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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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, 1, 1, 127, 966, 1701, 1050, 266, 28, 1, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1, 1, 1023, 28501, 145750, 246730, 179487, 63987, 11880, 1155, 55, 1
Offset: 1

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Keywords

Comments

Also known as Stirling set numbers and written {n, k}.
S2(n,k) counts partitions of an n-set into k nonempty subsets.
From Manfred Boergens, Apr 07 2025: (Start)
With regard to the preceding comment:
For disjoint collections of subsets see A256894.
For arbitrary collections of subsets see A163353.
For arbitrary collections of nonempty subsets see A055154. (End)
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 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
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) = (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
From Peter Bala, Oct 03 2008: (Start)
For tables of restricted Stirling numbers of the second kind see A143494 - A143496.
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).
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).
(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., 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
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
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
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
Number of words on k+1 unlabeled letters of length n+1 with no repeated letters. - Thomas Anton, Mar 14 2019
Also coefficients of moments of Poisson distribution about the origin expressed as polynomials in lambda. [Haight] (see also A331155). - N. J. A. Sloane, Jan 14 2020
k!*S2(n,k) is the number of surjections from an n-element set to a k-element set. - Jianing Song, Jun 01 2022

Examples

			The triangle S2(n, k) begins:
\ k    1       2       3        4         5         6         7         8        9
n \   10      11      12       13        14        15       ...
----------------------------------------------------------------------------------
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
14 |   1    8191  788970 10391745  40075035  63436373  49329280  20912320  5135130
  752752   66066    3367       91         1
15 |   1   16383 2375101 42355950 210766920 420693273 408741333 216627840 67128490
12662650 1479478  106470     4550       105         1
...
----------------------------------------------------------------------------------
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
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 103ff.
  • B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • G. Boole, Finite Differences, 5th ed. New York, NY: Chelsea, 1970.
  • C. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, 2002, Theorem 8.11, pp. 298-299.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 310.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Springer, p. 92.
  • F. N. David, M. G. Kendall, and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • S.N. Elaydi, An Introduction to Difference Equations, 3rd ed. Springer, 2005.
  • H. H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.7.
  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 244.
  • Frank Avery Haight, Handbook of the Poisson distribution, John Wiley, 1967. See pages 6,7.
  • A. D. Korshunov, Asymptotic behavior of Stirling numbers of the second kind. (Russian) Metody Diskret. Analiz. No. 39 (1983), 24-41.
  • E. Kuz'min and A. I. Shirshov: On the number e, pp. 111-119, eq.(6), in: Kvant Selecta: Algebra and Analysis, I, ed. S. Tabachnikov, Am.Math.Soc., 1999, p. 116, eq. (11).
  • J. Riordan, An Introduction to Combinatorial Analysis, p. 48.
  • J. Stirling, The Differential Method, London, 1749; see p. 7.

Crossrefs

Cf. A008275 (Stirling numbers of first kind), A048993 (another version of this triangle).
See also A331155.
Cf. A000110 (row sums), A102661 (partial row sums).

Programs

  • 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
    
  • J
    n ((] (1 % !)) * +/@((^~ * (] (1 ^ |.)) * (! {:)@]) i.@>:)) k NB. _Stephen Makdisi, Apr 06 2016
    
  • Magma
    [[StirlingSecond(n,k): k in [1..n]]: n in [1..12]]; // G. C. Greubel, May 22 2019
  • Maple
    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);
  • Mathematica
    Table[StirlingS2[n, k], {n, 11}, {k, n}] // Flatten (* Robert G. Wilson v, May 23 2006 *)
    BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
    rows = 12;
    B = BellMatrix[1&, rows];
    Table[B[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 28 2018, after Peter Luschny *)
    a[n_, n_] := 1; a[n_, 1] := 1;
    a[n_, k_] := a[n, k] = a[n-1, k-1] + k a[n-1, k]; Flatten@
    Table[a[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Jun 12 2024 *)
    With[{m = 11},
     Flatten@MapIndexed[Take[#, #2[[1]]] &,
       Transpose@
        Table[Range[1, m]! Coefficient[(E^x-1)^k/k! + O[x]^(m+1), x,
    Range[1, m]], {k, 1, m}]]] (* Oliver Seipel, Jun 12 2024 *)
  • Maxima
    create_list(stirling2(n+1,k+1),n,0,30,k,0,n); /* Emanuele Munarini, Jun 01 2012 */
    
  • 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
    
  • Sage
    stirling_number2 # Danny Rorabaugh, Oct 11 2015
    

Formula

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) = e^(y*e^x-y). E.g.f. for m-th column: (e^x-1)^m/m!.
S2(n, k) = (1/k!) * Sum_{i=0..k} (-1)^(k-i)*binomial(k, i)*i^n.
Row sums: Bell number A000110(n) = Sum_{k=1..n} S2(n, k), n>0.
S(n, k) = Sum (i_1*i_2*...*i_(n-k)) summed over all (n-k)-combinations {i_1, i_2, ..., i_k} with repetitions of the numbers {1, 2, ..., k}. Also S(n, k) = Sum (1^(r_1)*2^(r_2)*...* k^(r_k)) summed over integers r_j >= 0, for j=1..k, with Sum{j=1..k} r_j = n-k. [Charalambides]. - Wolfdieter Lang, Aug 15 2019.
A019538(n, k) = k! * S2(n, k).
A028248(n, k) = (k-1)! * S2(n, k).
For asymptotics see Hsu (1948), among other sources.
Sum_{n>=0} S2(n, k)*x^n = x^k/((1-x)(1-2x)(1-3x)...(1-kx)).
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
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
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
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
From Tom Copeland, Oct 10 2007: (Start)
Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..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; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..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=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)).
(Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (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(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
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
S2(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) = 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].
In particular, for f(y) = (1+y),
A) (1+y)^B(.,x) = e^(-x)*e^((1+y)*x) = e^(x*y) = e^[log(1+y)B(.,x)],
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
C) (1+dP)^(xD) = e^(dP:xD:) = P(:xD:) = e^[(e^M-I):xD:] = e^[M*xD] with action e^(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) = e^[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*e^(-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
From Daniel Forgues, Jan 16 2016: (Start)
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.
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)
Sum_{k=1..n} k*S2(n,k) = A138378(n). - Alois P. Heinz, Jan 07 2022
O.g.f. for the m-th column: x^m/(Product_{j=1..m} 1-j*x). - Daniel Checa, Aug 25 2022
S2(n,k) ~ (k^n)/k!, for fixed k as n->oo. - Daniel Checa, Nov 08 2022
S2(2n+k, n) ~ (2^(2n+k-1/2) * n^(n+k-1/2)) / (sqrt(Pi*(1-c)) * exp(n) * c^n * (2-c)^(n+k)), where c = -LambertW(-2 * exp(-2)). - Miko Labalan, Dec 21 2024
From Mikhail Kurkov, Mar 05 2025: (Start)
For a general proof of the formulas below via generating functions, see Mathematics Stack Exchange link.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(-k,j)*T(n,k+j-1) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link).
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} binomial(n,j)*T(n-j+1,k)*(-1)^j for 1 <= k < n with T(n,n) = 1. (End)

A000296 Set partitions without singletons: number of partitions of an n-set into blocks of size > 1. Also number of cyclically spaced (or feasible) partitions.

