     1, 0;

%e   15 |     1, 1, 1;

%e   16 |     1, 0, 0;

%e   17 |     1, 0, 0;

%e   18 |     1, 1, 0;

%e   19 |     1, 0, 0;

%e   20 |     1, 0, 1;

%e   21 |     1, 1, 0;

%e   22 |     1, 0, 0;

%e   23 |     1, 0, 0;

%e   24 |     1, 1, 0;

%e   25 |     1, 0, 1;

%e   26 |     1, 0, 0;

%e   27 |     1, 1, 0;

%e   28 |     1, 0, 0, 1;

%e   ...

%e For n = 15 the partitions of 15 into an odd number of consecutive parts are [15], [6, 5, 4] and [5, 4, 3, 2, 1]. There are a partition with only one part, a partition with three parts and a partition with five parts, so the 15th row of triangle is [1, 1, 1].

%t A351824[rowmax_]:=Table[Boole[Divisible[n,2k-1]],{n,rowmax},{k,Floor[(Sqrt[8n+1]+1)/4]}];A351824[50] (* _Paolo Xausa_, Apr 01 2023 *)

%Y Row sums give A082647.

%Y Row n has length A351846(n).

%Y Cf. A000384, A196020, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A244250, A262619, A262626, A279387, A280850, A280851, A286000, A286001, A296508, A299765, A335616, A347186, A347263, A347529, A348854, A351819, A352257, A352499.

%K nonn,tabf,easy

%O 1

%A _Omar E. Pol_, Feb 20 2022