A168019 Square array A(n,k) read by antidiagonals, in which row n lists the number of partitions of n into parts divisible by k+1.

1, 1, 1, 2, 0, 1, 3, 1, 0, 1, 5, 0, 0, 0, 1, 7, 2, 1, 0, 0, 1, 11, 0, 0, 0, 0, 0, 1, 15, 3, 0, 1, 0, 0, 0, 1, 22, 0, 2, 0, 0, 0, 0, 0, 1, 30, 5, 0, 0, 1, 0, 0, 0, 0, 1, 42, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 56, 7, 3, 2, 0, 1, 0, 0, 0, 0, 0, 1, 77, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset

0,4

Comments

Note that column k lists each partition number A000041 followed by k zeros. See also A168020 and A168021.

Let A(n,k) denote the number of partitions of n into parts divisible by k+1. Let p(n) denote the number of partitions of n. If k+1 is a divisor of n then A(n,k) = p(n/(k+1)) otherwise A(n,k) = 0. [Conjectured by Omar E. Pol, Nov 25 2009] - this is trivial, just divide each part size by k - Franklin T. Adams-Watters, May 14 2010.

Links

G. C. Greubel, Antidiagonals n = 0..50, flattened

Formula

From G. C. Greubel, Jan 13 2023: (Start)

A(n, k) = A000041(n/(k+1)) if (k+1)|n, otherwise 0 (array).

T(n, k) = A000041((n-k)/(k+1)) if (k+1)|(n-k), otherwise 0 (antidiagonals).

A(n, 0) = T(n, 0) = A000041(n).

T(2*n, n) = A(n, n) = A000007(n).

Sum_{k=0..n} T(n, k) = A083710(n+1). (End)

Example

The array, A(n, k), begins:
==================================================
... Column k: 0.. 1. 2. 3. 4. 5. 6. 7. 8. 9 10 11
. Row ...........................................
...n ............................................
==================================================
.. 0 ........ 1,  1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
.. 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;
   1, 1;
   2, 0, 1;
   3, 1, 0, 1;
   5, 0, 0, 0, 1;
   7, 2, 1, 0, 0, 1;
  11, 0, 0, 0, 0, 0, 1;
  15, 3, 0, 1, 0, 0, 0, 1;
  22, 0, 2, 0, 0, 0, 0, 0, 1;
  30, 5, 0, 0, 1, 0, 0, 0, 0, 1;
  42, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1;

Mathematica

T[n_, k_]:= If[IntegerQ[(n-k)/(k+1)], PartitionsP[(n-k)/(k+1)], 0];
Table[T[n, k], {n, 0, 15}, {k, 0, n}]//Flatten (* G. C. Greubel, Jan 13 2023 *)

Prog

(SageMath)
def A168019(n, k): return number_of_partitions((n-k)/(k+1)) if ((n-k)%(k+1))==0 else 0
flatten([[A168019(n, k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Jan 13 2023

Crossrefs

Cf. A000041, A035363, A083710, A135010.

Cf. A168016, A168120, A168021, A168121.

Sequence in context: A214576 A079217 A079221 * A026794 A137712 A194711

Adjacent sequences: A168016 A168017 A168018 * A168020 A168021 A168022

Keyword

easy,nonn,tabl

Author

Omar E. Pol, Nov 21 2009

Extensions

Edited by Charles R Greathouse IV, Mar 23 2010

Edited by Franklin T. Adams-Watters, May 14 2010

Status

approved