Original entry on oeis.org

1, 0, 1, 1, 4, 11, 41, 162, 715, 3425, 17722, 98253, 580317, 3633280, 24011157, 166888165, 1216070380, 9264071767, 73600798037, 608476008122, 5224266196935, 46499892038437, 428369924118314, 4078345814329009, 40073660040755337, 405885209254049952, 4232705122975949401
Offset: 0

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Comments

a(n+2) = p(n+1) where p(x) is the unique degree-n polynomial such that p(k) = A000110(k) for k = 0, 1, ..., n. - Michael Somos, Oct 07 2003
Number of complete rhyming schemes.
Whereas the Bell number B(n) (A000110(n)) is the number of terms in the polynomial that expresses the n-th moment of a probability distribution as a function of the first n cumulants, these numbers give the number of terms in the corresponding expansion of the central moment as a function of the first n cumulants. - Michael Hardy (hardy(AT)math.umn.edu), Jan 26 2005
a(n) is the number of permutations on [n] for which the left-to-right maxima coincide with the descents (entries followed by a smaller number). For example, a(4) counts 2143, 3142, 3241, 4123. - David Callan, Jul 20 2005
From Gus Wiseman, Feb 10 2019: (Start)
Also the number of stable partitions of an n-cycle, where a stable partition of a graph is a set partition of the vertex set such that no edge has both ends in the same block. A bijective proof is given in David Callan's article. For example, the a(5) = 11 stable partitions are:
{{1},{2},{3},{4},{5}}
{{1},{2},{3,5},{4}}
{{1},{2,4},{3},{5}}
{{1},{2,5},{3},{4}}
{{1,3},{2},{4},{5}}
{{1,4},{2},{3},{5}}
{{1},{2,4},{3,5}}
{{1,3},{2,4},{5}}
{{1,3},{2,5},{4}}
{{1,4},{2},{3,5}}
{{1,4},{2,5},{3}}
(End)
Also number of partitions of {1, 2, ..., n-1} with singletons. E.g., a(4) = 4: {1|2|3, 12|3, 13|2, 1|23}. Also number of cyclical adjacencies partitions of {1, 2, ..., n-1}. E.g., a(4) = 4: {12|3, 13|2, 1|23, 123}. The two partitions can be mapped by a Kreweras bijection. - Yuchun Ji, Feb 22 2021
Also the k-th central moment of a Poisson random variable with mean 1. a(n) = E[(X-1)^n, X~Poisson(1)]. - Thomas Dybdahl Ahle, Dec 14 2022

Examples

			a(4) = card({{{1, 2}, {3, 4}}, {{1, 4}, {2, 3}}, {{1, 3}, {2, 4}}, {{1, 2, 3, 4}}}) = 4.
		

References

  • Martin Gardner in Sci. Amer. May 1977.
  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.2.1.5 (p. 436).
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, NY, 2 vols., 1972, Vol. 1, p. 228.
  • J. Riordan, A budget of rhyme scheme counts, pp. 455-465 of Second International Conference on Combinatorial Mathematics, New York, April 4-7, 1978. Edited by Allan Gewirtz and Louis V. Quintas. Annals New York Academy of Sciences, 319, 1979.
  • J. Shallit, A Triangle for the Bell numbers, in V. E. Hoggatt, Jr. and M. Bicknell-Johnson, A Collection of Manuscripts Related to the Fibonacci Sequence, 1980, pp. 69-71.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

A diagonal of triangle in A106436.
Row sums of the triangle of associated Stirling numbers of second kind A008299. - Philippe Deléham, Feb 10 2005
Row sums of the triangle of basic multinomial coefficients A178866. - Johannes W. Meijer, Jun 21 2010
Row sums of A105794. - Peter Bala, Jan 14 2015
Row sums of A261139, main diagonal of A261137.
Column k=0 of A216963.
Column k=0 of A124323.

Programs

  • Magma
    [1,0] cat [ n le 1 select 1 else Bell(n)-Self(n-1) : n in [1..40]]; // Vincenzo Librandi, Jun 22 2015
    
  • Maple
    spec := [ B, {B=Set(Set(Z,card>1))}, labeled ]; [seq(combstruct[count](spec, size=n), n=0..30)];
    with(combinat): A000296 :=n->(-1)^n + add((-1)^(j-1)*bell(n-j),j=1..n): seq(A000295(n),n=0..30); # Emeric Deutsch, Oct 29 2006
    f:=exp(exp(x)-1-x): fser:=series(f, x=0, 31): 1, seq(n!*coeff(fser, x^n), n=1..23); # Zerinvary Lajos, Nov 22 2006
    G:={P=Set(Set(Atom,card>=2))}: combstruct[gfsolve](G,unlabeled,x): seq(combstruct[count]([P,G,labeled], size=i), i=0..23); # Zerinvary Lajos, Dec 16 2007
    # [a(0),a(1),..,a(n)]
    A000296_list := proc(n)
    local A, R, i, k;
    if n = 0 then return 1 fi;
    A := array(0..n-1);
    A[0] := 1; R := 1;
    for i from 0 to n-2 do
       A[i+1] := A[0] - A[i];
       A[i] := A[0];
       for k from i by -1 to 1 do
          A[k-1] := A[k-1] + A[k] od;
       R := R,A[i+1];
    od;
    R,A[0]-A[i] end:
    A000296_list(100);  # Peter Luschny, Apr 09 2011
  • Mathematica
    nn = 25; Range[0, nn]! CoefficientList[Series[Exp[Exp[x] - 1 - x], {x, 0, nn}], x]
    (* Second program: *)
    a[n_] := a[n] = If[n==0, 1, Sum[Binomial[n-1, i]*a[n-i-1], {i, 1, n-1}]]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Feb 06 2016, after Vladimir Kruchinin *)
    spsu[,{}]:={{}};spsu[foo,set:{i_,_}]:= Join@@Function[s,Prepend[#,s]&/@spsu[ Select[foo,Complement[#, Complement[set,s]]=={}&], Complement[set,s]]]/@Cases[foo,{i,_}];
    Table[Length[spsu[Select[Subsets[Range[n]],Select[Partition[Range[n],2,1,1], Function[ed,Complement[ed,#]=={}]]=={}&],Range[n]]],{n,8}] (* Gus Wiseman, Feb 10 2019 *)
    s = 1; Join[{1}, Table[s = BellB[n] - s, {n, 0, 25}]] (* Vaclav Kotesovec, Jun 20 2022 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(binomial(n-1,i)*a(n-i-1),i,1,n-1); /* Vladimir Kruchinin, Feb 22 2015 */
    
  • PARI
    a(n) = if(n<2, n==0, subst( polinterpolate( Vec( serlaplace( exp( exp( x+O(x^n)/x )-1 ) ) ) ), x, n) )
    
  • Python
    from itertools import accumulate, islice
    def A000296_gen():
        yield from (1,0)
        blist, a, b = (1,), 0, 1
        while True:
            blist = list(accumulate(blist, initial = (b:=blist[-1])))
            yield (a := b-a)
    A000296_list = list(islice(A000296_gen(),20)) # Chai Wah Wu, Jun 22 2022

