OFFSET
1,2
COMMENTS
In the square array, note that the column k starts with k-1 zeros. Then list each partition number of positive integers followed by k-1 zeros. See A000041, which is the main entry for this sequence.
LINKS
G. C. Greubel, Antidiagonals n = 1..50, flattened
FORMULA
EXAMPLE
The array, A(n, k), begins:
n | k = 1 2 3 4 5 6 7 8 9 10 11 12
---+--------------------------------------------------
1 | 1 0 0 0 0 0 0 0 0 0 0 0
2 | 2 1 0 0 0 0 0 0 0 0 0 0
3 | 3 0 1 0 0 0 0 0 0 0 0 0
4 | 5 2 0 1 0 0 0 0 0 0 0 0
5 | 7 0 0 0 1 0 0 0 0 0 0 0
6 | 11 3 2 0 0 1 0 0 0 0 0 0
7 | 15 0 0 0 0 0 1 0 0 0 0 0
8 | 22 5 0 2 0 0 0 1 0 0 0 0
9 | 30 0 3 0 0 0 0 0 1 0 0 0
10 | 42 7 0 0 2 0 0 0 0 1 0 0
11 | 56 0 0 0 0 0 0 0 0 0 1 0
12 | 77 11 5 3 0 2 0 0 0 0 0 1
...
Antidiagonal triangle, T(n,k), begins as:
1;
2, 0;
3, 1, 0;
5, 0, 0, 0;
7, 2, 1, 0, 0;
11, 0, 0, 0, 0, 0;
15, 3, 0, 1, 0, 0, 0;
22, 0, 2, 0, 0, 0, 0, 0;
30, 5, 0, 0, 1, 0, 0, 0, 0;
42, 0, 0, 0, 0, 0, 0, 0, 0, 0;
MATHEMATICA
T[n_, k_]:= If[IntegerQ[(n-k+1)/k], PartitionsP[(n-k+1)/k], 0];
Table[T[n, k], {n, 15}, {k, n}]//Flatten (* G. C. Greubel, Jan 12 2023 *)
PROG
(SageMath)
def A168020(n, k): return number_of_partitions((n-k+1)/k) if ((n-k+1)%k)==0 else 0
flatten([[A168020(n, k) for k in range(1, n+1)] for n in range(1, 16)]) # G. C. Greubel, Jan 12 2023
CROSSREFS
KEYWORD
AUTHOR
Omar E. Pol, Nov 20 2009
EXTENSIONS
Edited by Omar E. Pol, Nov 21 2009
Edited by Charles R Greathouse IV, Mar 23 2010
Edited by Max Alekseyev, May 07 2010
STATUS
approved