cp's OEIS Frontend

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.

Previous Showing 11-20 of 32 results. Next

A075497 Stirling2 triangle with scaled diagonals (powers of 2).

Original entry on oeis.org

1, 2, 1, 4, 6, 1, 8, 28, 12, 1, 16, 120, 100, 20, 1, 32, 496, 720, 260, 30, 1, 64, 2016, 4816, 2800, 560, 42, 1, 128, 8128, 30912, 27216, 8400, 1064, 56, 1, 256, 32640, 193600, 248640, 111216, 21168, 1848, 72, 1
Offset: 1

Views

Author

Wolfdieter Lang, Oct 02 2002

Keywords

Comments

This is a lower triangular infinite matrix of the Jabotinsky type. See the D. E. Knuth reference given in A039692 for exponential convolution arrays.
The row polynomials p(n,x) := Sum_{m=1..n} a(n,m)x^m, n >= 1, have e.g.f. J(x; z)= exp((exp(2*z) - 1)*x/2) - 1.
Subtriangle of (0, 2, 0, 4, 0, 6, 0, 8, 0, 10, 0, 12, ...) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 13 2013
Also the inverse Bell transform of the double factorial of even numbers Product_ {k=0..n-1} (2*k+2) (A000165). For the definition of the Bell transform see A264428 and for cross-references A265604. - Peter Luschny, Dec 31 2015
This is the exponential Riordan array [exp(2*x), (exp(2*x) - 1)/2] belonging to the derivative subgroup of the exponential Riordan group. In the notation of Corcino, this is the triangle of (2, 2)-Stirling numbers of the second kind. A factorization of the array as an infinite product is given in the example section. - Peter Bala, Feb 20 2025

Examples

			Triangle begins:
  [1];
  [2,1];
  [4,6,1]; p(3,x) = x*(4 + 6*x + x^2).
  ...;
Triangle (0, 2, 0, 4, 0, 6, 0, 8, ...) DELTA (1, 0, 1, 0, 1, 0, 1, 0, ...) begins:
  1
  0,  1
  0,  2,   1
  0,  4,   6,   1
  0,  8,  28,  12,  1
  0, 16, 120, 100, 20, 1. - _Philippe Deléham_, Feb 13 2013
From _Peter Bala_, Feb 23 2025: (Start)
The array factorizes as
/ 1               \       /1             \ /1             \ /1            \
| 2    1           |     | 2   1          ||0  1           ||0  1          |
| 4    6   1       |  =  | 4   4   1      ||0  2   1       ||0  0  1       | ...
| 8   28  12   1   |     | 8  12   6  1   ||0  4   4  1    ||0  0  2  1    |
|16  120 100  20  1|     |16  32  24  8  1||0  8  12  6  1 ||0  0  4  4  1 |
|...               |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 2*x), x/(1 - 2*x)) = P^2, where P denotes Pascal's triangle. See A038207. Cf. A143494. (End)
		

Crossrefs

Row sums are A004211.

Programs

  • Maple
    with(combinat):
    b:= proc(n, i) option remember; expand(`if`(n=0, 1,
           `if`(i<1, 0, add(x^j*multinomial(n, n-i*j, i$j)/j!*add(
            binomial(i, 2*k), k=0..i/2)^j*b(n-i*j, i-1), j=0..n/i))))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n$2)):
    seq(T(n), n=1..12);  # Alois P. Heinz, Aug 13 2015
    # Alternatively, giving the triangle in the form displayed in the Example section:
    gf := exp(x*exp(z)*sinh(z)):
    X := n -> series(gf, z, n+2):
    Z := n -> n!*expand(simplify(coeff(X(n), z, n))):
    A075497_row := n -> op(PolynomialTools:-CoefficientList(Z(n), x)):
    seq(A075497_row(n), n=0..9); # Peter Luschny, Jan 14 2018
  • Mathematica
    Table[(2^(n - m)) StirlingS2[n, m], {n, 9}, {m, n}] // Flatten (* Michael De Vlieger, Dec 31 2015 *)
  • PARI
    for(n=1, 11, for(m=1, n, print1(2^(n - m) * stirling(n, m, 2),", ");); print();) \\ Indranil Ghosh, Mar 25 2017
  • Sage
    # uses[inverse_bell_transform from A265605]
    multifact_2_2 = lambda n: prod(2*k + 2 for k in (0..n-1))
    inverse_bell_matrix(multifact_2_2, 9) # Peter Luschny, Dec 31 2015
    

Formula

a(n, m) = (2^(n-m)) * Stirling2(n, m).
a(n, m) = (Sum_{p=0..m-1} A075513(m, p)*((p+1)*2)^(n-m))/(m-1)! for n >= m >= 1, else 0.
a(n, m) = 2*m*a(n-1, m) + a(n-1, m-1), n >= m >= 1, else 0, with a(n, 0) := 0 and a(1, 1)=1.
G.f. for m-th column: (x^m)/Product_{k=1..m}(1-2*k*x), m >= 1.
E.g.f. for m-th column: (((exp(2*x)-1)/2)^m)/m!, m >= 1.
The row polynomials in t are given by D^n(exp(x*t)) evaluated at x = 0, where D is the operator (1+2*x)*d/dx. Cf. A008277. - Peter Bala, Nov 25 2011
From Peter Bala, Jan 13 2018: (Start)
n-th row polynomial R(n,x)= x o x o ... o x (n factors), where o is the deformed Hadamard product of power series defined in Bala, section 3.1.
R(n+1,x)/x = (x + 2) o (x + 2) o...o (x + 2) (n factors).
R(n+1,x) = x*Sum_{k = 0..n} binomial(n,k)*2^(n-k)*R(k,x).
Dobinski-type formulas: R(n,x) = exp(-x/2)*Sum_{i >= 0} (2*i)^n* (x/2)^i/i!; 1/x*R(n+1,x) = exp(-x/2)*Sum_{i >= 0} (2 + 2*i)^n* (x/2)^i/i!. (End)

A105794 Inverse of a generalized Stirling number triangle of first kind.

