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%I #10 Jan 04 2016 10:44:05
%S 1,1,1,1,1,1,1,462,1,1716,1,4719,1,11440,1,25883,1,56134,1,118456,
%T 2858856,1,245480,23279256,1,502588,124710300,1,1020680,551496660,1,
%U 2061709,2181183147,1,4149752,8021782197,1,8333153,28051272535
%N Triangle of Stirling numbers of order 6.
%C The number of partitions of the set N, |N|=n, into k blocks, all of cardinality greater than or equal to 6. This is the 6-associated Stirling number of the second kind.
%C This is entered as a triangular array. The entries S_6(n,k) are zero for 6k>n, so these values are omitted. Initial entry in sequence is S_6(6,1).
%C Rows are of lengths 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 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 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=6.
%F G.f.: Sum_{n>=0, k>=0} S_r(n,k)*u^k*t^n/n! = exp(u(e^t - Sum_{i=0..r-1} t^i/i!)).
%e There are 462 ways of partitioning a set N of cardinality 12 into 2 blocks each of cardinality at least 6, so S_6(12,2)=462.
%t S6[n_ /; 6 <= n <= 11, 1] = 1; S6[n_, k_] /; 1 <= k <= Floor[n/6] := S6[n, k] = k*S6[n-1, k] + Binomial[n-1, 5]*S6[n-6, k-1]; S6[_, _] = 0; Flatten[ Table[ S6[n, k], {n, 6, 24}, {k, 1, Floor[n/6]}]] (* _Jean-François Alcover_, Feb 21 2012 *)
%Y Cf. A008299, A059022, A059023, A059024.
%K nonn,tabf,nice
%O 6,8
%A Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000
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