OFFSET
3,5
COMMENTS
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 3-associated Stirling number of the second kind (Comtet) or the Stirling number of order 3 (Fekete).
This is entered as a triangular array. The entries S_3(n,k) are zero for 3k>n, so these values are omitted. The initial entry in the sequence is S_3(3,1).
Rows are of lengths 1,1,1,2,2,2,3,3,3,...
REFERENCES
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 222.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 76.
LINKS
Gilles Bonnet and Anna Gusakova, Concentration inequalities for Poisson U-statistics, arXiv:2404.16756 [math.PR], 2024. See p. 17.
Antal E. Fekete, Apropos two notes on notation, Amer. Math. Monthly, 101 (1994), 771-778.
Gergő Nemes, On the Coefficients of the Asymptotic Expansion of n!, J. Int. Seq. 13 (2010), 10.6.6.
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=3.
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!)).
T(n,k) = Sum_{j=0..min(n/2,k)} (-1)^j*B(n,j)*S_2(n-2j,k-j), where B are the Bessel numbers A100861 and S_2 are the 2-associated Stirling numbers of the second kind A008299. - Fabián Pereyra, Feb 20 2022
EXAMPLE
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.
From Wesley Ivan Hurt, Feb 24 2022: (Start)
Triangle starts:
1;
1;
1;
1, 10;
1, 35;
1, 91;
1, 210, 280;
1, 456, 2100;
1, 957, 10395;
1, 1969, 42735, 15400;
...
(End)
MAPLE
b:= proc(n) option remember; `if`(n=0, 1, add(
expand(x*b(n-j))*binomial(n-1, j-1), j=3..n))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n)):
seq(T(n), n=3..20); # Alois P. Heinz, Feb 21 2022
# alternative
A059022 := proc(n, k)
option remember;
if n<3 then
0;
elif n < 6 and k=1 then
1 ;
else
k*procname(n-1, k)+binomial(n-1, 2)*procname(n-3, k-1) ;
end if;
end proc: # R. J. Mathar, Apr 15 2022
MATHEMATICA
S3[3, 1] = S3[4, 1] = S3[5, 1] = 1; S3[n_, k_] /; 1 <= k <= Floor[n/3] := S3[n, k] = k*S3[n-1, k] + Binomial[n-1, 2]*S3[n-3, k-1]; S3[_, _] = 0; Flatten[ Table[ S3[n, k], {n, 3, 20}, {k, 1, Floor[n/3]}]] (* Jean-François Alcover, Feb 21 2012 *)
CROSSREFS
KEYWORD
nonn,tabf,nice
AUTHOR
Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000
STATUS
approved