Original entry on oeis.org

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

Views

Author

Paul Barry, Apr 20 2005

Keywords

Comments

Inverse of number triangle A105793. Row sums are the generalized Bell numbers A000296.
T(n,k)*(-1)^(n-k) gives the inverse Sheffer matrix of A094645. In the umbral notation (cf. Roman, p. 17, quoted in A094645) this is Sheffer for (1-t,-log(1-t)). - Wolfdieter Lang, Jun 20 2011
From Peter Bala, Jul 10 2013: (Start)
For a graph G and a positive integer k, the graphical Stirling number S(G;k) is the number of partitions of the vertex set of G into k nonempty independent sets (an independent set in G is a subset of the vertices, no two elements of which are adjacent). Omitting the first two rows and columns of this triangle produces the triangle of graphical Stirling numbers of cycles on n vertices (interpreting a 2-cycle as a single edge). In comparison, the classical Stirling numbers of the second kind, A008277, are the graphical Stirling numbers of path graphs on n vertices. See Galvin and Than. See also A227341.
This is the triangle of connection constants for expressing the polynomial sequence (x-1)^n in terms of the falling factorial polynomials, that is, (x-1)^n = Sum_{k = 0..n} T(n,k)*x_(k), where x_(k) = x*(x-1)*...*(x-k+1) denotes the falling factorial.
The row polynomials are a particular case of the Actuarial polynomials - see Roman 4.3.4. (End)

Examples

			The triangle starts with
  n=0:  1;
  n=1: -1,  1;
  n=2:  1, -1, 1;
  n=3: -1,  1, 0, 1;
  n=4:  1, -1, 1, 2, 1;
  n=5: -1,  1, 0, 5, 5, 1;
  ... - _Wolfdieter Lang_, Jun 20 2011
		

References

  • S. Roman, The umbral calculus, Pure and Applied Mathematics 111, Academic Press Inc., New York, 1984. Reprinted by Dover in 2005.

Crossrefs

Cf. A000296 (row sums), A105793 (matrix inverse), A227341.

Programs

  • Maple
    B:= Matrix(12,12,shape=triangular[lower],(n,k) -> combinat:-stirling1(n-1,k-1)+(n-1)*combinat:-stirling1(n-2,k-1)):
    A:= B^(-1):
    seq(seq(A[i,j],j=1..i),i=1..12); # Robert Israel, Jan 19 2015
    T := (n, k) -> add((-1)^(n - i)*binomial(n, i)*Stirling2(i, k), i=0..n):
    seq(seq(T(n, k), k=0..n), n=0..9);  # Peter Luschny, Feb 15 2025
  • Mathematica
    Table[Sum[(-1)^(n - i)*Binomial[n, i] StirlingS2[i, k], {i, 0, n}], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Oct 14 2019 *)

Formula

Term k in row n is given by {(-1)^(k+n) * (Sum_{j=0..k} (-1)^j * binomial(k,j) * (1-j)^n) / k! }; i.e., a finite difference. - Tom Copeland, Jun 05 2006
O.g.f. for row n = n-th finite difference of the Touchard (Bell) polynomials, T(x,j) and so the e.g.f. for these finite differences and therefore the sequence = exp{x*[exp(t)-1]-t} = exp{t*[T(x,.)-1]} umbrally. - Tom Copeland, Jun 05 2006
The e.g.f. A(x,t) = exp(x*(exp(t)-1)-t) satisfies the partial differential equation x*dA/dx - dA/dt = (1-x)*A.
Recurrence relation: T(n+1,k) = T(n,k-1) + (k-1)*T(n,k).
Let f(x) = ((x-1)/x)*exp(x). For n >= 1, the n-th row polynomial R(n,x) = x*exp(-x)*(x*d/dx)^(n-1)(f(x)) and satisfies the recurrence equation R(n+1,x) = (x-1)*R(n,x) + x*R'(n,x). - Peter Bala, Oct 28 2011
Let f(x) = exp(exp(x)-x). Then R(n,exp(x)) = 1/f(x)*(d/dx)^n(f(x)). Similar formulas hold for A008277, A039755, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012
From Peter Bala, Jul 10 2013: (Start)
T(n,k) = Sum_{i = 0..n-1} (-1)^i*Stirling2(n-1-i,k-1), for n >= 1, k >= 1.
The k-th column o.g.f. is (1/(1+x))*x^k/((1-x)*(1-2*x)*...*(1-(k-1)*x)) (the empty product occurring in the denominator when k = 0 and k = 1 is taken equal to 1).
Dobinski-type formula for the row polynomials: R(n,x) = exp(-x)*Sum_{k >= 0} (k-1)^n*x^k/k!.
Sum_{k = 0..n} binomial(n,k)*R(k,x) = n-th Bell polynomial (n-th row polynomial of A048993). (End)
From Peter Bala, Jan 13 2014: (Start)
T(n,k) = Sum_{i = 0..n} (-1)^(n-i)*binomial(n,i)*Stirling2(i,k).
T(n,k) = Sum_{i = 0..n} (-2)^(n-i)*binomial(n,i)*Stirling2(i+1,k+1).
Matrix product P^(-1) * S = P^(-2) * S1, where P = A007318, S = A048993 and S1 = A008277. (End)
From Werner Schulte, Feb 15 2025: (Start)
Conjecture 1: Sum_{k=0..n} (-1)^(n-k) * T(n, k) * (k+m)! / m! = (m+2)^n for m >= 0.
Conjecture 2: (-1)^n - B(n) = Sum_{k=1..n} (-1)^k * T(n, k) * (k-1)! / (k+1) where B(n) = B(n, 0) is n-th Bernoulli number. (End)

A143491 Unsigned 2-Stirling numbers of the first kind.

