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A008277 Triangle of Stirling numbers of the second kind, S2(n,k), n >= 1, 1 <= k <= n. 419

%I

%S 1,1,1,1,3,1,1,7,6,1,1,15,25,10,1,1,31,90,65,15,1,1,63,301,350,140,21,

%T 1,1,127,966,1701,1050,266,28,1,1,255,3025,7770,6951,2646,462,36,1,1,

%U 511,9330,34105,42525,22827,5880,750,45,1,1,1023,28501,145750,246730,179487,63987,11880,1155,55,1

%N Triangle of Stirling numbers of the second kind, S2(n,k), n >= 1, 1 <= k <= n.

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

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

%C 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.

%C Number of partitions of {1, ..., n+1} into k+1 nonempty 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

%C 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

%C Define f_1(x), f_2(x), ..., such that f_1(x)=e^x and for n = 2, 3, ..., f_{n+1}(x) = (d/dx)(x*f_n(x)). Then f_n(x) = e^x*Sum_{k=1..n} S2(n,k)*x^(k-1). - _Milan Janjic_, May 30 2008

%C From _Peter Bala_, Oct 03 2008: (Start)

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

%C S2(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).

%C Equivalently, S2(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 (restricted growth functions). 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).

%C (End)

%C Number of combinations of subsets in the plane. - _Mats Granvik_, Jan 13 2009

%C 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

%C 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

%C 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

%C 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

%C 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., binomial(Bell(.,x),n) = x^n/n! (cf. Copeland's 2007 formulas), implying binomial(xD,n) = binomial(Bell(.,:xD:),n) = :xD:^n/n! where D = d/dx and :xD:^n = x^n*D^n. - _Tom Copeland_, Apr 17 2014

%C S2(n,k) is the number of ways to nest n matryoshkas (Russian nesting dolls) so that exactly k matryoshkas are not contained in any other matryoshka. - _Carlo Sanna_, Oct 17 2015

%C The row polynomials R(n, x) = Sum_{k=1..n} S2(n, k)*x^k appear in the numerator of the e.g.f. of n-th powers, E(n, x) = Sum_{m>=0} m^n*x^m/m!, as E(n, x) = exp(x)*x*R(n, x), for n >= 1. - _Wolfdieter Lang_, Apr 02 2017

%C With offsets 0 for n and k this is the Sheffer product matrix A007318*A048993 denoted by (exp(t), (exp(t) - 1)) with e.g.f. exp(t)*exp(x*(exp(t) - 1)). - _Wolfdieter Lang_, Jun 20 2017

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%D J. Riordan, An Introduction to Combinatorial Analysis, p. 48.

%D J. Stirling, The Differential Method, London, 1749; see p. 7.

%H T. D. Noe, <a href="/A008277/b008277.txt">First 100 rows, flattened</a>

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%H H. S. Wilf, <a href="http://www.math.upenn.edu/~wilf/DownldGF.html">Generatingfunctionology</a>, 2nd edn., Academic Press, NY, 1994, pp. 17ff, 105ff.

%H M. C. Wolf, <a href="http://dx.doi.org/10.1215/S0012-7094-36-00253-3">Symmetric functions for non-commutative elements</a>, Duke Math. J., 2 (1936), 626-637.

%H <a href="/index/Cor#core">Index entries for "core" sequences</a>

%F 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.

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

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

%F Row sums: Bell number A000110(n) = Sum_{k=1..n} S2(n, k), n>0.

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

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

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

%F 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} (binomial(n-1, 2*q-2)*(n-2*q+2)^(k-1) - binomial(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} (binomial(3-1, 2*q-2)*(3-2*q+2)^(7-1) - binomial(3-1, 2*q-1)*(3-2*q+1)^(7-1)) = Sum_{q=1..2} (binomial(2, 2*q-2)*(5-2*q)^6 - binomial(2, 2*q-1)*(4-2*q)^6)/2! = (binomial(2, 0)*3^6 - binomial(2, 1)*2^6 + binomial(2, 2)*1^6 - binomial(2, 3)*0^6)/2! = (729 - 128 + 1 - 0)/2 = 301. - _André F. Labossière_, Jun 07 2004

%F Sum_{n>=0} S2(n, k)*x^n = x^k/((1-x)(1-2x)(1-3x)...(1-kx)).

%F Let P(n) = the number of integer partitions of n (A000041), p(i) = the number of parts of the i-th partition of n, d(i) = the number of distinct 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, and Sum_{i=1..P(n), p(i)=m} = sum running from i=1 to i=P(n) but taking only partitions with p(i)=m parts into account. Then S2(n, m) = Sum_{i=1..P(n), p(i)=m} n!/(Product_{j=1..p(i)} p(i, j)!) * 1/(Product_{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

%F Sum_{k=1..n} k*S2(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - _Emeric Deutsch_, Nov 01 2006

%F Recurrence: S2(n+1,k) = Sum_{i=0..n} binomial(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} binomial(n-1,i-1)*S2(i-1,k-1). - _Thomas Wieder_, Jan 27 2007

%F 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): S2(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: S2(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

%F From _Tom Copeland_, Oct 10 2007: (Start)(Initial index fix, Apr 17 2014)

%F 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) * binomial(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.

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

%F 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).

