A000587 -id:A000587 - cp's OEIS frontend

A213170 E.g.f.: exp(2*(1-exp(x))).

1, -2, 2, 2, -6, -14, 26, 178, 90, -2382, -9446, 13746, 287194, 998578, -3687782, -56264782, -208446118, 1017677490, 17194912282, 79540574642, -317691584294, -7577787031374, -47958156443238, 77252406086578, 4400217218583642, 39757699729476274, 54538870133137690

Offset: 0

Philippe Deléham, Feb 14 2013

Exponential self-convolution of complementary Bell numbers (A000587). - Vladimir Reshetnikov, Oct 07 2016

Seiichi Manyama, Table of n, a(n) for n = 0..580
Vaclav Kotesovec, Graph - asymptotic (1000 terms)

Cf. A000587, A001861, A007318, A048993.

b:= proc(n, m) option remember; `if`(n=0,
      (-2)^m, m*b(n-1, m)+b(n-1, m+1))
    end:
a:= n-> b(n, 0):
seq(a(n), n=0..27);  # Alois P. Heinz, Jul 17 2022

CoefficientList[Series[E^(2*(1-E^x)), {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Aug 04 2014 *)
Table[BellB[n, -2], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *)

x='x+O('x^50); Vec(serlaplace(exp(2*(1-exp(x))))) \\ G. C. Greubel, Nov 15 2017

a(n) = Sum_{k=0..n} A048993(n,k)*(-2)^k.
a(n) = Sum_{k=0..n} A000587(k)*A000587(n-k)*binomial(n,k).
G.f.: 1/(1+2*x/(1-x/(1+2*x/(1-2*x/(1+2*x/(1-3*x/(1+2*x/(1-4*x/(1+2*x/(1-...(continued fraction).
Sum_{k=0..n} binomial(n,k)*a(k) = a(n+1)/(-2). - Philippe Deléham, Feb 17 2013
G.f.: 1/Q(0) where Q(k) = 1 + x*(k+1) + x/(1 - 2*x*(k+1)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 07 2013
Lim sup n->infinity (abs(a(n))/n!)^(1/n) / abs(exp(1/LambertW(-n/2)) / LambertW(-n/2)) = 1. - Vaclav Kotesovec, Aug 04 2014
a(n) = B_n(-2), where B_n(x) is n-th Bell polynomial. - Vladimir Reshetnikov, Oct 20 2015
G.f.: Sum_{j>=0} (-2)^j*x^j / Product_{k=1..j} (1 - k*x). - Ilya Gutkovskiy, Apr 06 2019

A260876 Number of m-shape set partitions, square array read by ascending antidiagonals, A(m,n) for m, n >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 4, 5, 5, 1, 1, 11, 31, 15, 7, 1, 1, 36, 365, 379, 52, 11, 1, 1, 127, 6271, 25323, 6556, 203, 15, 1, 1, 463, 129130, 3086331, 3068521, 150349, 877, 22, 1, 1, 1717, 2877421, 512251515, 3309362716, 583027547, 4373461, 4140, 30

Offset: 0

Peter Luschny, Aug 02 2015

A set partition of m-shape is a partition of a set with cardinality m*n for some n >= 0 such that the sizes of the blocks are m times the parts of the integer partitions of n.
If m = 0, all possible sizes are zero. Thus the number of set partitions of 0-shape is the number of integer partitions of n (partition numbers A000041).
If m = 1, the set is {1, 2, ..., n} and the set of all possible sizes are the integer partitions of n. Thus the number of set partitions of 1-shape is the number of set partitions (Bell numbers A000110).
If m = 2, the set is {1, 2, ..., 2n} and the number of set partitions of 2-shape is the number of set partitions into even blocks A005046.
From Petros Hadjicostas, Aug 06 2019: (Start)
Irwin (1916) proved the following combinatorial result: Assume r_1, r_2, ..., r_n are positive integers and we have r_1*r_2*...*r_n objects. We divide them into r_1 classes of r_2*r_3*...*r_n objects each, then each class into r_2 subclasses of r_3*...*r_n objects each, and so on. We call each such classification, without reference to order, a "classification" par excellence. He proved that the total number of classifications is (r_1*r_2*...*r_n)!/( r1! * (r_2!)^(r_1) * (r_3!)^(r_1*r_2) * ... (r_n!)^(r_1*r_2*...*r_{n-1}) ).
Apparently, this problem appeared in Carmichael's "Theory of Numbers".
This result can definitely be used to prove some special cases of my conjecture below. (End)