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 24, 26, 9, 1, 120, 154, 71, 14, 1, 720, 1044, 580, 155, 20, 1, 5040, 8028, 5104, 1665, 295, 27, 1, 40320, 69264, 48860, 18424, 4025, 511, 35, 1, 362880, 663696, 509004, 214676, 54649, 8624, 826, 44, 1, 3628800, 6999840, 5753736, 2655764
Offset: 2

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

Essentially the same as A136124 but with column numbers differing by one. See A049444 for a signed version of this array. The unsigned 2-Stirling numbers of the first kind count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the elements 1 and 2 belong to distinct cycles. This is the particular case r = 2 of the unsigned r-Stirling numbers of the first kind, which count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the numbers 1, 2, ..., r belong to distinct cycles. The case r = 1 gives the usual unsigned Stirling numbers of the first kind, abs(A008275); for other cases see A143492 (r = 3) and A143493 (r = 4). The corresponding 2-Stirling numbers of the second kind can be found in A143494.
In general, the lower unitriangular array of unsigned r-Stirling numbers of the first kind (with suitable offsets in the row and column indexing) equals the matrix product St1 * P^(r-1), where St1 is the array of unsigned Stirling numbers of the first kind, abs(A008275) and P is Pascal's triangle, A007318. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For details of the related r-Lah numbers see A143497.
This sequence also represents the number of permutations in the alternating group An of length k, where the length is taken with respect to the generators set {(12)(ij)}. For a bijective proof of the relation between these numbers and the 2-Stirling numbers of the first kind see the Rotbart link. - Aviv Rotbart, May 05 2011
With offset n=0,k=0 : triangle T(n,k), read by rows, given by [2,1,3,2,4,3,5,4,6,5,...] DELTA [1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 29 2011
With offset n=0 and k=0, this is the Sheffer triangle (1/(1-x)^2,-log(1-x)) (in the umbral notation of S. Roman's book this would be called Sheffer for (exp(-2*t),1-exp(-t))). See the e.g.f. given below. Compare also with the e.g.f. for the signed version A049444. - Wolfdieter Lang, Oct 10 2011
Reversed rows correspond to the Betti numbers of the moduli space M(0,n) of smooth Riemann surfaces (see Murri link). - Tom Copeland, Sep 19 2012

Examples

			Triangle begins
  n\k|.....2.....3.....4.....5.....6.....7
  ========================================
  2..|.....1
  3..|.....2.....1
  4..|.....6.....5.....1
  5..|....24....26.....9.....1
  6..|...120...154....71....14.....1
  7..|...720..1044...580...155....20.....1
  ...
T(4,3) = 5. The permutations of {1,2,3,4} with 3 cycles such that 1 and 2 belong to different cycles are: (1)(2)(3 4), (1)(3)(24), (1)(4)(23), (2)(3)(14) and (2)(4)(13). The remaining possibility (3)(4)(12) is not allowed.
From _Aviv Rotbart_, May 05 2011: (Start)
Example of the alternating group permutations numbers:
Triangle begins
  n\k|.....0.....1.....2.....3.....4.....5.....6.....7
  ====================================================
  2..|.....1
  3..|.....1.....2
  4..|.....1.....5.....6
  5..|.....1.....9....26....24
  6..|.....1....14....71...154...120
  7..|.....1....20...155...580..1044..720
A(n,k) = number of permutations in An of length k, with respect to the generators set {(12)(ij)}. For example, A(2,0)=1 (only the identity is there), for A4, the generators are {(12)(13),(12)(14),(12,23),(12)(24),(12)(34)}, thus we have A(4,1)=5 (exactly 5 generators), the permutations of length 2 are:
   (12)(13)(12)(13) = (312)
   (12)(13)(12)(14) = (41)(23)
   (12)(13)(12)(24) = (432)(1)
   (12)(13)(12)(34) = (342)(1)
   (12)(23)(12)(24) = (13)(24)
   (12)(14)(12)(14) = (412)(3)
Namely, A(4,2)=6. Together with the identity [=(12)(12), of length 0. therefore A(4,0)=1] we have 12 permutations, comprising all A4 (4!/2=12). (End)
		

Crossrefs

Cf. A001705 - A001709 (column 3..7), A001710 (row sums), A008275, A049444 (signed version), A136124, A143492, A143493, A143494, A143497.
Cf. A094638.

Programs

  • Maple
    with combinat: T := (n, k) -> (n-2)! * add((n-j-1)*abs(stirling1(j,k-2))/j!,j = k-2..n-2): for n from 2 to 10 do seq(T(n, k), k = 2..n) end do;
  • Mathematica
    t[n_, k_] := (n-2)!*Sum[(n-j-1)*Abs[StirlingS1[j, k-2]]/j!, {j, k-2, n-2}]; Table[t[n, k], {n, 2, 11}, {k, 2, n}] // Flatten (* Jean-François Alcover, Apr 16 2013, after Maple *)

Formula

T(n,k) = (n-2)! * Sum_{j = k-2 .. n-2} (n-j-1)*|stirling1(j,k-2)|/j!.
Recurrence relation: T(n,k) = T(n-1,k-1) + (n-1)*T(n-1,k) for n > 2, with boundary conditions: T(n,1) = T(1,n) = 0, for all n; T(2,2) = 1; T(2,k) = 0 for k > 2.
Special cases: T(n,2) = (n-1)!; T(n,3) = (n-1)!*(1/2 + 1/3 + ... + 1/(n-1)).
T(n,k) = Sum_{2 <= i_1 < ... < i_(n-k) < n} (i_1*i_2*...*i_(n-k)). For example, T(6,4) = Sum_{2 <= i < j < 6} (i*j) = 2*3 + 2*4 + 2*5 + 3*4 + 3*5 + 4*5 = 71.
Row g.f.: Sum_{k = 2..n} T(n,k)*x^k = x^2*(x+2)*(x+3)*...*(x+n-1).
E.g.f. for column (k+2): Sum_{n>=k} T(n+2,k+2)*x^n/n! = (1/k!)*(1/(1-x)^2)*(log(1/(1-x)))^k.
E.g.f.: (1/(1-t))^(x+2) = Sum_{n>=0} Sum_{k = 0..n} T(n+2,k+2)*x^k*t^n/n! = 1 + (2+x)*t/1! + (6+5*x+x^2)*t^2/2! + ... .
This array is the matrix product St1 * P, where St1 denotes the lower triangular array of unsigned Stirling numbers of the first kind, abs(A008275) and P denotes Pascal's triangle, A007318. The row sums are n!/2 ( A001710 ). The alternating row sums are (n-2)!.
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-1,i) = |f(n,i,2)|, for n=1,2,...; i=0..n. - Milan Janjic, Dec 21 2008
From Gary W. Adamson, Jul 19 2011: (Start)
n-th row of the triangle = top row of M^(n-2), M = a reversed variant of the (1,2) Pascal triangle (Cf. A029635); as follows:
2, 1, 0, 0, 0, 0, ...
2, 3, 1, 0, 0, 0, ...
2, 5, 4, 1, 0, 0, ...
2, 7, 9, 5, 1, 0, ...
... (End)
The reversed, row polynomials of this entry multiplied by (1+x) are the row polynomials of A094638. E.g., (1+x)(1+5x+6x^2) = (1+6x+11x^2+6x^3). - Tom Copeland, Dec 11 2016

