1,5

Also known as Stirling set numbers and written {n, k}.

S2(n,k) counts partitions of an n-set into k nonempty subsets.

Triangle S2(n,k), 1<=k<=n, read by rows, given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938.

Number of partitions of {1, ...,n+1} into k+1 subsets of nonconsecutive integers, including the partition 1|2|...|n+1 if n=k. E.g., S2(3,2)=3 since the number of partitions of {1,2,3,4} into three subsets of nonconsecutive integers is 3, i.e., 13|2|4, 14|2|3, 1|24|3. - Augustine O. Munagi, Mar 20 2005

Draw n cards (with replacement) from a deck of k cards. Let prob(n,k) be the probability that each card was drawn at least once. Then prob(n,k) = S2(n,k)*k!/k^n (see A090582). - Rainer Rosenthal, Oct 22 2005

Define f_1(x),f_2(x),... such that f_1(x)=e^x and for n=2,3,... f_{n+1}(x)=diff(x*f_n(x),x). Then f_n(x)=e^x*sum(stirling2(n,k)*x^(k-1),k=1..n). - Milan Janjic, May 30 2008

From Peter Bala, Oct 03 2008: (Start)

For tables of restricted Stirling numbers of the second kind see A143494 - A143496.

Stirling2(n,k) gives the number of 'patterns' of words of length n using k distinct symbols - see [Cooper & Kennedy] for an exact definition of the term 'pattern'. As an example, the words AADCBB and XXEGTT, both of length 6, have the same pattern of letters. The five patterns of words of length 3 are AAA, AAB, ABA, BAA and ABC giving row 3 of this table as (1,3,1).

Equivalently, Stirling2(n,k) gives the number of sequences of positive integers (N_1,...,N_n) of length n, with k distinct entries, such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1. For example, Stirling(4,2) = 7 since the sequences of length 4 having 2 distinct entries that satisfy the conditions are (1,1,1,2), (1,1,2,1), (1,2,1,1), (1,1,2,2), (1,2,2,2), (1,2,2,1) and (1,2,1,2). (End)

Number of combinations of subsets in the plane. - Mats Granvik, Jan 13 2009

S2(n+1,k+1) is the number of size k collections of pairwise disjoint, nonempty subsets of [n]. For example: S2(4,3)=6 because there are six such collections of subsets of [3] that have cardinality two: {(1)(23)},{(12)(3)}, {(13)(2)}, {(1)(2)}, {(1)(3)}, {(2)(3)}. - Geoffrey Critzer, Apr 06 2009

Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1, k+1) equals the number of ways to choose 0 or more balls of each color in such a way that exactly (n-k) colors are chosen at least once, and no two colors are chosen the same positive number of times. - Matthew Vandermast, Nov 22 2010

S2(n,k) is the number of monotonic-labeled forests on n vertices with exactly k rooted trees, each of height one or less. See link "Counting forests with Stirling and Bell numbers" below. - Dennis P. Walsh, Nov 16 2011

If D is the operator d/dx, and E the operator xd/dx, Stirling numbers are given by:  E^n = sum_{k=1}^n s2(n,k) * x^k*D^k. - Hyunwoo Jang, Dec 13 2011

The Stirling polynomials of the second kind (a.k.a. the Bell / Touchard polynomials) are the umbral compositional inverses of the falling factorials (a.k.a the Pochhammer symbol or Stirling polynomials of the first kind), i.e., binom(Bell(.,x),n) = x^n/n! (cf. Copeland's 2007 formulas), implying binom(xD,n) = binom(Bell(.,:xD:),n) = :xD:^n/n! where D = d/dx and :xD:^n = x^n*D^n. - Tom Copeland, Apr 17 2014

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S2(n, k) = k*S2(n-1, k)+S2(n-1, k-1), n>1. S2(1, k) = 0, k>1. S2(1, 1)=1.

E.g.f.: A(x, y)=exp(y*exp(x)-y). E.g.f. for m-th column: ((exp(x)-1)^m)/m!.

S2(n, k) = (1/k!) * Sum_{i=0..k} (-1)^(k-i)*C(k, i)*i^n.