%F (End)

%F 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

%F The row polynomials are given by D^n(e^(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

%F Let f(x) = e^(e^x). Then for n >= 1, 1/f(x)*(d/dx)^n(f(x)) = 1/f(x)*(d/dx)^(n-1)(e^x*f(x)) = Sum_{k=1..n} S2(n,k)*e^(k*x). Similar formulas hold for A039755, A105794, A111577, A143494 and A154537. - _Peter Bala_, Mar 01 2012

%F S2(n,k) = A048993(n,k), 1 <= k <= n. - _Reinhard Zumkeller_, Mar 26 2012

%F 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

%F 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

%F 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

%F From _Tom Copeland_, Apr 17 2014: (Start)

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

%F 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) = e^(-x)*e^(f(y)*x). Then f(y)^B(.,:xD:)g(x) = [f(y)^(xD)]g(x) = e^[-(1-f(y)):xD:]g(x) = g[f(y)x].

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

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

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

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

%F 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).

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

%F (End)

%F 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

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

%F 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

%F From _Daniel Forgues_, Jan 16 2016: (Start)

%F Let x_(n), called a factorial term (Boole, 1970) or a factorial polynomial (Elaydi, 2005: p. 60), denote the falling factorial Product_{k=0..n-1} (x-k). Then, for n >= 1, x_(n) = Sum_{k=1..n} A008275(n,k) * x^k, x^n = Sum_{k=1..n} T(n,k) * x_(k), where A008275(n,k) are Stirling numbers of the first kind.

%F For n >= 1, the row sums yield the exponential numbers (or Bell numbers): Sum_{k=1..n} T(n,k) = A000110(n), and Sum_{k=1..n} (-1)^(n+k) * T(n,k) = (-1)^n * Sum_{k=1..n} (-1)^k * T(n,k) = (-1)^n * A000587(n), where A000587 are the complementary Bell numbers. (End)

%e The triangle S2(n, k) begins:

%e \ k 1 2 3 4 5 6 7 8 9

%e n \ 10 11 12 13 14 15 ...

%e ----------------------------------------------------------------------------------

%e 1 | 1

%e 2 | 1 1

%e 3 | 1 3 1

%e 4 | 1 7 6 1

%e 5 | 1 15 25 10 1

%e 6 | 1 31 90 65 15 1

%e 7 | 1 63 301 350 140 21 1

%e 8 | 1 127 966 1701 1050 266 28 1

%e 9 | 1 255 3025 7770 6951 2646 462 36 1

%e 10 | 1 511 9330 34105 42525 22827 5880 750 45

%e 1

%e 11 | 1 1023 28501 145750 246730 179487 63987 11880 1155

%e 55 1

%e 12 | 1 2047 86526 611501 1379400 1323652 627396 159027 22275

%e 1705 66 1

%e 13 | 1 4095 261625 2532530 7508501 9321312 5715424 1899612 359502

%e 39325 2431 78 1

%e 14 | 1 8191 788970 10391745 40075035 63436373 49329280 20912320 5135130

%e 752752 66066 3367 91 1

%e 15 | 1 16383 2375101 42355950 210766920 420693273 408741333 216627840 67128490

%e 12662650 1479478 106470 4550 105 1

%e ...

%e ----------------------------------------------------------------------------------

%e x^4 = 1 x_(1) + 7 x_(2) + 6 x_(3) + 1 x_(4), where x_(k) = P(x,k) = k!*C(x,k). - _Daniel Forgues_, Jan 16 2016

%p seq(seq(combinat[stirling2](n, k), k=1..n), n=1..10); # _Zerinvary Lajos_, Jun 02 2007

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

%t Table[StirlingS2[n, k], {n, 11}, {k, n}] // Flatten (* _Robert G. Wilson v_, May 23 2006 *)

%t BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];

%t rows = 12;

%t B = BellMatrix[1&, rows];

%t Table[B[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* _Jean-François Alcover_, Jun 28 2018, after _Peter Luschny_ *)

%o (PARI) for(n=1,22,for(k=1,n,print1(stirling(n,k,2),", "));print()); \\ _Joerg Arndt_, Apr 21 2013

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

%o (Haskell)

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

%o a008277_row n = a008277_tabl !! (n-1)

%o a008277_tabl = map tail $ a048993_tabl -- _Reinhard Zumkeller_, Mar 26 2012

%o (Maxima) create_list(stirling2(n+1,k+1),n,0,30,k,0,n); /* _Emanuele Munarini_, Jun 01 2012 */

%o (Sage) stirling_number2(n,k) # _Danny Rorabaugh_, Oct 11 2015

%o (J) n ((] (1 % !)) * +/@((^~ * (] (_1 ^ |.)) * (! {:)@]) i.@>:)) k NB. _Stephen Makdisi_, Apr 06 2016

%Y Cf. A008275 (Stirling numbers of first kind), A048993 (another version of this triangle).

%Y Cf. A000217, A001296, A001297, A001298, A007318, A028246, A039810-A039813, A048994, A087107-A087111, A087127, A094262, A127701.

%K nonn,easy,tabl,nice,core,changed

%O 1,5

%A _N. J. A. Sloane_

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Last modified September 18 21:35 EDT 2018. Contains 315153 sequences. (Running on oeis4.)