Formula

E.g.f.: exp(exp(x) - 1 - x).
B(n) = a(n) + a(n+1), where B = A000110 = Bell numbers [Becker].
Inverse binomial transform of Bell numbers (A000110).
a(n)= Sum_{k>=-1} (k^n/(k+1)!)/exp(1). - Vladeta Jovovic and Karol A. Penson, Feb 02 2003
a(n) = Sum_{k=0..n} ((-1)^(n-k))*binomial(n, k)*Bell(k) = (-1)^n + Bell(n) - A087650(n), with Bell(n) = A000110(n). - Wolfdieter Lang, Dec 01 2003
O.g.f.: A(x) = 1/(1-0*x-1*x^2/(1-1*x-2*x^2/(1-2*x-3*x^2/(1-... -(n-1)*x-n*x^2/(1- ...))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
a(n) = Sum_{k = 0..n} {(-1)^(n-k) * Sum_{j = 0..k}((-1)^j * binomial(k,j) * (1-j)^n)/ k!} = sum over row n of A105794. - Tom Copeland, Jun 05 2006
a(n) = (-1)^n + Sum_{j=1..n} (-1)^(j-1)*B(n-j), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Oct 29 2006
Let A be the upper Hessenberg matrix of order n defined by: A[i, i-1] = -1, A[i,j] = binomial(j-1, i-1), (i <= j), and A[i, j] = 0 otherwise. Then, for n >= 2, a(n) = (-1)^(n)charpoly(A,1). - Milan Janjic, Jul 08 2010
From Sergei N. Gladkovskii, Sep 20 2012, Oct 11 2012, Dec 19 2012, Jan 15 2013, May 13 2013, Jul 20 2013, Oct 19 2013, Jan 25 2014: (Start)
Continued fractions:
G.f.: (2/E(0) - 1)/x where E(k) = 1 + 1/(1 + 2*x/(1 - 2*(k+1)*x/E(k+1))).
G.f.: 1/U(0) where U(k) = 1 - x*k - x^2*(k+1)/U(k+1).
G.f.: G(0)/(1+2*x) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k-x-1) - x*(2*k+1)*(2*k+3)*(2*x*k-x-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k-1)/G(k+1))).
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1+x-k*x)/(1-x/(x-1/G(k+1))).
G.f.: 1 + x^2/(1+x)/Q(0) where Q(k) = 1-x-x/(1-x*(2*k+1)/(1-x-x/(1-x*(2*k+2)/Q(k+1)))).
G.f.: 1/(x*Q(0)) where Q(k) = 1 + 1/(x + x^2/(1 - x - (k+1)/Q(k+1))).
G.f.: -(1+(2*x+1)/G(0))/x where G(k) = x*k - x - 1 - (k+1)*x^2/G(k+1).
G.f.: T(0) where T(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1-x*k)*(1-x-x*k)/T(k+1)).
G.f.: (1 + x * Sum_{k>=0} (x^k / Product_{p=0..k}(1 - p*x))) / (1 + x). (End)
a(n) = Sum_{i=1..n-1} binomial(n-1,i)*a(n-i-1), a(0)=1. - Vladimir Kruchinin, Feb 22 2015
G.f. A(x) satisfies: A(x) = (1/(1 + x)) * (1 + x * A(x/(1 - x)) / (1 - x)). - Ilya Gutkovskiy, May 21 2021
a(n) ~ exp(n/LambertW(n) - n - 1) * n^(n-1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n-1)). - Vaclav Kotesovec, Jun 28 2022

Extensions

More terms, new description from Christian G. Bower, Nov 15 1999
a(23) corrected by Sean A. Irvine, Jun 22 2015

A039755 Triangle of B-analogs of Stirling numbers of the second kind.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 13, 9, 1, 1, 40, 58, 16, 1, 1, 121, 330, 170, 25, 1, 1, 364, 1771, 1520, 395, 36, 1, 1, 1093, 9219, 12411, 5075, 791, 49, 1, 1, 3280, 47188, 96096, 58086, 13776, 1428, 64, 1, 1, 9841, 239220, 719860, 618870, 209622, 32340, 2388, 81, 1, 1
Offset: 0

Views

Author

Ruedi Suter (suter(AT)math.ethz.ch)

Keywords

Comments

Let M be an infinite lower triangular bidiagonal matrix with (1,3,5,7,...) in the main diagonal and (1,1,1,...) in the subdiagonal. n-th row = M^n * [1,0,0,0,...]. - Gary W. Adamson, Apr 13 2009
From Peter Bala, Aug 08 2011: (Start)
A type B_n set partition is a partition P of the set {1, 2, ..., n, -1, -2, ..., -n} such that for any block B of P, -B is also a block of P, and there is at most one block, called a zero-block, satisfying B = -B. We call (B, -B) a block pair of P if B is not a zero-block. Then T(n,k) is the number of type B_n set partitions with k block pairs. See [Wang].
For example, T(2,1) = 4 since the B_2 set partitions with 1 block pair are {1,2}{-1,-2}, {1,-2}{-1,2}, {1,-1}{2}{-2} and {2,-2}{1}{-1} (the last two partitions contain a zero block).
(End)
Exponential Riordan array [exp(x), (1/2)*(exp(2*x) - 1)]. Triangle of connection constants for expressing the monomial polynomials x^n as a linear combination of the basis polynomials (x-1)*(x-3)*...*(x-(2*k-1)) of A039757. An example is given below. Inverse array is A039757. Equals matrix product A008277 * A122848. - Peter Bala, Jun 23 2014
T(n, k) also gives the (dimensionless) volume of the multichoose(k+1, n-k) = binomial(n, k) polytopes of dimension n-k with side lengths from the set {1, 3, ..., 1+2*k}. See the column g.f.s and the complete homogeneous symmetric function formula for T(n, k) below. - Wolfdieter Lang, May 26 2017
T(n, k) is the number of k-dimensional subspaces (i.e., sets of fixed points like rotation axes and symmetry planes) of the n-cube. See "Sets of fixed points..." in LINKS section. - Tilman Piesk, Oct 26 2019

Examples

			Triangle T(n,k) begins:
  n\k 0     1       2        3       4       5      6     7    8   9 10 ...
  0:  1
  1:  1     1
  2:  1     4       1
  3:  1    13       9        1
  4:  1    40      58       16       1
  5:  1   121     330      170      25       1
  6:  1   364    1771     1520     395      36      1
  7:  1  1093    9219    12411    5075     791     49     1
  8:  1  3280   47188    96096   58086   13776   1428    64    1
  9:  1  9841  239220   719860  618870  209622  32340  2388   81   1
 10:  1 29524 1205941  5278240 6289690 2924712 630042 68160 3765 100  1
 ... reformatted and extended by _Wolfdieter Lang_, May 26 2017
The sequence of row polynomials of A214406 begins [1, 1+x, 1+8*x+3*x^2, ...]. The o.g.f.'s for the diagonals of this triangle thus begin
1/(1-x) = 1 + x + x^2 + x^3 + ...
(1+x)/(1-x)^3 = 1 + 4*x + 9*x^2 + 16*x^3 + ...
(1+8*x+3*x^2)/(1-x)^5 = 1 + 13*x + 58*x^2 + 170*x^3 + ... . - _Peter Bala_, Jul 20 2012
Connection constants: x^3 = 1 + 13*(x-1) + 9*(x-1)*(x-3) + (x-1)*(x-3)*(x-5). Hence row 3 = [1,13,9,1]. - _Peter Bala_, Jun 23 2014
Complete homogeneous symmetric functions: T(3, 1) = h^{(2)}_2 = 1^2 + 3^2 + 1^1*3^1 = 13. The three 2D polytopes are two squares and a rectangle. T(3, 2) = h^{(3)}_1 = 1^1 + 3^1 + 5^1 = 9. The 1D polytopes are three lines. - _Wolfdieter Lang_, May 26 2017
T(4, 3) = 16 is the number of 3-dimensional subspaces (mirror hyperplanes) of the 4-cube. (These are 4 cubes and 12 cuboids.) See "Sets of fixed points..." in LINKS section. - _Tilman Piesk_, Oct 26 2019
		

Crossrefs

Programs

  • Magma
    [[(&+[(-1)^(k-j)*(2*j+1)^n*Binomial(k, j): j in [0..k]])/( 2^k*Factorial(k)): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 14 2019
    
  • Maple
    A039755 := proc(n,k) if k < 0 or k > n then 0 ; elif n <= 1 then 1; else procname(n-1,k-1)+(2*k+1)*procname(n-1,k) ; end if; end proc:
    seq(seq(A039755(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Oct 30 2009
  • Mathematica
    t[n_, k_] = Sum[(-1)^(k-j)*(2j+1)^n*Binomial[k, j], {j, 0, k}]/(2^k*k!); Flatten[Table[t[n, k], {n, 0, 10}, {k, 0, n}]][[1 ;; 56]]
    (* Jean-François Alcover, Jun 09 2011, after Peter Bala *)
  • PARI
    T(n,k)=if(k<0 || k>n,0,n!*polcoeff(polcoeff(exp(x+y/2*(exp(2*x+x*O(x^n))-1)),n),k))
    