A143496 4-Stirling numbers of the second kind.

Original entry on oeis.org

1, 4, 1, 16, 9, 1, 64, 61, 15, 1, 256, 369, 151, 22, 1, 1024, 2101, 1275, 305, 30, 1, 4096, 11529, 9751, 3410, 545, 39, 1, 16384, 61741, 70035, 33621, 7770, 896, 49, 1, 65536, 325089, 481951, 305382, 95781, 15834, 1386, 60, 1, 262144, 1690981, 3216795
Offset: 4

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 4 of the r-Stirling numbers of the second kind. The 4-Stirling numbers of the second kind count the ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1, 2, 3 and 4 belong to distinct subsets. For remarks on the general case see A143494 (r = 2). The corresponding array of 4-Stirling numbers of the first kind is A143493. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For 4-Lah numbers refer to A143499.
From Wolfdieter Lang, Sep 29 2011: (Start)
T(n,k) = S(n,k,4), n >= k >= 4, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column k from (A20) with k->4, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(4*x),exp(x)-1) with e.g.f. of column no. m >= 0: exp(4*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|.....4.....5.....6.....7.....8.....9
========================================
4..|.....1
5..|.....4.....1
6..|....16.....9.....1
7..|....64....61....15.....1
8..|...256...369...151....22.....1
9..|..1024..2101..1275...305....30.....1
...
T(6,5) = 9. The set {1,2,3,4,5,6} can be partitioned into five subsets such that 1, 2, 3 and 4 belong to different subsets in 9 ways: {{1,5}{2}{3}{4}{6}}, {{1,6}{2}{3}{4}{5}}, {{2,5}{1}{3}{4}{6}}, {{2,6}{1}{3}{4}{5}}, {{3,5}{1}{2}{4}{6}}, {{3,6}{1}{2}{4}{5}}, {{4,5}{1}{2}{3}{6}}, {{4,6}{1}{2}{3}{5}} and {{5,6}{1}{2}{3}{4}}.
		

Crossrefs

Cf. A003468 (column 7), A005060 (column 5), A008277, A016103 (column 6), A045379 (row sums), A049459 (matrix inverse), A143493, A143494, A143495, A143499.

Programs

  • Maple
    with combinat: T := (n, k) -> 1/(k-4)!*add ((-1)^(k-i)*binomial(k-4,i)*(i+4)^(n-4),i = 0..k-4): for n from 4 to 13 do seq(T(n, k), k = 4..n) end do;
  • Mathematica
    t[n_, k_] := StirlingS2[n, k] - 6*StirlingS2[n-1, k] + 11*StirlingS2[n-2, k] - 6*StirlingS2[n-3, k]; Flatten[ Table[ t[n, k], {n, 4, 13}, {k, 4, n}]] (* Jean-François Alcover, Dec 02 2011 *)

Formula

T(n+4,k+4) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*(i+4)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - 6*Stirling2(n-1,k) + 11*Stirling2(n-2,k) - 6*Stirling2(n-3,k) for n,k >= 4.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 4 with boundary conditions: T(n,3) = T(3,n) = 0 for all n; T(4,4) = 1; T(4,k) = 0 for k > 4. Special cases: T(n,4) = 4^(n-4); T(n,5) = 5^(n-4) - 4^(n-4).
E.g.f. (k+4)-th column (with offset 4): (1/k!)*exp(4*x)*(exp(x)-1)^k.
O.g.f. k-th column: Sum_{n>=k} T(n,k)*x^n = x^k/((1-4*x)*(1-5*x)*...*(1-k*x)).
E.g.f.: exp(4*t + x*(exp(t)-1)) = Sum_{n = 0..infinity} Sum_(k = 0..n) T(n+4,k+4)*x^k*t^n/n! = Sum_{n = 0..infinity} B_n(4;x)*t^n/n! = 1 + (4+x)*t/1! + (16+9*x+x^2)*t^2/2! + ..., where the row polynomials, B_n(4;x) := Sum_{k = 0..n} T(n+4,k+4)*x^k, may be called the 4-Bell polynomials.
Dobinski-type identities: Row polynomial B_n(4;x) = exp(-x)*Sum_{i = 0..infinity} (i+4)^n*x^i/i!; Sum_{k = 0..n} k!*T(n+4,k+4)*x^k = Sum_{i = 0..infinity} (i+4)^n*x^i/(1+x)^(i+1).
The T(n,k) are the connection coefficients between the falling factorials and the shifted monomials (x+4)^(n-4). For example, 16 + 9*x + x*(x-1) = (x+4)^2; 64 + 61*x + 15*x*(x-1) + x*(x-1)*(x-2) = (x+4)^3.
This array is the matrix product P^3 * S, where P denotes Pascal's triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
The inverse array is A049459, the signed 4-Stirling numbers of the first kind.
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^(-4)*E^n*x^4 = Sum_{k = 0..n} T(n+4,k+4)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k=4..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_4(x) = x^4. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 4-Eulerian numbers E_4(n,j) := A144698(n,j): T(n,k) = 4!/k!*Sum_{j = n-k..n-4} E_4(n,j)*binomial(j,n-k) for n >= k >= 4.
(End)

A038719 Triangle T(n,k) (0 <= k <= n) giving number of chains of length k in partially ordered set formed from subsets of n-set by inclusion.