Row sums: Bell number A000110(n) = sum(S2(n, k)) k=1..n, n>0.

A019538(n, k) = k! * S2(n, k).

A028248(n, k) = (k-1)! * S2(n, k).

For asymptotics see Hsu (1948), among other sources.

The k-th row (k>=1) contains a(n, k) for n=1 to k where a(n, k) = (1/(n-1)!) * Sum_{q=1..[2*n+1+(-1)^(n-1)]/4} [ C(n-1, 2*q-2)*(n-2*q+2)^(k-1) -C(n-1, 2*q-1)*(n-2*q+1)^(k-1) ]. E.g. Row 7 contains S2(7, 3)=301 { A001298, S2(n+4, n) } and will be computed as the following: S2(7, 3) = a(3, 7) = 1/(3-1)! * Sum_{q=1..2} [ C(3-1, 2*q-2)*(3-2*q+2)^(7-1) -C(3-1, 2*q-1)*(3-2*q+1)^(7-1) ] = Sum_{q=1..2} [ C(2, 2*q-2)*(5-2*q)^6 -C(2, 2*q-1)*(4-2*q)^6 ]/2! = [ C(2, 0)*3^6 -C(2, 1)*2^6 +C(2, 2)*1^6 -C(2, 3)*0^6 ]/2! = [ 729 -128 +1 -0 ]/2 = 301. - André F. Labossière, Jun 07 2004

sum(n>=0, T(n, k)*x^n ) = x^k/((1-x)(1-2x)(1-3x)...(1-kx)).

With P(n) = the number of integer partitions of n, p(i) = the number of parts of the i-th partition of n, d(i) = the number of different parts of the i-th partition of n, p(j, i) = the j-th part of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, sum_[p(i)=m]_{i=1}^{P(n)} = sum running from i=1 to i=p(n) but taking only partitions with p(i)=m parts into account, prod_{j=1}^{p(i)} = product running from j=1 to j=p(i), prod_{j=1}^{d(i)} = product running from j=1 to j=d(i) one has S2(n, m) = sum_[p(i)=m]_{i=1}^{P(n)} (n!/prod_{j=1}^{p(i)} p(i, j)!) (1/prod_{j=1}^{d(i)} m(i, j)!). For example, S2(6, 3) = 90 because n=6 has the following partitions with m=3 parts: (114), (123), (222). Their complexions are: (114): (6!/1!*1!*4!)*(1/2!*1!) = 15, (123): (6!/1!*2!*3!)*(1/1!*1!*1!) = 60, (222): (6!/2!*2!*2!)*(1/3!) = 15. The sum of the complexions is 15+60+15=90=S2(6, 3). - Thomas Wieder, Jun 02 2005

Sum(k*S2(n,k), k=1..n)=B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Nov 01 2006

Recurrence: S2(n+1,k) = sum_{i=0}^n C(n,i) S2(i, k-1). With the starting conditions S2(n,k) = 1 for n = 0 or k = 1 and S2(n,k) = 0 for k = 0 we have the closely related recurrence S2(n,k) = sum_{i=k}^n C(n-1,i-1) S2(i-1, k-1) where C(n,m) is the binomial coefficient. - Thomas Wieder, Jan 27 2007

Representation of Stirling numbers of the second kind S2(n,k), n=1,2..., k=1,2...n, as special values of hypergeometric function of type (n)F(n-1): stirling2(n,k)= (-1)^(k-1)*hypergeom([ -k+1,2,2...2],[1,1...1],1)/(k-1)!, i.e. having n parameters in the numerator: one equal to -k+1 and n-1 parameters all equal to 2 ; and having n-1 parameters in the denominator all equal to 1 and the value of the argument equal to 1. Example: stirling2(6,k)= seq(evalf((-1)^(k-1)*hypergeom([ -k+1,2,2,2,2,2],[1,1,1,1,1],1)/(k-1)!),k=1..6)=1,31,90,65,15,1. - Karol A. Penson, Mar 28 2007

From Tom Copeland, Oct 10 2007: (Start)(Initial index fix, Apr 17 2014)

Bell(n,x) = sum(j=1,...,n) S2(n,j) * x^j = sum(j=1,...,n) E(n,j) * Lag(n,-x, j-n) = sum(j=1,...,n) [ E(n,j)/n! ] * [ n!*Lag(n,-x, j-n) ] = sum(j=1,...,n) E(n,j) * C(Bell(.,x)+j,n) umbrally where Bell(n,x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; Lag(n,x,m), the associated Laguerre polynomials of order m; and C(x,y) = x!/[ y!*(x-y)! ].

