A327884 Number T(n,k) of set partitions of [n] such that at least one of the block sizes is k or k=0; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

1, 1, 1, 2, 1, 1, 5, 4, 3, 1, 15, 11, 9, 4, 1, 52, 41, 35, 20, 5, 1, 203, 162, 150, 90, 30, 6, 1, 877, 715, 672, 455, 175, 42, 7, 1, 4140, 3425, 3269, 2352, 1015, 280, 56, 8, 1, 21147, 17722, 17271, 13132, 6237, 1890, 420, 72, 9, 1, 115975, 98253, 97155, 76540, 39480, 12978, 3150, 600, 90, 10, 1

Offset: 0

Author

Alois P. Heinz, Sep 28 2019

Keywords

• nonn
• tabl

Examples

			T(4,1) = 11: 123|4, 124|3, 12|3|4, 134|2, 13|2|4, 1|234, 1|23|4, 14|2|3, 1|24|3, 1|2|34, 1|2|3|4.
T(4,2) = 9: 12|34, 12|3|4, 13|24, 13|2|4, 14|23, 1|23|4, 14|2|3, 1|24|3, 1|2|34.
T(4,3) = 4: 123|4, 124|3, 134|2, 1|234.
T(4,4) = 1: 1234.
T(5,1) = 41: 1234|5, 1235|4, 123|4|5, 1245|3, 124|3|5, 12|34|5, 125|3|4, 12|35|4, 12|3|45, 12|3|4|5, 1345|2, 134|2|5, 13|24|5, 135|2|4, 13|25|4, 13|2|45, 13|2|4|5, 14|23|5, 1|2345, 1|234|5, 15|23|4, 1|235|4, 1|23|45, 1|23|4|5, 145|2|3, 14|25|3, 14|2|35, 14|2|3|5, 15|24|3, 1|245|3, 1|24|35, 1|24|3|5, 15|2|34, 1|25|34, 1|2|345, 1|2|34|5, 15|2|3|4, 1|25|3|4, 1|2|35|4, 1|2|3|45, 1|2|3|4|5.
Triangle T(n,k) begins:
      1;
      1,     1;
      2,     1,     1;
      5,     4,     3,     1;
     15,    11,     9,     4,    1;
     52,    41,    35,    20,    5,    1;
    203,   162,   150,    90,   30,    6,   1;
    877,   715,   672,   455,  175,   42,   7,  1;
   4140,  3425,  3269,  2352, 1015,  280,  56,  8, 1;
  21147, 17722, 17271, 13132, 6237, 1890, 420, 72, 9, 1;
  ...
		

Links

• Alois P. Heinz, Rows n = 0..140, flattened
• Wikipedia, Iverson bracket
• Wikipedia, Partition of a set

Crossrefs

Columns k=0-3 give: A000110, A000296(n+1), A327885, A328153.

T(2n,n) gives A276961.

Cf. A080510, A327869.

Programs

Maple

b:= proc(n, k) option remember; `if`(n=0, 1, add(
      `if`(j=k, 0, b(n-j, k)*binomial(n-1, j-1)), j=1..n))
    end:
T:= (n, k)-> b(n, 0)-`if`(k=0, 0, b(n, k)):
seq(seq(T(n, k), k=0..n), n=0..11);

Mathematica

b[n_, k_] := b[n, k] = If[n == 0, 1, Sum[If[j == k, 0, b[n - j,k] Binomial[ n - 1, j - 1]], {j, 1, n}]];
T[n_, k_] := b[n, 0] - If[k == 0, 0, b[n, k]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 30 2020, after Alois P. Heinz *)

Formula

E.g.f. of column k: exp(exp(x)-1) - [k>0] * exp(exp(x)-1-x^k/k!).

T(n,0) - T(n,1) = A000296(n).