Original entry on oeis.org

1, 2, 1, 4, 5, 2, 8, 19, 18, 6, 16, 65, 110, 84, 24, 32, 211, 570, 750, 480, 120, 64, 665, 2702, 5460, 5880, 3240, 720, 128, 2059, 12138, 35406, 57120, 52080, 25200, 5040, 256, 6305, 52670, 213444, 484344, 650160, 514080, 221760, 40320, 512, 19171
Offset: 0

Views

Author

N. J. A. Sloane, May 02 2000

Keywords

Comments

The relation of this triangle to A143494 given in the Formula section leads to the following combinatorial interpretation: T(n,k) gives the number of partitions of the set {1,2,...,n+2} into k + 2 blocks where 1 and 2 belong to two distinct blocks and the remaining k blocks are labeled from a fixed set of k labels. - Peter Bala, Jul 10 2014
Also, the number of distinct k-level fuzzy subsets of a set consisting of n elements ordered by set inclusion. - Rajesh Kumar Mohapatra, Mar 16 2020

Examples

			Triangle begins
   1;
   2,   1;
   4,   5,   2;
   8,  19,  18,   6;
  16,  65, 110,  84,  24;
  ...
From _Peter Bala_, Feb 02 2022: (Start)
Table of successive differences of k^2 starting at k = 2
4   9   16
  5   7
    2
gives [4, 5, 2] as row 2 of this triangle.
Table of successive differences of k^3 starting at k = 2
8   27   64   125
  19   37   61
     18   24
        6
gives [8, 19, 8, 6] as row 3 of this triangle. (End)
		

Crossrefs

Row sums give A007047. Columns give A000079, A001047, A038721. Next-to-last diagonal gives A038720.
Diagonal gives A000142. - Rajesh Kumar Mohapatra, Mar 16 2020

Programs

  • Haskell
    a038719 n k = a038719_tabl !! n !! k
    a038719_row n = a038719_tabl !! n
    a038719_tabl = iterate f [1] where
       f row = zipWith (+) (zipWith (*) [0..] $ [0] ++ row)
                           (zipWith (*) [2..] $ row ++ [0])
    -- Reinhard Zumkeller, Jul 08 2012
  • Maple
    T:= proc(n, k) option remember;
          `if` (n=0, `if`(k=0, 1, 0), k*T(n-1, k-1) +(k+2)*T(n-1, k))
        end:
    seq(seq(T(n, k), k=0..n), n=0..10); # Alois P. Heinz, Aug 02 2011
  • Mathematica
    t[n_, k_] := Sum[ (-1)^(k-i)*Binomial[k, i]*(2+i)^n, {i, 0, k}]; Flatten[ Table[ t[n, k], {n, 0, 9}, {k, 0, n}]] (* Jean-François Alcover, after Pari *)
  • PARI
    T(n,k)=sum(i=0,k,(-1)^(k-i)*binomial(k,i)*(2+i)^n)
    

Formula

T(n, k) = Sum_{j=0..k} (-1)^j*C(k, j)*(k+2-j)^n.
T(n+1, k) = k*T(n, k-1) + (k+2)*T(n, k), T(0,0) = 1, T(0,k) = 0 for k>0.
E.g.f.: exp(2*x)/(1+y*(1-exp(x))). - Vladeta Jovovic, Jul 21 2003
A038719 as a lower triangular matrix is the binomial transform of A028246. - Gary W. Adamson, May 15 2005
Binomial transform of n-th row = 2^n + 3^n + 4^n + ...; e.g., binomial transform of [8, 19, 18, 6] = 2^3 + 3^3 + 4^3 + 5^3 + ... = 8, 27, 64, 125, ... - Gary W. Adamson, May 15 2005
From Peter Bala, Jul 09 2014: (Start)
T(n,k) = k!*( Stirling2(n+2,k+2) - Stirling2(n+1,k+2) ).
T(n,k) = k!*A143494(n+2,k+2).
n-th row polynomial = 1/(1 + x)*( sum {k >= 0} (k + 2)^n*(x/(1 + x))^k ). Cf. A028246. (End)
The row polynomials have the form (2 + x) o (2 + x) o ... o (2 + x), where o denotes the black diamond multiplication operator of Dukes and White. See example E12 in the Bala link. - Peter Bala, Jan 18 2018
Z(P,m) = Sum_{k=0..n} T(n,k)Binomial(m-2,k) = m^n, the zeta polynomial of the poset B_n. Each length m multichain from 0 to 1 in B_n corresponds to a function from [n] into [m]. - Geoffrey Critzer, Dec 25 2020
The entries in row n are the first terms in a table of the successive differences of the sequence [2^n, 3^n, 4^n, ...]. Examples are given below. - Peter Bala, Feb 02 2022

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), May 09 2000

A193685 5-Stirling numbers of the second kind.