			[ n ] [0  1   2       3        4           5              6]
[ m ] ------------------------------------------------------
[ 0 ] [1, 1,  2,      3,       5,          7,            11]  A000041
[ 1 ] [1, 1,  2,      5,      15,         52,           203]  A000110
[ 2 ] [1, 1,  4,     31,     379,       6556,        150349]  A005046
[ 3 ] [1, 1, 11,    365,   25323,    3068521,     583027547]  A291973
[ 4 ] [1, 1, 36,   6271, 3086331, 3309362716, 6626013560301]  A291975
        A260878,A309725, ...
For example the number of set partitions of {1,2,...,9} with sizes in [9], [6,3] and [3,3,3] is 1, 84 and 280 respectively. Thus A(3,3) = 365.
Formatted as a triangle:
[1]
[1, 1]
[1, 1,   2]
[1, 1,   2,    3]
[1, 1,   4,    5,     5]
[1, 1,  11,   31,    15,    7]
[1, 1,  36,  365,   379,   52,  11]
[1, 1, 127, 6271, 25323, 6556, 203, 15]
.
From _Peter Luschny_, Aug 14 2019: (Start)
For example consider the case n = 4. There are five integer partitions of 4:
  P = [[4], [3, 1], [2, 2], [2, 1, 1], [1, 1, 1, 1]]. The shapes are m times the parts of the integer partitions: S(m) = [[4m], [3m, m], [2m, 2m], [2m, m, m], [m, m, m, m]].
* In the case m = 1 we look at set partitions of {1, 2, 3, 4} with sizes in  [[4], [3, 1], [2, 2], [2, 1, 1], [1, 1, 1, 1]] which gives rise to [1, 4, 3, 6, 1] with sum 15.
* In the case m = 2 we look at set partitions of {1, 2, .., 8} with sizes in [[8], [6, 2], [4, 4], [4, 2, 2], [2, 2, 2, 2]] which gives rise to [1, 28, 35, 210, 105] with sum 379.
* In the case m = 0 we look at set partitions of {} with sizes in [[0], [0, 0], [0, 0], [0, 0, 0], [0, 0, 0, 0]] which gives rise to [1, 1, 1, 1, 1] with sum 5 (because the only partition of the empty set is the set that contains the empty set, thus from the definition T(0,4) = Sum_{S(0)} card({0}) = A000041(4) = 5).
If n runs through 0, 1, 2,... then the result is an irregular triangle in which the n-th row lists multinomials for partitions of [m*n] which have only parts which are multiples of m. These are the triangles A080575 (m = 1), A257490 (m = 2), A327003 (m = 3), A327004 (m = 4). In the case m = 0 the triangle is A000012 subdivided into rows of length A000041. See the cross references how this case integrates into the full picture.
(End)

Alois P. Heinz, Antidiagonals n = 0..54, flattened
Frank Irwin, Solution to Problem 223 proposed by T. E. Mason in October 1914, Amer. Math. Monthly 23(9) (1916), 352-353.

-----------------------------------------------------------------
[m] |  multi-   |  sum of   |  main     |  by       |  comple-  |
    |  nomials  |  rows     |  diagonal |  size     |  mentary  |
-----------------------------------------------------------------
[0] |  A000012  |  A000041  |  A000012  |  A008284  |  A081362  |
[1] |  A036040  |  A000110  |  A000012  |  A048993  |  A000587  |
[2] |  A257490  |  A005046  |  A001147  |  A156289  |  A260884  |
[3] |  A327003  |  A291973  |  A025035  |  A291451  |  A291974  |
[4] |  A327004  |  A291975  |  A025036  |  A291452  |  A291976  |
Cf. A326996 (main diagonal), A260883 (ordered), A260875 (complementary).
Columns include A000012, A260878, A309725.