  • Sage
    [[sum((-1)^(k-j)*(2*j+1)^n*binomial(k, j) for j in (0..k))/( 2^k*factorial(k)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Feb 14 2019

Formula

E.g.f. row polynomials: exp(x + y/2 * (exp(2*x) - 1)).
T(n,k) = T(n-1,k-1) + (2*k+1)*T(n-1,k) with T(0,k) = 1 if k=0 and 0 otherwise. Sum_{k=0..n} T(n,k) = A007405(n). - R. J. Mathar, Oct 30 2009; corrected by Joshua Swanson, Feb 14 2019
T(n,k) = (1/(2^k*k!)) * Sum_{j=0..k} (-1)^(k-j)*C(k,j)*(2*j+1)^n.
T(n,k) = (1/(2^k*k!)) * A145901(n,k). - Peter Bala
The row polynomials R(n,x) satisfy the Dobinski-type identity:
R(n,x) = exp(-x/2)*Sum_{k >= 0} (2*k+1)^n*(x/2)^k/k!, as well as the recurrence equation R(n+1,x) = (1+x)*R(n,x)+2*x*R'(n,x). The polynomial R(n,x) has all real zeros (apply [Liu et al., Theorem 1.1] with f(x) = R(n,x) and g(x) = R'(n,x)). The polynomials R(n,2*x) are the row polynomials of A154537. - Peter Bala, Oct 28 2011
Let f(x) = exp((1/2)*exp(2*x)+x). Then the row polynomials R(n,x) are given by R(n,exp(2*x)) = (1/f(x))*(d/dx)^n(f(x)). Similar formulas hold for A008277, A105794, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012
From Peter Bala, Jul 20 2012: (Start)
The o.g.f. for the n-th diagonal (with interpolated zeros) is the rational function D^n(x), where D is the operator x/(1-x^2)*d/dx. For example, D^3(x) = x*(1+8*x^2+3*x^4)/(1-x^2)^5 = x + 13*x^3 + 58*x^5 + 170*x^7 + ... . See A214406 for further details.
An alternative formula for the o.g.f. of the n-th diagonal is exp(-x/2)*(Sum_{k >= 0} (2*k+1)^(k+n-1)*(x/2*exp(-x))^k/k!).
(End)
From Tom Copeland, Dec 31 2015: (Start)
T(n,m) = Sum_{i=0..n-m} 2^(n-m-i)*binomial(n,i)*St2(n-i,m), where St2(n,k) are the Stirling numbers of the second kind, A048993 (also A008277). See p. 755 of Dolgachev and Lunts.
The relation of this entry's e.g.f. above to that of the Bell polynomials, Bell_n(y), of A048993 establishes this formula from a binomial transform of the normalized Bell polynomials, NB_n(y) = 2^n Bell_n(y/2); that is, e^x exp[(y/2)(e^(2x)-1)] = e^x exp[x*2*Bell.(y/2)] = exp[x(1+NB.(y))] = exp(x*P.(y)), so the row polynomials of this entry are given by P_n(y) = [1+NB.(y)]^n = Sum_{k=0..n} C(n,k) NB_k(y) = Sum_{k=0..n} 2^k C(n,k) Bell_k(y/2).
The umbral compositional inverses of the Bell polynomials are the falling factorials Fct_n(y) = y! / (y-n)!; i.e., Bell_n(Fct.(y)) = y^n = Fct_n(Bell.(y)). Since P_n(y) = [1+2Bell.(y/2)]^n, the umbral inverses are determined by [1 + 2 Bell.[ 2 Fct.[(y-1)/2] / 2 ] ]^n = [1 + 2 Bell.[ Fct.[(y-1)/2] ] ]^n = [1+y-1]^n = y^n. Therefore, the umbral inverse sequence of this entry's row polynomials is the sequence IP_n( y) = 2^n Fct_n[(y-1)/2] = (y-1)(y-3) .. (y-2n+1) with IP_0(y) = 1 and, from the binomial theorem, with e.g.f. exp[x IP.(y)]= exp[ x 2Fct.[(y-1)/2] ] = (1+2x)^[(y-1)/2] = exp[ [(y-1)/2] log(1+2x) ].
(End)
Let B(n,k) = T(n,k)*((2*k)!)/(2^k*k!) and P(n,x) = Sum_{k=0..n} B(n,k)*x^(2*k+1). Then (1) P(n+1,x) = (x+x^3)*P'(n,x) for n >= 0, and (2) Sum_{n>=0} B(n,k)/(n!)*t^n = binomial(2*k,k)*exp(t)*(exp(2*t)-1)^k/4^k for k >= 0, and (3) Sum_{n>=0} t^n* P(n,x)/(n!) = x*exp(t)/sqrt(1+x^2-x^2*exp(2*t)). - Werner Schulte, Dec 12 2016
From Wolfdieter Lang, May 26 2017: (Start)
G.f. column k: x^k/Product_{j=0..k} (1 - (1+2*j)*x), k >= 0.
T(n, k) = h^{(k+1)}_{n-k}, the complete homogeneous symmetric function of degree n-k of the k+1 symbols a_j = 1 + 2*j, j = 0, 1, ..., k. (End)
With p(n, x) = Sum_{k=0..n} A001147(k) * T(n, k) * x^k for n >= 0 holds:
(1) Sum_{i=0..n} p(i, x)*p(n-i, x) = 2^n*(Sum_{k=0..n} A028246(n+1, k+1)*x^k);
(2) p(n, -1/2) = (n!) * ([t^n] sqrt(2 / (1 + exp(-2*t)))). - Werner Schulte, Feb 16 2024

A143494 Triangle read by rows: 2-Stirling numbers of the second kind.

Original entry on oeis.org

1, 2, 1, 4, 5, 1, 8, 19, 9, 1, 16, 65, 55, 14, 1, 32, 211, 285, 125, 20, 1, 64, 665, 1351, 910, 245, 27, 1, 128, 2059, 6069, 5901, 2380, 434, 35, 1, 256, 6305, 26335, 35574, 20181, 5418, 714, 44, 1, 512, 19171, 111645, 204205, 156660, 58107, 11130, 1110, 54, 1
Offset: 2

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 2 of the r-Stirling numbers of the second kind. The 2-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1 and 2 belong to distinct subsets.
More generally, the r-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the numbers 1, 2, ..., r belong to distinct subsets. The case r = 1 gives the usual Stirling numbers of the second kind A008277; for other cases see A143495 (r = 3) and A143496 (r = 4).
The lower unitriangular array of r-Stirling numbers of the second kind equals the matrix product P^(r-1) * S (with suitable offsets in the row and column indexing), where P is Pascal's triangle, A007318 and S is the array of Stirling numbers of the second kind, A008277.
For the definition of and entries relating to the corresponding r-Stirling numbers of the first kind see A143491. For entries on r-Lah numbers refer to A143497. The theory of r-Stirling numbers of both kinds is developed in [Broder].
From Peter Bala, Sep 19 2008: (Start)
Let D be the derivative operator d/dx and E the Euler operator x*d/dx. Then x^(-2)*E^n*x^2 = Sum_{k = 0..n} T(n+2,k+2)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k= 2..n} T(n,k)*x^k satisfy the recurrence R_(n+1)(x) = x*R_n(x) + x*d/dx(R_n(x)) with R_2(x) = x^2. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 2-Eulerian numbers E_2(n,j) := A144696(n,j): T(n,k) = 2!/k!*Sum_ {j = n-k..n-2} E_2(n,j)*binomial(j,n-k) for n >= k >= 2. (End)
From Wolfdieter Lang, Sep 29 2011: (Start)
T(n,k) = S(n,k,2), n>=k>=2, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column no. k from (A20) with k->2, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(2*x),exp(x)-1) with e.g.f. of column no. m>=0: exp(2*x)*((exp(x)-1)^m)/m!. See one of the formulas given below. For Sheffer matrices see the W. Lang link under A006232 with the S. Roman reference, also found in A132393. (End)