Original entry on oeis.org

1, 5, 1, 25, 11, 1, 125, 91, 18, 1, 625, 671, 217, 26, 1, 3125, 4651, 2190, 425, 35, 1, 15625, 31031, 19981, 5590, 740, 45, 1, 78125, 201811, 170898, 64701, 12250, 1190, 56, 1, 390625, 1288991, 1398097, 688506, 174951, 24150, 1806, 68, 1, 1953125, 8124571, 11075670, 6906145, 2263065, 416451, 44016, 2622, 81, 1, 9765625, 50700551, 85654261, 66324830, 27273730, 6427575, 900627, 75480, 3675, 95, 1
Offset: 0

Views

Author

Wolfdieter Lang, Oct 06 2011

Keywords

Comments

This is the lower triangular Sheffer matrix (exp(5*x),exp(x)-1). For Sheffer matrices see the W. Lang link under A006232 with references, and the rules for the conversion to the umbral notation of S. Roman's book.
The general case is Sheffer (exp(r*x),exp(x)-1), r=0,1,..., corresponding to r-Stirling2 numbers with row and column offsets 0. See the Broder link for r-Stirling2 numbers with offset [r,r].
a(n,m), n >= m >= 0, gives the number of partitions of the set {1.2....,n+5} into m+5 nonempty distinct subsets such that 1,2,3,4 and 5 belong to distinct subsets.
a(n,m) appears in the following normal ordering of Bose operators a and a* satisfying the Lie algebra [a,a*]=1: (a*a)^n (a*)^5 = Sum_{m=0..n} a(n,m)*(a*)^(5+m)*a^m, n >= 0. See the Mikhailov papers, where a(n,m) = S(n+5,m+5,5).
With a->D=d/dx and a*->x we also have
(xD)^n x^5 = Sum_{m=0..n} a(n,m)*x^(5+m)*D^m, n >= 0.

Examples

			n\m  0    1    2   3  4  5 ...
0    1
1    5    1
2   25   11    1
3  125   91   18   1
4  625  671  217  26  1
5 3125 4651 2190 425 35  1
...
5-restricted S2: a(1,0)=5 from 1,6|2|3|4|5, 2,6|1|3|4|5,
3,6|1|2|4|5, 4,6|1|2|3|5 and 5,6|1|2|3|4.
Recurrence: a(4,2) = (5+2)*a(3,2)+ a(3,1) = 7*18 + 91 = 217.
Normal ordering (n=1): (xD)^1 x^5 = Sum_{m=0..1} a(1,m)*x^(5+m)*D^m = 5*x^5 + 1*x^6*D.
a(2,1) = Sum_{j=0..1} S1(5,5-j)*S2(7-j,6) = 1*21 - 10*1 = 11.
		

Crossrefs

Cf. A196834 (row sums), A196835 (alternating row sums).
Columns: A000351 (m=0), A005062 (m=1), A019757 (m=2), A028165 (m=3), ...

Programs

  • Mathematica
    a[n_, m_] := Sum[ StirlingS1[5, 5-j]*StirlingS2[n+5-j, m+5], {j, 0, Min[5, n-m]}]; Flatten[ Table[ a[n, m], {n, 0, 10}, {m, 0, n}] ] (* Jean-François Alcover, Dec 02 2011, after Wolfdieter Lang *)

Formula

E.g.f. of row polynomials s(n,x):=Sum_{m=0..n} a(n,m)*x^m: exp(5*z + x(exp(z)-1)).
E.g.f. of column no. m (with leading zeros):
exp(5*x)*((exp(x)-1)^m)/m!, m >= 0 (Sheffer).
O.g.f. of column no. m (without leading zeros):
1/Product_{j=0..m} (1-(5+j)*x), m >= 0. (Compute the first derivative of the column e.g.f. and compare its Laplace transform with the partial fraction decomposition of the o.g.f. x^(m-1)/Product_{j=0..m} (1-(5+j)*x). This works for every r-restricted Stirling2 triangle.)
Recurrence: a(n,m) = (5+m)*a(n-1,m) + a(n-1,m-1), a(0,0)=1, a(n,m)=0 if n < m, a(n,-1)=0.
a(n,m) = Sum_{j=0..min(5,n-m)} S1(5,5-j)*S2(n+5-j,m+5), n >= m >= 0, with S1 and S2 the Stirling1 and Stirling2 numbers A008275 and A048993, respectively (see the Mikailov papers).
Dobinski-type formula for the row polynomials: R(n,x) = exp(-x)*Sum_{k>=0} k*(4+k)^(n-1)*x^(k-1)/k!. - Peter Bala, Jun 23 2014

A025211 Expansion of 1/((1-2x)(1-3x)(1-4x)(1-5x)).

Original entry on oeis.org

1, 14, 125, 910, 5901, 35574, 204205, 1132670, 6129101, 32566534, 170691885, 885423630, 4556561101, 23305343894, 118631189165, 601616805790, 3042056477901, 15346559343654, 77279066272045, 388583895311150, 1951684190615501, 9793511186181814, 49108010998116525
Offset: 0

Views

Author

Keywords

Comments

This gives the fourth column of the Sheffer triangle A143494 (2-restricted Stirling2 numbers). See the e.g.f. given below, and comments on the general case under A193685. - Wolfdieter Lang, Oct 08 2011

Crossrefs

Programs

Formula

If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*stirling2(k,j)*x^(m-k) then a(n-3) = (-1)^(n-1)*f(n,3,-5), (n >= 3). - Milan Janjic, Apr 26 2009
If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*stirling2(k,j)*x^(m-k) then a(n-3) = f(n,3,2), (n >= 3). - Milan Janjic, Apr 26 2009
a(n) = 3^(n+3)/2 - 2*4^(n+2) - 2^(n+2)/3 + 5^(n+3)/6. - R. J. Mathar, Mar 22 2011
O.g.f.: 1/((1-2*x)*(1-3*x)*(1-4*x)*(1-5*x)).
E.g.f.: (d^3/dx^3) (exp(2*x)*((exp(x)-1)^3)/3!). See the Sheffer comment given above. - Wolfdieter Lang, Oct 08 2011

A074051 For each n there are uniquely determined numbers a(n) and b(n) and a polynomial p_n(x) such that for all integers m we have Sum_{i=1..m}i^n(i+1)! = a(n)*Sum_{i=1..m} (i+1)! + p_n(m)*(m+2)! + b(n). The sequence b(n) is A074052.

