%I
%S 1,1,1,1,10,1,35,1,91,1,210,280,1,456,2100,1,957,10395,1,1969,42735,
%T 15400,1,4004,158301,200200,1,8086,549549,1611610,1,16263,1827826,
%U 10335325,1401400,1,32631,5903898,57962905,28028000,1,65382,18682014
%N Triangle of Stirling numbers of order 3.
%C The number of partitions of the set N, N=n, into k blocks, all of cardinality greater than or equal to 3. This is the 3associated Stirling number of the second kind (Comtet) or the Stirling number of order 3 (Fekete).
%C This is entered as a triangular array. The entries S_3(n,k) are zero for 3k>n, so these values are omitted. Initial entry in sequence is S_3(3,1).
%C Rows are of lengths 1,1,1,2,2,2,3,3,3,...
%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), 771778.
%H G. Nemes, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Nemes/nemes2.html">On the Coefficients of the Asymptotic Expansion of n!</a>, J. Int. Seq. 13 (2010), 10.6.6.
%F S_r(n+1,k) = k S_r(n,k) + binomial(n,r1)*S_r(nr+1,k1); for this sequence, r=3.
%F G.f.: Sum_{n>=0, k>=0} S_r(n,k)*u^k*t^n/n! = exp(u(e^t  Sum_{i=0..r1} t^i/i!)).
%e There are 10 ways of partitioning a set N of cardinality 6 into 2 blocks each of cardinality at least 3, so S_3(6,2) = 10.
%t S3[3, 1] = S3[4, 1] = S3[5, 1] = 1; S3[n_, k_] /; 1 <= k <= Floor[n/3] := S3[n, k] = k*S3[n1, k] + Binomial[n1, 2]*S3[n3, k1]; S3[_, _] = 0; Flatten[ Table[ S3[n, k], {n, 3, 20}, {k, 1, Floor[n/3]}]] (* _JeanFrançois Alcover_, Feb 21 2012 *)
%Y Cf. A008299, A059023, A059024, A059025.
%K nonn,tabf,nice
%O 3,5
%A Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000