Examples

			Triangle begins
  n\k|...2....3....4....5....6....7
  =================================
  2..|...1
  3..|...2....1
  4..|...4....5....1
  5..|...8...19....9....1
  6..|..16...65...55...14....1
  7..|..32..211..285..125...20....1
  ...
T(4,3) = 5. The set {1,2,3,4} can be partitioned into three subsets such that 1 and 2 belong to different subsets in 5 ways: {{1}{2}{3,4}}, {{1}{3}{2,4}}, {{1}{4}{2,3}}, {{2}{3}{1,4}} and {{2}{4}{1,3}}; the remaining possibility {{1,2}{3}{4}} is not allowed.
From _Peter Bala_, Feb 23 2025: (Start)
The array factorizes as
/ 1               \       /1             \ /1             \ /1            \
| 2    1           |     | 2   1          ||0  1           ||0  1          |
| 4    5   1       |  =  | 4   3   1      ||0  2   1       ||0  0  1       | ...
| 8   19   9   1   |     | 8   7   4  1   ||0  4   3  1    ||0  0  2  1    |
|16   65  55  14  1|     |16  15  11  6  1||0  8   7  4  1 ||0  0  4  3  1 |
|...               |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 2*x), x/(1 - x)). See A055248. (End)
		

Crossrefs

A001047 (column 3), A005493 (row sums), A008277, A016269 (column 4), A025211 (column 5), A049444 (matrix inverse), A074051 (alt. row sums).

Programs

  • Maple
    with combinat: T := (n, k) -> (1/(k-2)!)*add ((-1)^(k-i)*binomial(k-2,i)*(i+2)^(n-2),i = 0..k-2): for n from 2 to 11 do seq(T(n, k), k = 2..n) end do;
  • Mathematica
    t[n_, k_] := StirlingS2[n, k] - StirlingS2[n-1, k]; Flatten[ Table[ t[n, k], {n, 2, 11}, {k, 2, n}]] (* Jean-François Alcover, Dec 02 2011 *)
  • Sage
    @CachedFunction
    def stirling2r(n, k, r) :
        if n < r: return 0
        if n == r: return 1 if k == r else 0
        return stirling2r(n-1,k-1,r) + k*stirling2r(n-1,k,r)
    A143494 = lambda n,k: stirling2r(n, k, 2)
    for n in (2..6):
        [A143494(n, k) for k in (2..n)] # Peter Luschny, Nov 19 2012

Formula

T(n+2,k+2) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*(i+2)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - Stirling2(n-1,k) for n, k >= 2.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 2, with boundary conditions T(n,1) = T(1,n) = 0 for all n, T(2,2) = 1 and T(2,k) = 0 for k > 2. Special cases: T(n,2) = 2^(n-2); T(n,3) = 3^(n-2) - 2^(n-2).
As a sum of monomial functions of degree m: T(n+m,n) = Sum_{2 <= i_1 <= ... <= i_m <= n} (i_1*i_2*...*i_m). For example, T(6,4) = Sum_{2 <= i <= j <= 4} (i*j) = 2*2 + 2*3 + 2*4 + 3*3 + 3*4 + 4*4 = 55.
E.g.f. column k+2 (with offset 2): 1/k!*exp(2*x)*(exp(x) - 1)^k.
O.g.f. k-th column: Sum_{n >= k} T(n,k)*x^n = x^k/((1-2*x)*(1-3*x)*...*(1-k*x)).
E.g.f.: exp(2*t + x*(exp(t) - 1)) = Sum_{n >= 0} Sum_{k = 0..n} T(n+2,k+2) *x^k*t^n/n! = Sum_{n >= 0} B_n(2;x)*t^n/n! = 1 + (2 + x)*t/1! + (4 + 5*x + x^2)*t^2/2! + ..., where the row polynomial B_n(2;x) := Sum_{k = 0..n} T(n+2,k+2)*x^k denotes the 2-Bell polynomial.
Dobinski-type identities: Row polynomial B_n(2;x) = exp(-x)*Sum_{i >= 0} (i + 2)^n*x^i/i!. Sum_{k = 0..n} k!*T(n+2,k+2)*x^k = Sum_{i >= 0} (i + 2)^n*x^i/(1 + x)^(i+1).
The T(n,k) are the connection coefficients between falling factorials and the shifted monomials (x + 2)^(n-2). For example, from row 4 we have 4 + 5*x + x*(x - 1) = (x + 2)^2, while from row 5 we have 8 + 19*x + 9*x*(x - 1) + x*(x - 1)*(x - 2) = (x + 2)^3.
The row sums of the array are the 2-Bell numbers, B_n(2;1), equal to A005493(n-2). The alternating row sums are the complementary 2-Bell numbers, B_n(2;-1), equal to (-1)^n*A074051(n-2).
This array is the matrix product P * S, where P denotes the Pascal triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
Also, this array equals the transpose of the upper triangular array A126351. The inverse array is A049444, the signed 2-Stirling numbers of the first kind. See A143491 for the unsigned version of the inverse.
Let f(x) = exp(exp(x)). Then for n >= 1, the row polynomials R(n,x) are given by R(n+2,exp(x)) = 1/f(x)*(d/dx)^n(exp(2*x)*f(x)). Similar formulas hold for A008277, A039755, A105794, A111577 and A154537. - Peter Bala, Mar 01 2012

A111577 Galton triangle T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 21, 12, 1, 1, 85, 105, 22, 1, 1, 341, 820, 325, 35, 1, 1, 1365, 6081, 4070, 780, 51, 1, 1, 5461, 43932, 46781, 14210, 1596, 70, 1, 1, 21845, 312985, 511742, 231511, 39746, 2926, 92, 1, 1, 87381, 2212740, 5430405, 3521385, 867447, 95340, 4950, 117, 1
Offset: 1

Views

Author

Gary W. Adamson, Aug 07 2005

Keywords

Comments

In triangles of analogs to Stirling numbers of the second kind, the multipliers of T(n-1,k) in the recurrence are terms in arithmetic sequences: in Pascal's triangle A007318, the multiplier = 1. In triangle A008277, the Stirling numbers of the second kind, the multipliers are in the set (1,2,3...). For this sequence here, the multipliers are from A016777.
Riordan array [exp(x), (exp(3x)-1)/3]. - Paul Barry, Nov 26 2008
From Peter Bala, Jan 27 2015: (Start)
Working with an offset of 0, this is the triangle of connection constants between the polynomial basis sequences {x^n}, n>=0 and {n!*3^n*binomial((x - 1)/3,n)}, n>=0. An example is given below.
Call this array M and let P denote Pascal's triangle A007318, then P * M = A225468, P^2 * M = A075498. Also P^(-1) * M is a shifted version of A075498.
This triangle is the particular case a = 3, b = 0, c = 1 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. (End)
Named after the English scientist Francis Galton (1822-1911). - Amiram Eldar, Jun 13 2021
This is the array of (r, β)-Stirling numbers for r = 1 and β = 3. See Corcino. - Peter Bala, Feb 26 2025