A:= proc(m, n) option remember; `if`(m=0, combinat[numbpart](n),
      `if`(n=0, 1, add(binomial(m*n-1, m*k-1)*A(m, n-k), k=1..n)))
    end:
seq(seq(A(d-n, n), n=0..d), d=0..10);  # Alois P. Heinz, Aug 14 2019

A[m_, n_] := A[m, n] = If[m==0, PartitionsP[n], If[n==0, 1, Sum[Binomial[m n - 1, m k - 1] A[m, n - k], {k, 1, n}]]];
Table[Table[A[d - n, n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Dec 06 2019, after Alois P. Heinz *)

def A260876(m, n):
    shapes = ([x*m for x in p] for p in Partitions(n))
    return sum(SetPartitions(sum(s), s).cardinality() for s in shapes)
for m in (0..4): print([A260876(m,n) for n in (0..6)])

From Petros Hadjicostas, Aug 02 2019: (Start)
A(m, 2) = 1 + (1/2) * binomial(2*m, m) for m >= 1.
A(m, 3) = 1 + binomial(3*m, m) + (3*m)!/(6 * (m!)^3) for m >= 1.
A(m, 4) = (1/4!) * multinomial(4*m, [m, m, m, m]) + (1/2) * multinomial(4*m, [2*m, m, m]) + multinomial(4*m, [m, 3*m]) + (1/2) * multinomial(4*m, [2*m, 2*m]) + 1 for m >= 1.
Conjecture: For n >= 0, let P be the set of all possible lists (a_1,...,a_n) of nonnegative integers such that a_1*1 + a_2*2 + ... + a_n*n = n. Consider terms of the form multinomial(n*m, m*[1,..., 1,2,..., 2,..., n,..., n])/(a_1! * a_2! * ... * a_n!), where in the list [1,...,1,2,...,2,...,n,...,n] the number 1 occurs a_1 times, 2 occurs a_2 times, ..., and n occurs a_n times. (Here a_n = 0 or 1.) Summing these terms over P we get A(m, n) provided m >= 1. (End)
Conjecture for a recurrence: A(m, n) = Sum_{k = 0..n-1} binomial(m*n - 1, m*k) * A(m, k) with A(m, 0) = 1 for m >= 1 and n >= 0. (Unfortunately, the recurrence does not hold for m = 0.) - Petros Hadjicostas, Aug 12 2019

A291451 Triangle read by rows, expansion of e.g.f. exp(x(exp(z)/3 + 2exp(-z/2)* cos(z*sqrt(3)/2)/3 - 1)), nonzero coefficients of z.

Original entry on oeis.org

1, 0, 1, 0, 1, 10, 0, 1, 84, 280, 0, 1, 682, 9240, 15400, 0, 1, 5460, 260260, 1401400, 1401400, 0, 1, 43690, 7128576, 99379280, 285885600, 190590400, 0, 1, 349524, 193360720, 6600492080, 42549306800, 76045569600, 36212176000

Offset: 0

Peter Luschny, Sep 07 2017

			Triangle starts:
[1]
[0, 1]
[0, 1,    10]
[0, 1,    84,     280]
[0, 1,   682,    9240,    15400]
[0, 1,  5460,  260260,  1401400,   1401400]
[0, 1, 43690, 7128576, 99379280, 285885600, 190590400]

Cf. A048993 (m=1), A156289 (m=2), this seq. (m=3), A291452 (m=4).
Diagonal: A000012 (m=1), A001147 (m=2), A025035 (m=3), A025036 (m=4).
Row sums: A000110 (m=1), A005046 (m=2), A291973 (m=3), A291975 (m=4).
Alternating row sums: A000587 (m=1), A260884 (m=2), A291974 (m=3), A291976 (m=4).