Original entry on oeis.org

1, -1, 0, 3, -7, 0, 59, -217, 146, 2593, -15551, 32802, 160709, -1856621, 7971872, 1299951, -287113779, 2262481448, -7275903849, -36989148757, 698330745002, -4867040141851, 10231044332629, 184216198044034, -2679722886596295, 17971204188130391, -17976259717948832
Offset: 0

Views

Author

Jan Fricke, Aug 14 2002

Keywords

Comments

If a(n)=0 then Sum_{i>=1}i^n(i+1)! = b(n) in the p-adic numbers. The only known numbers n with a(n)=0 are 2 and 5.
a(n)*(-1)^n gives the alternating row sums of the Sheffer triangle A143494 (2-restricted Stirling2). - Wolfdieter Lang, Oct 06 2011

Examples

			a(2)=0 because Sum_{i=1..m}i^2(i+1)! = (m-1)(m+2)!+2.
a(3)=3 because Sum_{i=1..m}i^3(i+1)! = 3*Sum_{i=1..m}(i+1)!+(m^2-m-1)(m+2)!+2.
		

Crossrefs

Programs

  • Maple
    alias(S2 = combinat[stirling2]);
    A074051 := proc(n) local k;
    1 + add((-1)^(n+k) * (S2(n+1, k+1) - S2(n+2, k+1)), k = 0..n) end:
    seq(A074051(i), i = 0..26); # Peter Luschny, Apr 17 2011
  • Mathematica
    A[a_] := Module[{p, k}, p[n_] = 0; For[k = a - 1, k >= 0, k--, p[n_] = Expand[p[n] + n^k Coefficient[n^a - (n + 2)p[n] + p[n - 1], n^(k + 1)]] ]; Expand[n^a - (n + 2)p[n] + p[n - 1]] ]
    (* Second program: *)
    a[n_] := (-1)^n (BellB[n+2, -1] - BellB[n+1, -1]);
    Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jun 21 2018, after Peter Luschny *)
  • Python
    from itertools import accumulate
    def A074051_list(size):
        if size < 1: return []
        L, accu = [], [1]
        for n in range(size-1):
            accu = list(accumulate([-accu[-1]] + accu))
            L.append(-(-1)**n*accu[-2])
        return L
    print(A074051_list(28)) # Peter Luschny, Apr 25 2016

Formula

From Vladeta Jovovic, Jan 27 2005: (Start)
Second inverse binomial transform of A000587.
E.g.f.: exp(1 - 2*x - exp(-x)).
G.f.: Sum_{k >= 0}((x/(1+2*x))^k/Product_{l=0..k}(1 + l*x/(1+2*x)))/(1+2*x).
a(n) = Sum_{k=0..n} (-1)^(n-k)*(k^2-3*k+1)*Stirling2(n, k). (End)
a(n) = (-1)^n*(A000587(n+2)-A000587(n+1)). - Peter Luschny, Apr 17 2011
From Sergei N. Gladkovskii, Sep 28 2012 to Apr 22 2013: (Start)
Continued fractions:
G.f.: 1/U(0) where U(k)= x*k + 1 + x + x^2*(k+1)/U(k+1).
G.f.: -1/U(0) where U(k)= -x*k - 1 - x + x^2*(k+1)/U(k+1).
G.f.: 1/(U(0) - x) where U(k)= 1 + x + x*(k+1)/(1 - x/U(k+1)).
G.f.: 1/(U(0) + x) where U(k)= 1 + x*(2*k+1) - x*(k+1)/(1 + x/U(k+1)).
G.f.: 1/G(0) where G(k)= 1 + 2*x/(1 + 1/(1 + 2*x*(k+1)/G(k+1))).
G.f.: 1 - 2*x/(G(0) + 2*x) where G(k)= 1 + 1/(1 + 2*x*(k+1)/(1 + 2*x/G(k+1))).
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1+k*x+2*x)/(1-x/(x-1/G(k+1))).
G.f.: (G(0)-2-2*x)/x^2 where G(k) = 1 + 1/(1+k*x)/(1-x/(x+1/G(k+1) )).
G.f.: (S-2-2*x)/x^2 where S = sum(k>=0, (2 + x*k)*x^k/prod(i=0..k, (1+x*i))).
G.f.: (G(0)-2)/x where G(k) = 1 + 1/(1+k*x+x)/(1-x/(x+1/G(k+1))).
G.f.: (1+x)/x/Q(0) - 1/x, where Q(k)= 1 + x - x/(1 + x*(k+1)/Q(k+1)). (End)
a(n) = exp(1) * (-1)^n * Sum_{k>=0} (-1)^k * (k+2)^n / k!. - Ilya Gutkovskiy, Sep 02 2021

Extensions

More terms from Vladeta Jovovic, Jan 27 2005

A330804 Number of chains in partitions of [n] ordered by refinement.

Original entry on oeis.org

1, 1, 3, 15, 127, 1743, 36047, 1051039, 41082783, 2073110239, 131183712063, 10171782421727, 948427290027807, 104693416370374783, 13502772386271932927, 2011983769934772172799, 343000542276546601893439, 66334607666382842941084991, 14444628785932359077548728255, 3518072269888902413311442552511
Offset: 0

Views

Author

S. R. Kannan, Rajesh Kumar Mohapatra, Jan 01 2020

Keywords

Comments

Also the number of fuzzy equivalence matrices of order n.
Number of chains of equivalence relations on a set of n-elements.
Number of chains in Stirling numbers of the second kind.
Number of chains in the unordered partition of {1,...,n}.