Examples

			T(5,3) = T(4,2) + 7*T(4,3) = 21 + 7*12 = 105.
The triangle starts in row n = 1 as:
  1;
  1,  1;
  1,  5,   1;
  1, 21,  12,  1;
  1, 85, 105, 22, 1;
Connection constants: Row 4: [1, 21, 12, 1] so
x^3 = 1 + 21*(x - 1) + 12*(x - 1)*(x - 4) + (x - 1)*(x - 4)*(x - 7). - _Peter Bala_, Jan 27 2015
From _Peter Bala_, Feb 26 2025: (Start)
The array factorizes as
/1                \     /1               \/1              \/1             \
|1   1            |     |1   1           ||0  1           ||0  1          |
|1   5    1       |  =  |1   4   1       ||0  1   1       ||0  0  1       | ...
|1  21   12   1   |     |1  13   7   1   ||0  1   4  1    ||0  0  1  1    |
|1  85  105  22  1|     |1  44  34  10  1||0  1  13  7  1 ||0  0  1  4  1 |
|...              |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - x), x/(1 - 3*x)). Cf. A193843. (End)
		

Crossrefs

Programs

  • Maple
    A111577 := proc(n,k) option remember; if k = 1 or k = n then 1; else procname(n-1,k-1)+(3*k-2)*procname(n-1,k) ; fi; end:
    seq( seq(A111577(n,k),k=1..n), n=1..10) ; # R. J. Mathar, Aug 22 2009
  • Mathematica
    T[, 1] = 1; T[n, n_] = 1;
    T[n_, k_] := T[n, k] = T[n-1, k-1] + (3k-2) T[n-1, k];
    Table[T[n, k], {n, 1, 10}, {k, 1, n}] (* Jean-François Alcover, Jun 13 2019 *)

Formula

T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k).
E.g.f.: exp(x)*exp((y/3)*(exp(3x)-1)). - Paul Barry, Nov 26 2008
Let f(x) = exp(1/3*exp(3*x) + x). Then, with an offset of 0, the row polynomials R(n,x) are given by R(n,exp(3*x)) = 1/f(x)*(d/dx)^n(f(x)). Similar formulas hold for A008277, A039755, A105794, A143494 and A154537. - Peter Bala, Mar 01 2012
T(n, k) = 1/(3^k*k!)*Sum_{j=0..k}((-1)^(k-j)*binomial(k,j)*(3*j+1)^n). - Peter Luschny, May 20 2013
From Peter Bala, Jan 27 2015: (Start)
T(n,k) = Sum_{i = 0..n-1} 3^(i-k+1)*binomial(n-1,i)*Stirling2(i,k-1).
O.g.f. for n-th diagonal: exp(-x/3)*Sum_{k >= 0} (3*k + 1)^(k+n-1)*((x/3*exp(-x))^k)/k!.
O.g.f. column k (with offset 0): 1/( (1 - x)*(1 - 4*x)*...*(1 - (3*k + 1)*x) ). (End)

Extensions

Edited and extended by R. J. Mathar, Aug 22 2009

A094645 Triangle of generalized Stirling numbers of the first kind.

Original entry on oeis.org

1, -1, 1, 0, -1, 1, 0, -1, 0, 1, 0, -2, -1, 2, 1, 0, -6, -5, 5, 5, 1, 0, -24, -26, 15, 25, 9, 1, 0, -120, -154, 49, 140, 70, 14, 1, 0, -720, -1044, 140, 889, 560, 154, 20, 1, 0, -5040, -8028, -64, 6363, 4809, 1638, 294, 27, 1, 0, -40320, -69264, -8540, 50840, 44835, 17913, 3990, 510, 35, 1
Offset: 0

Views

Author

Vladeta Jovovic, May 17 2004

Keywords

Comments

From Wolfdieter Lang, Jun 20 2011: (Start)
The row polynomials s(n,x) := Sum_{j=0..n} T(n,k)*x^k satisfy risefac(x-1,n) = s(n,x), with the rising factorials risefac(x-1,n) := Product_{j=0..n-1} (x-1+j), n >= 1, risefac(x-1,0) = 1. Compare with the formula risefac(x,n) = s1(n,x), with the row polynomials s1(n,x) of A132393 (unsigned Stirling1).
This is the lower triangular Sheffer array with e.g.f.
T(x,z) = (1-z)*exp(-x*log(1-z)) (the rewritten e.g.f. from the formula section). See the W. Lang link under A006232 for Sheffer matrices and the Roman reference. In the notation which indicates the column e.g.f.s this is Sheffer (1-z,-log(1-z)). In the umbral notation (cf. Roman) this is called Sheffer for (exp(t),1-exp(-t)).
The row polynomials satisfy s(n,x) = (x+n-1)*s(n-1,x), s(0,x)=1, and s(n,x) = (x-1)*s1(n-1,x), n >= 1, s1(0,x) = 1, with the unsigned Stirling1 row polynomials s1(n,x).
The row polynomials also satisfy
s(n,x) - s(n,x-1) = n*s(n-1,x), n > 1, s(0,x) = 1
(from the Meixner identity, see the Meixner reference given at A060338).
The row polynomials satisfy as well (from corollary 3.7.2. p. 50 of the Roman reference)
s(n,x) = (x-1)*s(n-1,x+1), n >= 1, s(0,n) = 1.
The exponential convolution identity is
s(n,x+y) = Sum_{k=0..n} binomial(n,k)*s(k,y)*s1(n-k,x),
n >= 0, with symmetry x <-> y.
The row sums are 1 for n=0 and 0 otherwise, and the alternating row sums are 1,-2,2, followed by zeros, with e.g.f. (1-x)^2.
The Sheffer a-sequence Sha(n) = A164555(n)/A027642(n) with e.g.f. x/(1-exp(-x)), and the z-sequence is Shz(n) = -1 with e.g.f. -exp(x).
The inverse Sheffer matrix is ((-1)^(n-k))*A105794(n,k) with e.g.f. exp(z)*exp(x*(1-exp(-z))). (End)
Triangle T(n,k), read by rows, given by (-1, 1, 0, 2, 1, 3, 2, 4, 3, 5, ...) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 16 2012
Also coefficients of t in t*(t-1)*Sum[(-1)^(n+m) t^(m-1) StirlingS1[n,m], {m,n}] in which setting t^k equal to k gives n!, from this follows that the dot product of row n with [0,...,n] equals (n-1)!. - Wouter Meeussen, May 15 2012

Examples

			Triangle begins
   1;
  -1,   1;
   0,  -1,   1;
   0,  -1,   0,   1;
   0,  -2,  -1,   2,   1;
   0,  -6,  -5,   5,   5,   1;
   0, -24, -26,  15,  25,   9,   1;
   ...
Recurrence:
  -2 = T(4,1) = T(3,0) + (4-2)*T(3,1) = 0 + 2*(-1).
Row polynomials:
  s(3,x) = -x+x^3 = (x-1)*s1(2,x) = (x-1)*(x+x^2).
  s(3,x) = (x-1)*s(2,x+1) = (x-1)*(-(x+1)+(x+1)^2).
  s(3,x) - s(3,x-1) = -x+x^3 -(-(x-1)+(x-1)^3) = 3*(-x+x^2) = 3*s(2,x).
		

References

  • S. Roman, The Umbral Calculus, Academic Press, New York, 1984.