CL := (f, x) -> PolynomialTools:-CoefficientList(f, x):
A291451_row := proc(n) exp(x*(exp(z)/3+2*exp(-z/2)*cos(z*sqrt(3)/2)/3-1)):
series(%, z, 66): CL((3*n)!*coeff(series(%,z,3*(n+1)),z,3*n),x) end:
for n from 0 to 7 do A291451_row(n) od;
# Alternative:
A291451row := proc(n) local P; P := proc(m, n) option remember;
if n = 0 then 1 else add(binomial(m*n, m*k)*P(m, n-k)*x, k=1..n) fi end:
CL(P(3, n), x); seq(%[k+1]/k!, k=0..n) end: # Peter Luschny, Sep 03 2018

P[m_, n_] := P[m, n] = If[n == 0, 1, Sum[Binomial[m*n, m*k]*P[m, n - k]*x, {k, 1, n}]];
row[n_] := Module[{cl = CoefficientList[P[3, n], x]}, Table[cl[[k + 1]]/k!, {k, 0, n}]];
Table[row[n], {n, 0, 7}] // Flatten (* Jean-François Alcover, Jul 23 2019, after Peter Luschny *)

A291452 Triangle read by rows, expansion of e.g.f. exp(x*(cos(z) + cosh(z) - 2)/2), nonzero coefficients of z.

Original entry on oeis.org

1, 0, 1, 0, 1, 35, 0, 1, 495, 5775, 0, 1, 8255, 450450, 2627625, 0, 1, 130815, 35586525, 727476750, 2546168625, 0, 1, 2098175, 2941884000, 181262956875, 1932541986375, 4509264634875

Offset: 0

Peter Luschny, Sep 07 2017

			Triangle starts:
[1]
[0, 1]
[0, 1,      35]
[0, 1,     495,       5775]
[0, 1,    8255,     450450,      2627625]
[0, 1,  130815,   35586525,    727476750,    2546168625]
[0, 1, 2098175, 2941884000, 181262956875, 1932541986375, 4509264634875]

Cf. A048993 (m=1), A156289 (m=2), A291451 (m=3), this seq. (m=4).
Diagonal: A000012 (m=1), A001147 (m=2), A025035 (m=3), A025036 (m=4).
Row sums: A000110 (m=1), A005046 (m=2), A291973 (m=3), A291975 (m=4).
Alternating row sums: A000587 (m=1), A260884 (m=2), A291974 (m=3), A291976 (m=4).

CL := (f,x) -> PolynomialTools:-CoefficientList(f,x):
A291452_row := proc(n) exp(x*(cos(z)+cosh(z)-2)/2):
series(%, z, 88): CL((4*n)!*coeff(series(%,z,4*(n+1)),z,4*n),x) end:
for n from 0 to 7 do A291452_row(n) od;
# Alternative:
A291452row := proc(n) local P; P := proc(m, n) option remember;
if n = 0 then 1 else add(binomial(m*n, m*k)*P(m, n-k)*x, k=1..n) fi end:
CL(P(4, n), x); seq(%[k+1]/k!, k=0..n) end: # Peter Luschny, Sep 03 2018

P[m_, n_] := P[m, n] = If[n == 0, 1, Sum[Binomial[m*n, m*k]*P[m, n - k]*x, {k, 1, n}]];
row[n_] := Module[{cl = CoefficientList[P[4, n], x]}, Table[cl[[k + 1]]/k!, {k, 0, n}]];
Table[row[n], {n, 0, 6}] // Flatten (* Jean-François Alcover, Jul 23 2019, after Peter Luschny *)

A260884 Number of set partitions of a 2n-set into even blocks which have even length minus the number of partitions into even blocks which have odd length.

Original entry on oeis.org

1, -1, 2, -1, -43, 254, 4157, -70981, -1310398, 40933619, 1087746617, -43668096946, -1926040182823, 74089958942999, 6019198050509282, -160614025972447321, -28821072878928603043, 164386116957516330494, 190177688405403480505877, 5097857816569586800024019

Offset: 0

Peter Luschny, Sep 18 2015

sign

			a(5) = 254. Consider the shapes [10], [8, 2], [6, 4], [6, 2, 2], [4, 4, 2], [4, 2, 2, 2], [2, 2, 2, 2, 2]. Computing the number of associated set partitions gives -1 + 45 + 210 - 630 - 1575 + 3150 - 945 = 254.

