A352499 Irregular triangle read by rows: T(n,k) is the sum of all parts of the partition of n into consecutive parts that contains 2*k-1 parts, and the first element of the column k is in row A000384(k).

1, 2, 3, 4, 5, 6, 6, 7, 0, 8, 0, 9, 9, 10, 0, 11, 0, 12, 12, 13, 0, 14, 0, 15, 15, 15, 16, 0, 0, 17, 0, 0, 18, 18, 0, 19, 0, 0, 20, 0, 20, 21, 21, 0, 22, 0, 0, 23, 0, 0, 24, 24, 0, 25, 0, 25, 26, 0, 0, 27, 27, 0, 28, 0, 0, 28, 29, 0, 0, 0, 30, 30, 30, 0, 31, 0, 0, 0, 32, 0, 0, 0

Offset

1,2

Comments

This triangle is formed from the odd-indexed columns of the triangle A285891.

Links

Paolo Xausa, Table of n, a(n) for n = 1..10490 (rows 1..800 of triangle, flattened).

Formula

T(n,k) = n*A351824(n,k).

T(n,k) = n*[(2*k-1)|n], where 1 <= k <= floor((sqrt(8*n+1)+1)/4) and [] is the Iverson bracket. - Paolo Xausa, Apr 12 2023

Example

Triangle begins:
   1;
   2;
   3;
   4;
   5;
   6,  6;
   7,  0;
   8,  0;
   9,  9;
  10,  0;
  11,  0;
  12, 12;
  13,  0;
  14,  0;
  15, 15, 15;
  16,  0,  0;
  17,  0,  0;
  18, 18,  0;
  19,  0,  0;
  20,  0, 20;
  21, 21,  0;
  22,  0,  0;
  23,  0,  0;
  24, 24,  0;
  25,  0, 25;
  26,  0,  0;
  27, 27,  0;
  28,  0,  0, 28;
  ...
For n = 21 the partitions of 21 into on odd number of consecutive parts are [21] and [8, 7, 6], so T(21,1) = 1 and T(21,2) = 8 + 7 + 6 = 21. There is no partition of 21 into five consecutive parts so T(21,3) = 0.

Mathematica

A352499[rowmax_]:=Table[Boole[Divisible[n, 2k-1]]n, {n, rowmax}, {k, Floor[(Sqrt[8n+1]+1)/4]}]; A352499[50] (* Paolo Xausa, Apr 12 2023 *)

Crossrefs

Row sums give A352257.

Row n has A351846(n) terms.

The number of nonzero terms in row n equals A082647(n).

Cf. A000384, A237048, A245579, A285891, A299765, A351824, A352425.

Sequence in context: A069754 A097622 A236561 * A110010 A091987 A357149

Adjacent sequences: A352496 A352497 A352498 * A352500 A352501 A352502

Keyword

nonn,tabf

Author

Omar E. Pol, Mar 19 2022

Status

approved