Examples

			Consider the set S = {1, 2, 3}. The a(3) = 5+ 7+ 3 = 15 in the lattice of set partitions of {1,2,3}:
{{1},{2},{3}}  {{1},{2},{3}} < {{1,2},{3}}  {{1},{2},{3}} < {{1,2},{3}} < {{1,2,3}}
{{1,2},{3}}    {{1},{2},{3}} < {{1,3},{2}}  {{1},{2},{3}} < {{1,3},{2}} < {{1,2,3}}
{{1,3},{2}}    {{1},{2},{3}} < {{1},{2,3}}  {{1},{2},{3}} < {{1},{2,3}} < {{1,2,3}}
{{1},{2,3}}    {{1},{2},{3}} < {{1,2,3}}
{{1,2,3}}      {{1,2},{3}} < {{1,2,3}}
               {{1,3},{2}} < {{1,2,3}}
               {{1},{2,3}} < {{1,2,3}}
		

Crossrefs

Programs

  • Maple
    b:= proc(n, k, t) option remember; `if`(k<0, 0, `if`({n, k}={0}, 1,
          add(`if`(k=1, 1, b(v, k-1, 1))*Stirling2(n, v), v=k..n-t)))
        end:
    a:= n-> add(b(n, k, 0), k=0..n):
    seq(a(n), n=0..20);  # Alois P. Heinz, Feb 07 2020
    # second Maple program:
    a:= proc(n) option remember; uses combinat;
          bell(n) + add(stirling2(n, i)*a(i), i=1..n-1)
        end:
    seq(a(n), n=0..20);  # Alois P. Heinz, Sep 03 2020
  • Mathematica
    b[n_, k_, t_] := b[n, k, t] = If[k < 0, 0, If[Union@{n, k} == {0}, 1, Sum[If[k == 1, 1, b[v, k - 1, 1]]*StirlingS2[n, v], {v, k, n - t}]]];
    a[n_] := Sum[b[n, k, 0], {k, 0, n}];
    a /@ Range[0, 20] (* Jean-François Alcover, Feb 08 2020, after Alois P. Heinz *)
  • PARI
    b(n, k, t) = {if (k < 0, return(0)); if ((n==0) && (k==0), return (1)); sum(v = k, n - t, if (k==1, 1, b(v, k-1, 1))*stirling(n, v, 2));}
    a(n) = sum(k=0, n, b(n, k, 0);); \\ Michel Marcus, Feb 08 2020

Formula

a(n) = Sum_{k=0..n} A331955(n,k).
a(n) = Bell(n) + Sum_{i=1..n-1} Stirling2(n,i)*a(i). - Alois P. Heinz, Sep 03 2020
a(n) ~ A086053 * n!^2 / (2^(n-2) * log(2)^n * n^(1 + log(2)/3)). - Vaclav Kotesovec, Jul 01 2025
a(n) = 2 * A331957(n) - 1 = 4 * A005121(n) - 1 for n > 1. - Rajesh Kumar Mohapatra, Jul 01 2025

Extensions

More terms from Michel Marcus, Feb 07 2020

A232472 2-Fubini numbers.

Original entry on oeis.org

2, 10, 62, 466, 4142, 42610, 498542, 6541426, 95160302, 1520385010, 26468935022, 498766780786, 10114484622062, 219641848007410, 5085371491003502, 125055112347154546, 3255163896227709422, 89416052656071565810, 2584886208925055791982, 78447137202259689678706, 2493719594804686310662382
Offset: 2

Views

Author

N. J. A. Sloane, Nov 27 2013

Keywords

Examples

			G.f.: 2*x^2 + 10*x^3 + 62*x^4 + 466*x^5 + 4142*x^6 + 42610*x^7 + 498542*x^8 + ...
		

Crossrefs

Programs

  • Magma
    r:=2; r_Fubini:=func;
    [r_Fubini(n, r): n in [r..22]]; // Bruno Berselli, Mar 30 2016
  • Maple
    # r-Stirling numbers of second kind (e.g., A008277, A143494, A143495):
    T := (n,k,r) -> (1/(k-r)!)*add ((-1)^(k+i+r)*binomial(k-r,i)*(i+r)^(n-r),i = 0..k-r):
    # r-Bell numbers (e.g. A000110, A005493, A005494):
    B := (n,r) -> add(T(n,k,r),k=r..n);
    SB := r -> [seq(B(n,r),n=r..30)];
    SB(2);
    # r-Fubini numbers (e.g., A000670, A232472, A232473, A232474):
    F := (n,r) -> add((k)!*T(n,k,r),k=r..n);
    SF := r -> [seq(F(n,r),n=r..30)];
    SF(2);
  • Mathematica
    Rest[max=20; t=Sum[n^(n - 1) x^n / n!, {n, 1, max}]; 2 Range[0, max]!CoefficientList[Series[D[1 / (1 - y (Exp[x] - 1)), y] /.y->1, {x, 0, max}], x]] (* Vincenzo Librandi Jan 03 2016 *)
    Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i+k+r)*(i+r)^(n-r)/(i!*(k-i-r)!), {i, 0, k-r}], {k, r, n}]; Table[Fubini[n, 2], {n, 2, 22}] (* Jean-François Alcover, Mar 30 2016 *)

Formula

Let A(x) be the g.f. A232472, B(x) the g.f. A000670, then A(x) = (1-x)*B(x) - 1. - Sergei N. Gladkovskii, Nov 29 2013
a(n) = Sum_{k>=2} T_k*k^(n-2)/2^k where T_k is the (k-1)-st triangular number (i.e., T_k = k*(k-1)/2). - Derek Orr, Jan 01 2016
a(n) = 2*A069321(n-1). - Vincenzo Librandi, Jan 03 2016, corrected by Vaclav Kotesovec, Jul 01 2018
a(n) ~ n! / (2 * (log(2))^(n+1)). - Vaclav Kotesovec, Jul 01 2018
From Peter Bala, Dec 08 2020: (Start)
a(n+2) = Sum_{k = 0..n} (k+2)!/k!*( Sum{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i+2)^n ).
a(n+2) = Sum_{k = 0..n} 2^(n-k)*binomial(n,k)*( Sum_{i = 0..k} Stirling2(k,i)*(i+2)! ).
a(n) = 2*A069321(n-1) = A000670(n) - A000670(n-1).
a(n+1)= (1/2)*Sum_{k = 0..n} binomial(n,k)*A000670(k+1) for n >= 1.
E.g.f. with offset 0: 2*exp(2*z)/(2 - exp(z))^3 = 2 + 10*z + 62*z^2/2! + 466*z^3/3! + .... (End)
Previous Showing 11-20 of 32 results. Next