Alois P. Heinz, Table of n, a(n) for n = 0..300

This is case m=2 of A260875. Case m=1 are the complementary Bell numbers A000587.

b:= proc(n, t) option remember; `if`(n=0, 1-2*t, add(
      b(n-2*j, 1-t)*binomial(n-1, 2*j-1), j=1..n/2))
    end:
a:= n-> b(2*n, 0):
seq(a(n), n=0..20);  # Alois P. Heinz, Jun 28 2016

Table[Sum[BellY[2 n, k, Mod[Range[2 n], 2] - 1], {k, 0, 2 n}], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 09 2016 *)

a(n) = {polcoef(serlaplace(exp(1 - cosh(x + O(x^(2*n + 1))))), 2*n)} \\ Andrew Howroyd, Jan 27 2020

def ComplementaryBell(m, n):
    shapes = ([x*m for x in p] for p in Partitions(n))
    return sum((-1)^len(s)*SetPartitions(sum(s), s).cardinality() for s in shapes)
A260884 = lambda n: ComplementaryBell(2,n)
[A260884(n) for n in (0..19)]

E.g.f.: exp(1 - cosh(x)) (even powers only). - Ilya Gutkovskiy, Jan 27 2020

A292861 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of e.g.f. exp(k*(1 - exp(x))).

Original entry on oeis.org

1, 1, 0, 1, -1, 0, 1, -2, 0, 0, 1, -3, 2, 1, 0, 1, -4, 6, 2, 1, 0, 1, -5, 12, -3, -6, -2, 0, 1, -6, 20, -20, -21, -14, -9, 0, 1, -7, 30, -55, -20, 24, 26, -9, 0, 1, -8, 42, -114, 45, 172, 195, 178, 50, 0, 1, -9, 56, -203, 246, 370, 108, -111, 90, 267, 0, 1, -10, 72, -328, 679, 318, -1105, -2388, -3072, -2382, 413, 0

Offset: 0

Seiichi Manyama, Sep 25 2017

			Square array begins:
   1,  1,   1,   1,   1,     1,     1, ...
   0, -1,  -2,  -3,  -4,    -5,    -6, ...
   0,  0,   2,   6,  12,    20,    30, ...
   0,  1,   2,  -3, -20,   -55,  -114, ...
   0,  1,  -6, -21, -20,    45,   246, ...
   0, -2, -14,  24, 172,   370,   318, ...
   0, -9,  26, 195, 108, -1105, -4074, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0..4 give A000007, A000587, A213170, A309084, A309085.
Rows n=0..1 give A000012, (-1)*A001477.
Main diagonal gives A292866.
Cf. A292860, A309386.

A:= proc(n, k) option remember; `if`(n=0, 1,
      -(1+add(binomial(n-1, j-1)*A(n-j, k), j=1..n-1))*k)
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);  # Alois P. Heinz, Sep 25 2017

A[n_, k_] := Sum[(-k)^j StirlingS2[n, j], {j, 0, n}];
Table[A[n, d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Feb 10 2021 *)
A292861[n_, k_] := BellB[k, k - n];
Table[A292861[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Peter Luschny, Dec 23 2021 *)

A(0,k) = 1 and A(n,k) = -k * Sum_{j=0..n-1} binomial(n-1,j) * A(j,k) for n > 0.
A(n,k) = Sum_{j=0..n} (-k)^j * Stirling2(n,j). - Seiichi Manyama, Jul 27 2019
A(n,k) = BellPolynomial(n, -k). - Peter Luschny, Dec 23 2021

A350263 Triangle read by rows. T(n, k) = BellPolynomial(n, -k).

