login
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).
4
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).
Sequence in context: A376367 A097622 A236561 * A110010 A091987 A357149
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Mar 19 2022
STATUS
approved