Crossrefs

Programs

Formula

E.g.f.: (1-y)^(1-x).
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000142(n+1), A001710(n+2), A001715(n+3), A001720(n+4), A001725(n+5), A001730(n+6), A049388(n), A049389(n), A049398(n), A051431(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 respectively. - Philippe Deléham, Nov 13 2007
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then |T(n,i)| = |f(n,i,-1)|, for n=1,2,...; i=0..n. - Milan Janjic, Dec 21 2008
From Wolfdieter Lang, Jun 20 2011: (Start)
T(n,k) = |S1(n-1,k-1)| - |S1(n-1,k)|, n >= 1, k >= 1, with |S1(n,k)| = A132393(n,k) (unsigned Stirling1).
Recurrence: T(n,k) = T(n-1,k-1) + (n-2)*T(n-1,k) if n >= k >= 0; T(n,k) = 0 if n < k; T(n,-1) = 0; T(0,0) = 1.
E.g.f. column k: (1-x)*((-log(1-x))^k)/k!. (End)
T(n,k) = Sum_{i=0..n} binomial(n,i)*(n-i)!*Stirling1(i,k)*TC(m,n,i) where TC(m,n,k) = Sum_{i=0..n-k} binomial(n+1,n-k-i)*Stirling2(i+m+1,i+1)*(-1)^i, m = 1 for n >= 0. See A130534, A370518 for m=0 and m=2. - Igor Victorovich Statsenko, Feb 27 2024

A261139 S'_t(n) is the number of set partitions of {1,2,...,t} into exactly n parts such that no part contains both 1 and t or both i and i+1 for some i with 1 <= i < t; triangle S'_t(n), t >= 0, 0 <= n <= t, read by rows.

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 2, 1, 0, 0, 0, 5, 5, 1, 0, 0, 1, 10, 20, 9, 1, 0, 0, 0, 21, 70, 56, 14, 1, 0, 0, 1, 42, 231, 294, 126, 20, 1, 0, 0, 0, 85, 735, 1407, 924, 246, 27, 1, 0, 0, 1, 170, 2290, 6363, 6027, 2400, 435, 35, 1
Offset: 0

Views

Author

Mark Wildon, Aug 10 2015

Keywords

Comments

S'A261137%20may%20be%20defined%20by%20B'_t(n)%20=%20Sum">t(n) is the number of sequences of t non-identity top-to-random shuffles of a deck of n cards that move each card at some time, and overall leave the deck invariant. (See link below.) A261137 may be defined by B'_t(n) = Sum{m=0..n} S'_t(m).

Examples

			Triangle starts:
  1;
  0, 0;
  0, 0, 1;
  0, 0, 0,  1;
  0, 0, 1,  2,   1;
  0, 0, 0,  5,   5,    1;
  0, 0, 1, 10,  20,    9,   1;
  0, 0, 0, 21,  70,   56,  14,   1;
  0, 0, 1, 42, 231,  294, 126,  20,  1;
  0, 0, 0, 85, 735, 1407, 924, 246, 27,  1;
  ...
		

Crossrefs

Columns n=3,4 give: A000975, A243869.
Row sums give A000296.
Cf. A261137.
The same as A105794, except for the first two columns.

Programs

  • Maple
    g:= proc(t, l, h) option remember; `if`(t=0, `if`(l=1, 0, x^h),
           add(`if`(j=l, 0, g(t-1, j, max(h,j))), j=1..h+1))
        end:
    S:= t-> (p-> seq(coeff(p, x, i), i=0..t))(g(t, 0$2)):
    seq(S(t), t=0..12);  # Alois P. Heinz, Aug 10 2015
  • Mathematica
    StirPrimedGF[n_, x_] := x^n/(1 + x)*Product[1/(1 - j*x), {j, 1, n - 1}]; T[0, 0] = 1; T[, 0] = T[, 1] = 0; T[n_, k_] := SeriesCoefficient[ StirPrimedGF[k, x], {x, 0, n}]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* script completed by Jean-François Alcover, Jan 31 2016 *)
  • PARI
    a(n,k)=if(k==0, n==0, sum(j=0, k, binomial(k, j) * (-1)^(k-j) * ((j-1)^n + (-1)^n * (j-1))) / k!);
    for(n=0, 10, for(k=0, n, print1( a(n, k), ", "); ); print(); ); \\ Andrew Howroyd, Apr 08 2017

Formula

G.f. for column n > 1: x^n/((1+x)*Product_{j=1..n-1} (1-j*x)).
S'_t(n) ~ (n-1)^t/n! as t tends to infinity.
Recurrence: S't(n) = S'{t-1}(n-1) + (n-1)*S'_{t-1}(n) for n >= 3.
S't(n) = (1/n!) * Sum{j=0..n} (-1)^(n-j) * binomial(n, j) * ((j-1)^t + (-1)^t * (j-1)) for t>0. - Andrew Howroyd, Apr 08 2017
Sum_{n=0..t} (n-1)*S'{t-1}(n) + n*S'{t-2}(n) = A000296(t) for t >= 3. - Yuchun Ji, Feb 23 2021
T(m, k) = Sum_{i=k..m} Stirling2(i-1, k-1)*(-1)^(i+m), for k >= 2. (See Peter Bala's original formula at A105794 dated Jul 10 2013.) - Igor Victorovich Statsenko, May 31 2024
T(m, k) = (Sum_{i=0..m} Stirling2(i, k)*binomial(m,i)*(-1)^(m-i))*I(m,k), where I(m,k) = (1-Sum_{i=0..m} Stirling1(k, i))^(m+k) for k >= 0. (See Peter Bala's original formula at A105794 dated Jul 10 2013.) - Igor Victorovich Statsenko, Jun 01 2024

A269953 Triangle read by rows: T(n, k) = Sum_{j=0..n} binomial(-j-1, -n-1)*S1(j, k) where S1 are the Stirling cycle numbers A132393.

Original entry on oeis.org

1, -1, 1, 1, -1, 1, -1, 2, 0, 1, 1, 0, 5, 2, 1, -1, 9, 15, 15, 5, 1, 1, 35, 94, 85, 40, 9, 1, -1, 230, 595, 609, 315, 91, 14, 1, 1, 1624, 4458, 4844, 2779, 924, 182, 20, 1, -1, 13209, 37590, 43238, 26817, 9975, 2310, 330, 27, 1
Offset: 0

Views

Author

Peter Luschny, Apr 12 2016

Keywords

Comments

Replacing the Stirling cycle numbers in the definition by the Stirling set numbers leads to A105794.
From Wolfdieter Lang, Jun 19 2017: (Start)
The triangle t(n, k) = (-1)^(n-k)*T(n, k) is the matrix product of P = A007318 (Pascal) and s1 = A048994 (signed Stirling1). This is Sheffer (exp(t), log(1+t)).
The present triangle T is therefore the Sheffer triangle (exp(-t), -log(1-t)). Note that P is Sheffer (exp(t), t) (of the Appell type). (End)
The triangle T(n,k) is a representative of the parametric family of triangles T(m,n,k), whose columns are the coefficients of the standard expansion of the function f(x) = (-log(1-x))^(k)*exp(-m*x)/k! for the case m=1. See A381082. - Igor Victorovich Statsenko, Feb 14 2025

Examples

			Triangle starts:
   1;
  -1,  1;
   1, -1,  1;
  -1,  2,  0,  1;
   1,  0,  5,  2,  1;
  -1,  9, 15, 15,  5,  1;
   1, 35, 94, 85, 40,  9,  1.
		

Crossrefs

Columns k=0..4 give A033999, A002741, A381064, A381065, A381066.
Cf. A000166 (row sums), A080956 (diag n,n-1).
KummerU(-n,1-n-x,z): this sequence (z=-1), A094816 (z=1), |A137346| (z=2), A327997 (z=3).