Original entry on oeis.org

1, 0, -1, 0, 0, 2, 0, 1, 2, -3, 0, 1, -6, -21, -20, 0, -2, -14, 24, 172, 370, 0, -9, 26, 195, 108, -1105, -4074, 0, -9, 178, -111, -2388, -4805, 2046, 34293, 0, 50, 90, -3072, -3220, 23670, 87510, 111860, -138312, 0, 267, -2382, -4053, 47532, 121995, -115458, -1193157, -2966088, -2932533

Offset: 0

Peter Luschny, Dec 23 2021

			[0] 1
[1] 0,  -1
[2] 0,   0,     2
[3] 0,   1,     2,    -3
[4] 0,   1,    -6,   -21,   -20
[5] 0,  -2,   -14,    24,   172,    370
[6] 0,  -9,    26,   195,   108,  -1105, -  4074
[7] 0,  -9,   178,  -111, -2388,  -4805,    2046,    34293
[8] 0,  50,    90, -3072, -3220,  23670,   87510,   111860,  -138312
[9] 0, 267, -2382, -4053, 47532, 121995, -115458, -1193157, -2966088, -2932533

Main diagonal: A292866, column 1: A000587, variant: A292861.

A350263 := (n, k) -> ifelse(n = 0, 1, BellB(n, -k)):
seq(seq(A350263(n, k), k = 0..n), n = 0..9);

T[n_, k_] := BellB[n, -k]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten

A028575 Row sums of triangle A011801.

Original entry on oeis.org

1, 5, 49, 721, 14177, 349141, 10334689, 357361985, 14137664833, 629779342213, 31195027543505, 1700812505769169, 101218448336028193, 6528869281965115541, 453720852957751220353, 33796334125623555379969, 2686138908337714715560577, 226908450494953996837748869

Offset: 1

Wolfdieter Lang

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.

Sequences with e.g.f. exp(1-(1-m*x)^(1/m)) - 1: A000012 (m=1), A001515 (m=2), A015735 (m=3), A016036 (m=4), this sequence (m=5), A028844 (m=6).

Cf. A011801.

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!(Laplace( Exp(1-(1-5*x)^(1/5)) - 1 ))); // G. C. Greubel, Oct 02 2023

Rest[With[{nn=20},CoefficientList[Series[Exp[1-(1-5x)^(1/5)]-1, {x,0,nn}], x] Range[0,nn]!]] (* Harvey P. Dale, Aug 02 2016 *)

def A028575_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( exp(1-(1-5*x)^(1/5)) -1 ).egf_to_ogf().list()
a=A028575_list(40); a[1:] # G. C. Greubel, Oct 02 2023

E.g.f.: exp(1 - (1-5*x)^(1/5)) - 1.
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator 1/(1-x)^4*d/dx. Cf. A001515, A015735 and A016036. - Peter Bala, Nov 25 2011
D-finite with recurrence: a(n) -20*(n-3)*a(n-1) +30*(5*n^2-35*n +62)*a(n-2) -100*(n-4)*(5*n^2-40*n+81)*a(n-3) +(5*n-22)*(5*n-21)*(5*n-24)*(5*n-23)*a(n-4) -a(n-5) = 0. - R. J. Mathar, Jan 28 2020
From Seiichi Manyama, Jan 20 2025: (Start)
a(n) = Sum_{k=0..n} (-1)^k * 5^(n-k) * |Stirling1(n,k)| * A000587(k).
a(n) = e * (-5)^n * n! * Sum_{k>=0} (-1)^k * binomial(k/5,n)/k!. (End)

A101851 a(n) = Sum_{k=0..n} (-1)^(n-k)kStirling2(n,k).

Original entry on oeis.org

0, 1, 1, -2, -1, 11, -18, -41, 317, -680, -1767, 19911, -68264, -59643, 2076973, -11905466, 18577387, 269836343, -2819431570, 12357816867, 17355428041, -752675321800, 6318046208653, -21416130683133, -152569023028272, 3016508107668601, -23667435182395287

Offset: 0

Vladeta Jovovic, Jan 27 2005

Robert Israel, Table of n, a(n) for n = 0..580
Bérénice Delcroix-Oger and Clément Dupont, Lie-operads and operadic modules from poset cohomology, arXiv:2505.06094 [math.CO], 2025. See p. 28, Table 2, operad NAP.

