A059024 - OEIS

%I #27 Apr 08 2022 21:02:35

%S 1,1,1,1,1,1,126,1,462,1,1254,1,3003,1,6721,1,14443,126126,1,30251,

%T 1009008,1,62322,5309304,1,127024,23075052,1,257108,89791416,1,518092,

%U 325355316,488864376,1,1041029,1122632043,6844101264,1,2088043

%N Triangle of Stirling numbers of order 5.

%C 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.

%C 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).

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

%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 222.

%D J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 76.

%H Alois P. Heinz, <a href="/A059024/b059024.txt">Rows n = 5..320, flattened</a>

%H A. E. Fekete, <a href="http://www.jstor.org/stable/2974533">Apropos two notes on notation</a>, Amer. Math. Monthly, 101 (1994), 771-778.

%F 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.

%F 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))).

%F 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

%e 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.

%e Triangle begins:

%e 1;

%e 1;

%e 1;

%e 1;

%e 1;

%e 1,    126;

%e 1,    462;

%e 1,   1254;

%e 1,   3003;

%e 1,   6721;

%e 1,  14443,    126126;

%e 1,  30251,   1009008;

%e 1,  62322,   5309304;

%e 1, 127024,  23075052;

%e 1, 257108,  89791416;

%e 1, 518092, 325355316, 488864376;

%e ...

%p T:= proc(n,k) option remember; `if`(k<1 or k>n/5, 0,

%p       `if`(k=1, 1, k*T(n-1, k)+binomial(n-1, 4)*T(n-5, k-1)))

%p     end:

%p seq(seq(T(n, k), k=1..n/5), n=5..25);  # _Alois P. Heinz_, Aug 18 2017

%t 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 *)

%Y Row sums give A057814.

%Y Cf. A008299, A059022, A059023, A059025.

%K nonn,tabf,nice

%O 5,7

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