Programs

  • Maple
    A269953 := (n,k) -> add(binomial(-j-1,-n-1)*abs(Stirling1(j,k)), j=0..n):
    seq(print(seq(A269953(n, k), k=0..n)), n=0..9);
    # Alternative:
    egf := exp(-t)*(1-t)^(-x): ser := series(egf, t, 12): p := n -> coeff(ser, t, n):
    seq(n!*seq(coeff(p(n), x, k), k=0..n), n=0..9); # Peter Luschny, Oct 28 2019
  • Mathematica
    Flatten[Table[Sum[Binomial[-j-1,-n-1] Abs[StirlingS1[j,k]], {j,0,n}], {n,0,9},{k,0,n}]]
    (* Or: *)
    p [n_] := HypergeometricU[-n, 1 - n - x, -1];
    Table[CoefficientList[p[n], x], {n, 0, 9}] (* Peter Luschny, Oct 28 2019 *)

Formula

From Wolfdieter Lang, Jun 19 2017: (Start)
E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n,k)*x^k: exp(-t)/(1 - t)^x.
E.g.f. of column k sequence: exp(-x)*(-log(1-x))^k/k!, k >= 0. (End)
From Peter Bala, Oct 26 2019: (Start)
Let R(n, x) = (-1)^n*Sum_{k >= 0} binomial(n,k)*k!* binomial(-x,k) the n-th row polynomial of this triangle.
R(n, x) = c_n(-x;-1), where c_n(x;a) denotes the n-th Poisson Charlier polynomial.
The series representation e = Sum_{k >= 0} 1/k! is the case n = 0 of the more general result e = n!*Sum_{k >= 0} 1/(k!*R(n,k)*R(n,k+1)), n = 0,2,3,4,.... (End)
R(n, x) = KummerU(-n, 1-n-x, -1). - Peter Luschny, Oct 28 2019

A105793 Expansion of e.g.f. (1 + y)^(1 + x).

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, -1, 0, 1, 0, 2, -1, -2, 1, 0, -6, 5, 5, -5, 1, 0, 24, -26, -15, 25, -9, 1, 0, -120, 154, 49, -140, 70, -14, 1, 0, 720, -1044, -140, 889, -560, 154, -20, 1, 0, -5040, 8028, -64, -6363, 4809, -1638, 294, -27, 1, 0, 40320, -69264, 8540, 50840, -44835, 17913, -3990, 510, -35, 1
Offset: 0

Views

Author

Paul Barry, Apr 20 2005

Keywords

Comments

Generalized Stirling number triangle of first kind. Row sums are (1,2,2,0,0,0,...) = 2C(2,n) - 2C(1,n) + C(0,n). Inverse is A105794.
Triangle T(n,k), 0 <= k <= n, read by rows, given by [1, -1, 0, -2, -1, -3, -2, -4, -3, -5, -4, -6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 23 2006

Examples

			From _Wolfdieter Lang_, Jun 19 2017: (Start)
The triangle T(n, k) starts
  n\k  0     1      2    3     4      5     6     7   8   9 10 ...
  0:   1
  1:   1     1
  2:   0     1      1
  3:   0    -1      0    1
  4:   0     2     -1   -2     1
  5:   0    -6      5    5    -5      1
  6:   0    24    -26  -15    25     -9     1
  7:   0  -120    154   49  -140     70   -14     1
  8:   0   720  -1044 -140   889   -560   154   -20   1
  9:   0 -5040   8028  -64 -6363   4809 -1638   294 -27   1
  10:  0 40320 -69264 8540 50840 -44835 17913 -3990 510 -35  1
  ... reformatted
Recurrence from a- and z-sequence (see above): T(1, 0) = T(0, 0) = 1; T(1, 1) = (1/1)*(1*T(0, 0)) = 1, T(2, 0) = 2*(T(1, 0) - T(1, 1)) = 0, T(2, 1) = (2/1)*(T(1,0) + (-1/2)*T(1, 1)) = 1. T(3, 1) = (3/1)*(0 + (-1/2)*T(2, 1) + (1/6)*T(2, 2)) = -1. (End)
		

Crossrefs

Programs

  • Mathematica
    t[0, 0] = 1; t[n_, k_] := StirlingS1[n, k] + n*StirlingS1[n-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 04 2013, after Giuliano Cabrele *)

Formula

E.g.f.: (1+y)^(1+x); rows have g.f. k!*binomial(x+1, k); Columns have g.f. (1+x)*log(1+x)^k.
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then T(n,i) = f(n,i,-1), for n=1,2,...; i=0...n. - Milan Janjic, Dec 21 2008
So T(n,k) = Stirling1(n,k) + n*Stirling1(n-1,k), Stirling1 being the (signed) Stirling numbers of first kind A048994. In terms of lower triangular matrices, 0<= k <= n, T is also the product [Stirling1] * [Pascal] = [A048994] * [A007318], i.e., T(n,k) = Sum_{j=0..n} Stirling1(n,j) * binomial(j,k). - Giuliano Cabrele, Jan 19 2009
This is the triangle of connection constants for expressing the basis of falling factorial polynomials x_(k) := x*(x-1)*...*(x-k+1) in terms of the polynomial sequence (x-1)^n, that is, x_(n) = Sum_{k = 0..n} T(n,k)*(x-1)^k. - Peter Bala, Jul 10 2013
From Wolfdieter Lang, Jun 19 2017: (Start)
Triangle T is the (infinite) matrix product of A048994 (Stirling1) and A007318 (Pascal): T(n,k) = Sum_{m=k..n} Stirling1(n, m)*Pascal(m, k), n >= k >= 0, and 0 for n < k. Note that the Pascal matrix is Sheffer (e^t, t) of the Appell type.
T is the Sheffer (aka exponential Riordan) matrix (1+t, log(1+t)).
E.g.f. column k: (1+x)*(log(1+x))^k/k!, k >= 0.
The a-sequence for T is A027641/A027642 (Bernoulli), and the z-sequence is A033999 (repeat(1,-1)) (see a W. Lang link under A006232 for a- and z-sequences for Sheffer matrices, also for references).
Therefore the combined recurrence is: T(n, 0) = n*Sum_{j=0..n-1} (-1)^j*T(n-1, j), n >= 1, T(0, 0) = 1, and T(n, m) = (n/m)*Sum_{j=0..n-m} binomial(m-1+j, m-1)*Bernoulli(j)*T(n-1, m-1+j), n >= m >= 1. (End)

A243869 Expansion of x^4/[(1+x)*Product_{k=1..3} (1-k*x)].

Original entry on oeis.org

1, 5, 20, 70, 231, 735, 2290, 7040, 21461, 65065, 196560, 592410, 1782691, 5358995, 16098830, 48340180, 145107921, 435498525, 1306845100, 3921234350, 11765101151, 35298099655, 105899891370, 317710858920, 953154946381, 2859509578385, 8578618213640
Offset: 4

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Author

R. J. Mathar, Jun 13 2014

Keywords

Comments

The number of ways to partition a set of n people around a circular table into 4 affinity groups with no two members of a group seated next to each other [Knuth].
The first two primes of the sequence are a(5) and a(96). - Bruno Berselli, Jun 13 2014

Crossrefs

Cf. A000975 (3 affinity groups).
Column k=4 of A261139.

Programs

  • Magma
    [(3^n-4*2^n+(-1)^n+6)/24: n in [4..30]]; // Bruno Berselli, Jun 13 2014
    
  • Mathematica
    Table[(3^n - 4 2^n + (-1)^n + 6)/24, {n, 4, 30}] (* Bruno Berselli, Jun 13 2014 *)
  • PARI
    for(n=4,50, print1(( 3^n - 4*2^n + (-1)^n + 6 )/24, ", ")) \\ G. C. Greubel, Oct 11 2017

Formula

a(n) - 3*a(n-1) = A000975(n-3).
From Bruno Berselli, Jun 13 2014: (Start)
G.f.: x^4/(1 - 5*x + 5*x^2 + 5*x^3 - 6*x^4).
a(n) = ( 3^n - 4*2^n + (-1)^n + 6 )/24. (End)
a(n) = 5*a(n-1) - 5*a(n-2) - 5*a(n-3) + 6*a(n-4). - Wesley Ivan Hurt, May 27 2021
a(n) = Sum_{i=0..n-1} Stirling2(i,3)*(-1)^(i+n-1). (See Peter Bala's original formula at A105794 dated Jul 10 2013.) - Igor Victorovich Statsenko, May 31 2024
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