By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*C(x,n), for x in the equation, the equation becomes x^n = sum(j=1,...,n) S2(n,j) * j! * C(x,j).

Note that E(n,j)/n! = E(n,j)/ {sum(k=1,..,n)E(n,k)}. Also [n!*Lag(n,-1, j-n)] is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x = 1; n!*Bell(n,1) = n!*sum(j=1,...,n) S2(n,j) = sum(j=1,...,n) {E(n,j) * [n!*Lag(n,-1, j-n)]}. (End)

n-th row = leftmost column of nonzero terms of A127701^(n-1). Also, (n+1)-th row of the triangle = A127701 * n-th row; deleting the zeros. Example: A127701 * [1, 3, 1, 0, 0, 0,...] = [1, 7, 6, 1, 0, 0, 0,...]. - Gary W. Adamson, Nov 21 2007

The row polynomials are given by D^n(exp(x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A147315 and A094198. See also A185422. - Peter Bala, Nov 25 2011

Let f(x) = exp(exp(x)). Then for n >= 1, 1/f(x)*(d/dx)^n(f(x)) = 1/f(x)*(d/dx)^(n-1)(exp(x)*f(x)) = sum {k = 1..n} Stirling2(n,k)*exp(k*x). Similar formulas hold for A039755, A105794, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012

T(n,k) = A048993(n,k), 1 <= k <= n. - Reinhard Zumkeller, Mar 26 2012

O.g.f. for the n-th diagonal is D^n(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012

n*i!*S2(n-1,i) = sum_{j=i+1..n} (-1)^(j-i+1)*j!/(j-i)*S2(n,j). - Leonid Bedratyuk, Aug 19 2012

G.f.: (1/Q(0)-1)/(x*y), where Q(k) = 1 -(y+k)*x - (k+1)*y*x^2/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013

From Tom Copeland, Apr 17 2014: (Start)

Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x).

With Bell(n,x)=B(n,x) defined above, D = d/dx, and :xD:^n = x^n*D^n, a Dobinski formula gives umbrally f(y)^B(.,x) = exp(-x)exp(f(y)*x). Then f(y)^B(.,:xD:)g(x)=[f(y)^(xD)]g(x)=exp[-(1-f(y)):xD:]g(x)=g[f(y)x].

In particular, for f(y) = (1+y),

A)(1+y)^B(.,x) = exp(-x)exp((1+y)*x) = exp(x*y) = exp[log(1+y)B(.,x)],

B)(I+dP)^B(.,x)= exp(x*dP)= P(x) = exp[x*(e^M-I)]= exp[M*B(.,x)] with dP = A132440, M = A238385-I = log(I+dP), and I = identity matrix, and

C)(1+dP)^(xD)= exp(dP:xD:)= P(x)= exp[(e^M-I):xD:] = exp[M*xD] with action exp(dP:xD:)g(x) = g[(I+dP)*x].

D) P(x)^m = P(m*x), which implies [sum(k=1,..,m, a_k)]^j=B(j,m*x) where the sum is umbrally evaluated only after exponentiation with (a_k)^q = B(.,x)^q = B(q,x). E.g., (a1+a2+a3)^2=a1^2+a2^2+a3^2+2(a1*a2+a1*a3+a2*a3) = 3*B(2,x)+6*B(1,x)^2 = 9x^2+3x = B(2,3x)

E) P(x)^2 = P(2x) = exp[M*B(.,2x)] = A038207(x), the face vectors of the n-Dim hypercubes. (End)

As a matrix equivalent of some inversions mentioned above, A008277*A008275 = I, the identity matrix, regarded as lower triangular matrices. - Tom Copeland, Apr 26 2014

O.g.f. for the n-th diagonal of the triangle (n = 0,1,2,...): sum {k >= 0} k^(k+n)*(x*exp(-x))^k/k!. Cf. the generating functions of the diagonals of A039755. Also cf. A112492. - Peter Bala, Jun 22 2014

Floor[1/(-1 + Sum_{n>=k} 1/S2(n,k))] = A034856(k-1), for k>=2. The fractional portion goes to zero at large k. - Richard R. Forberg, Jan 17 2015

The triangle S2(n, k) begins:

n\k 1    2      3       4       5       6       7       8      9    10   11 12 13 ...