Cf. A005493, A000587.

A101851 := proc(n) local k;
add((-1)^(n-k)*k*combinat[stirling2](n,k), k = 0..n) end:
seq(A101851(n),n = 0..26); # Peter Luschny, Apr 17 2011

Table[Sum[(-1)^(n-k) k StirlingS2[n,k],{k,0,n}],{n,0,30}] (* Harvey P. Dale, Aug 09 2013 *)
Table[(-1)^n (BellB[n, -1] + BellB[n + 1, -1]), {n, 0, 25}] (* Vladimir Reshetnikov, Oct 21 2015 *)

a(n) = sum(k=0, n, (-1)^(n-k)*k*stirling(n, k, 2)); \\ Michel Marcus, Oct 22 2015

E.g.f.: (1-exp(-x))*exp(1-exp(-x)): G.f.: Sum(k*x^k/Product(1+l*x, l = 1 .. k), k = 1 .. infinity).
a(n) = Sum_{k=0..n} (-1)^(k+1)*binomial(n,k)*A000587(k+2). - Peter Luschny, Apr 17 2011
G.f.: x*G(0)/(1+x) where G(k) = 1 + 2*x*(k+1)/((2*k+1)*(2*x*k+2*x+1) - x*(2*k+1)*(2*k+3)*(2*x*k+2*x+1)/(x*(2*k+3) + 2*(k+1)*(2*x*k+3*x+1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 19 2012
G.f.: 1/x - G(0)/x, where G(k) = 1 - x^2*(k+1)/(x^2*(k+1) + (x*k + 1 - x)*(x*k + 1)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Feb 06 2014
a(n) = (-1)^n*(A000587(n)+A000587(n+1)). - Vladimir Reshetnikov, Oct 21 2015

A130410 Alternating row sums of triangle A130191 (Stirling2)^2.

Original entry on oeis.org

1, -1, -1, 0, 6, 32, 115, 172, -2030, -29013, -250051, -1587556, -5178877, 52922256, 1435509569, 20813187553, 230664704969, 1884809758791, 5120430335582, -216605840330716, -6440821191934686, -122368984222010397, -1842986108839510180, -21473141673616814694

Offset: 0

Wolfdieter Lang Jun 01 2007

Stirling2 transform of A000587. 2nd Stirling2 transform of A033999. - Vladimir Reshetnikov, Oct 22 2015

			E.g.f.: 1 - x - (1/2)*x^2 + (1/4)*x^4+(4/15)*x^5 + (23/144)*x^6 + (43/1260)*x^7 - (29/576)*x^8 - (9671/120960)*x^9 ...
G.f. = 1 - x - x^2 + 6*x^4 + 32*x^5 + 115*x^6 + 172*x^7 - 2030*x^8 - 29013*x^9 + ...

Robert Israel, Table of n, a(n) for n = 0..470
Eric Weisstein's MathWorld, Stirling Transform.
Eric Weisstein's MathWorld, Bell Polynomial.

Cf. A048993, A000258 (row sums of A130191), A000587, A033999, A130191.

Egf:= 1/exp(exp(exp(x)-1)-1):
S:= series(Egf,x,101):
seq(coeff(S,x,j)*j!, j=0..100); # Robert Israel, Oct 22 2015

Table[Sum[BellY[n, k, -BellB[Range[n]]], {k, 0, n}], {n, 0, 23}] (* Vladimir Reshetnikov, Nov 09 2016 *)

a(n) = sum(A130191(n,m)*(-1)^m,m=0..n), n>=0.
E.g.f.: 1/exp(f(x)) with f(x):=exp(exp(x)-1)-1.
a(n) = sum(k=0..n, A000587(k)*stirling2(n,k)) = sum(k=0..n, B_k(-1)*stirling2(n,k)), where B_k(x) is k-th Bell polynomial.

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A101851 a(n) = Sum_{k=0..n} (-1)^(n-k)kStirling2(n,k).