A059024 Triangle of Stirling numbers of order 5.

1, 1, 1, 1, 1, 1, 126, 1, 462, 1, 1254, 1, 3003, 1, 6721, 1, 14443, 126126, 1, 30251, 1009008, 1, 62322, 5309304, 1, 127024, 23075052, 1, 257108, 89791416, 1, 518092, 325355316, 488864376, 1, 1041029, 1122632043, 6844101264, 1, 2088043

Offset: 5

Author

Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000

Keywords

• nonn
• tabf
• nice

Comments

The number of partitions of the set N, |N|=n, into k blocks, all of cardinality greater than or equal to 5. This is the 5-associated Stirling number of the second kind.

This is entered as a triangular array. The entries S_5(n,k) are zero for 5k>n, so these values are omitted. Initial entry in sequence is S_5(5,1).

Rows are of lengths 1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,...

Examples

			There are 126 ways of partitioning a set N of cardinality 10 into 2 blocks each of cardinality at least 5, so S_5(10,2) = 126.
Triangle begins:
1;
1;
1;
1;
1;
1,    126;
1,    462;
1,   1254;
1,   3003;
1,   6721;
1,  14443,    126126;
1,  30251,   1009008;
1,  62322,   5309304;
1, 127024,  23075052;
1, 257108,  89791416;
1, 518092, 325355316, 488864376;
...
		

References

• L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 222.
• J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 76.

Links

• Alois P. Heinz, Rows n = 5..320, flattened
• A. E. Fekete, Apropos two notes on notation, Amer. Math. Monthly, 101 (1994), 771-778.

Crossrefs

Row sums give A057814.

Cf. A008299, A059022, A059023, A059025.

Programs

Maple

T:= proc(n,k) option remember; `if`(k<1 or k>n/5, 0,
      `if`(k=1, 1, k*T(n-1, k)+binomial(n-1, 4)*T(n-5, k-1)))
    end:
seq(seq(T(n, k), k=1..n/5), n=5..25);  # Alois P. Heinz, Aug 18 2017

Mathematica

S5[n_ /; 5 <= n <= 9, 1] = 1; S5[n_, k_] /; 1 <= k <= Floor[n/5] := S5[n, k] = k*S5[n-1, k] + Binomial[n-1, 4]*S5[n-5, k-1]; S5[, ] = 0; Flatten[ Table[ S5[n, k], {n, 5, 25}, {k, 1, Floor[n/5]}]] (* Jean-François Alcover, Feb 21 2012 *)

Formula

S_r(n+1, k) = k*S_r(n, k) + binomial(n, r-1)*S_r(n-r+1, k-1); for this sequence, r=5.

G.f.: Sum_{n>=0, k>=0} S_r(n,k)*u^k*t^n/n! = exp(u(e^t-sum(t^i/i!, i=0..r-1))).

T(n,k) = Sum_{j=0..min(n/4,k)} (-1)^j*n!/(24^j*j!*(n-4j)!)*S_4(n-4j,k-j), where S_4 are the 4-associated Stirling numbers of the second kind A059023. - Fabián Pereyra, Feb 21 2022