1:  1

2:  1    1

3:  1    3      1

4:  1    7      6       1

5:  1   15     25      10       1

6:  1   31     90      65      15       1

7:  1   63    301     350     140      21       1

8:  1  127    966    1701    1050     266      28       1

9:  1  255   3025    7770    6951    2646     462      36      1

10: 1  511   9330   34105   42525   22827    5880     750     45     1

11: 1 1023  28501  145750  246730  179487   63987   11880   1155    55    1

12: 1 2047  86526  611501 1379400 1323652  627396  159027  22275  1705   66  1

13: 1 4095 261625 2532530 7508501 9321312 5715424 1899612 359502 39325 2431 78  1

Row n=14:  1 8191 788970 10391745 40075035 63436373 49329280 20912320 5135130 752752 66066 3367 91 1.

Row n=15: 1 16383 2375101 42355950 210766920 420693273 408741333 216627840 67128490 12662650 1479478 106470 4550 105 1.

... reformatted and extended. - Wolfdieter Lang, Aug 01 2014

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seq(seq(combinat[stirling2](n, k), k=1..n), n=1..10); # Zerinvary Lajos, Jun 02 2007

stirling_2 := (n, k) -> (1/k!) * add((-1)^(k-i)*binomial(k, i)*i^n, i=0..k);

Stirling2rec := proc(n::integer, k::integer) # Calculate the Stirling numbers of the second kind S2(n, k) by a recurrence. local i, Result; if k > n or n < 0 or k < 0 then return fail end if; if n = 0 or k = 1 then Result := 1; return Result end if; if k = 0 then Result := 0; return Result end if; Result := 0; for i from k to n do Result := Result + binomial(n - 1, i - 1) * Stirling2rec(i - 1, k - 1); end do; return Result; end proc; # Thomas Wieder, Jan 27 2007

Table[StirlingS2[n, k], {n, 11}, {k, n}] // Flatten (* Robert G. Wilson v, May 23 2006 *)

p[x_, n_] = Sum[m^n*x^m/m!, {m, 0, Infinity}]/(x*Exp[x]);

Table[FullSimplify[ExpandAll[p[x, n]]], {n, 1, 10}];

Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 1, 10}];

Flatten[%] (* Roger L. Bagula, Jan 11 2009 *)

(PARI) for(n=1, 22, for(k=1, n, print1(stirling(n, k, 2), ", ")); print()); \\ Joerg Arndt, Apr 21 2013

(PARI) Stirling2(n, k)=sum(i=0, k, (-1)^i*binomial(k, i)*i^n)*(-1)^k/k!  \\ M. F. Hasler, Mar 06 2012

(Haskell)

a008277 n k = a008277_tabl !! (n-1) !! (k-1)

a008277_row n = a008277_tabl !! (n-1)

a008277_tabl = map tail $ a048993_tabl  -- Reinhard Zumkeller, Mar 26 2012

(Maxima) create_list(stirling2(n+1, k+1), n, 0, 30, k, 0, n); /* Emanuele Munarini, Jun 01 2012 */

Cf. A008275 (Stirling numbers of first kind), A039810-A039813, A048993 (another version of this triangle), A048994, A028246.

Cf. A094262, A000217, A001296, A001297, A001298, A087127, A087107-A087111.

Cf. A127701.

Sequence in context: A250119 A154959 A080417 * A218577 A193387 A185982

Adjacent sequences:  A008274 A008275 A008276 * A008278 A008279 A008280

nonn,easy,tabl,nice,core,changed

N. J. A. Sloane

approved