A000587 -id:A000587 - cp's OEIS frontend

A293037 E.g.f.: exp(1 + x - exp(x)).

1, 0, -1, -1, 2, 9, 9, -50, -267, -413, 2180, 17731, 50533, -110176, -1966797, -9938669, -8638718, 278475061, 2540956509, 9816860358, -27172288399, -725503033401, -5592543175252, -15823587507881, 168392610536153, 2848115497132448, 20819319685262839

Offset: 0

Seiichi Manyama, Sep 28 2017

sign

Seiichi Manyama, Table of n, a(n) for n = 0..593

Column k=1 of A293051.
Column k=1 of A335977.
Cf. A000587 (k=0), this sequence (k=1), A293038 (k=2), A293039 (k=3), A293040 (k=4).
Cf. A000296, A014182, A193683, A193684, A196835.

f:= series(exp(1 + x - exp(x)), x= 0, 101): seq(factorial(n) * coeff(f, x, n), n = 0..30); # Muniru A Asiru, Oct 31 2017
# second Maple program:
b:= proc(n, t) option remember; `if`(n=0, 1-2*t,
      add(b(n-j, 1-t)*binomial(n-1, j-1), j=1..n))
    end:
a:= n-> b(n+1, 1):
seq(a(n), n=0..35);  # Alois P. Heinz, Dec 01 2021

m = 26; Range[0, m]! * CoefficientList[Series[Exp[1 + x - Exp[x]], {x, 0, m}], x] (* Amiram Eldar, Jul 06 2020 *)
Table[Sum[Binomial[n, k] * BellB[k, -1], {k, 0, n}], {n, 0, 30}] (* Vaclav Kotesovec, Jul 06 2020 *)

my(N=40, x='x+O('x^N)); Vec(serlaplace(exp(-exp(x)+1+x)))

a(n) = if(n==0, 1, -sum(k=0, n-2, binomial(n-1, k)*a(k))); \\ Seiichi Manyama, Aug 02 2021

a(n) = exp(1) * Sum_{k>=0} (-1)^k*(k + 1)^n/k!. - Ilya Gutkovskiy, Jun 13 2019
a(n) = Sum_{k=0..n} binomial(n,k) * Bell(k, -1). - Vaclav Kotesovec, Jul 06 2020
a(0) = 1; a(n) = - Sum_{k=0..n-2} binomial(n-1,k) * a(k). - Seiichi Manyama, Aug 02 2021

A014182 Expansion of e.g.f. exp(1-x-exp(-x)).

Original entry on oeis.org

1, 0, -1, 1, 2, -9, 9, 50, -267, 413, 2180, -17731, 50533, 110176, -1966797, 9938669, -8638718, -278475061, 2540956509, -9816860358, -27172288399, 725503033401, -5592543175252, 15823587507881, 168392610536153, -2848115497132448, 20819319685262839

Offset: 0

Noam D. Elkies

E.g.f. A(x) = y satisfies (y + y' + y'') * y - y'^2 = 0. - Michael Somos, Mar 11 2004
The 10-adic sum: B(n) = Sum_{k>=0} k^n*k! simplifies to: B(n) = A014182(n)*B(0) + A014619(n) for n>=0, where B(0) is the 10-adic sum of factorials (A025016); a result independent of base. - Paul D. Hanna, Aug 12 2006
Equals row sums of triangle A143987 and (shifted) = right border of A143987. [Gary W. Adamson, Sep 07 2008]
From Gary W. Adamson, Dec 31 2008: (Start)
Equals the eigensequence of the inverse of Pascal's triangle, A007318.
Binomial transform shifts to the right: (1, 1, 0, -1, 1, 2, -9, ...).
Double binomial transform = A109747. (End)
Convolved with A154107 = A000110, the Bell numbers. - Gary W. Adamson, Jan 04 2009

			G.f. = 1 - x^2 + x^3 + 2*x^4 - 9*x^5 + 9*x^6 + 50*x^7 - 267*x^8 + 413*x^9 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Mohammad Ghorbani, Mehdi Hassani, and Hossein Moshtagh, Moments and asymptotic expansion of derangement polynomials in terms of Touchard polynomials, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 4, 832-842. See pp. 834, 839.

Essentially same as A000587. See also A014619.
Cf. A025016.
Cf. A143987, A109747, A154107, A000110.

With[{nn=30},CoefficientList[Series[Exp[1-x-Exp[-x]],{x,0,nn}],x] Range[0,nn]!]  (* Harvey P. Dale, Jan 15 2012 *)
a[ n_] := SeriesCoefficient[ (1 - Sum[ k / Pochhammer[ 1/x + 1, k], {k, n}]) / (1 - x), {x, 0, n} ]; (* Michael Somos, Nov 07 2014 *)

{a(n)=sum(j=0,n,(-1)^(n-j)*Stirling2(n+1,j+1))}
{Stirling2(n,k)=(1/k!)*sum(i=0,k,(-1)^(k-i)*binomial(k,i)*i^n)} \\ Paul D. Hanna, Aug 12 2006

{a(n) = if( n<0, 0, n! * polcoeff( exp( 1 - x - exp( -x + x * O(x^n))), n))} /* Michael Somos, Mar 11 2004 */

def A014182_list(len):  # len>=1
    T = [0]*(len+1); T[1] = 1; R = [1]
    for n in (1..len-1):
        a,b,c = 1,0,0
        for k in range(n,-1,-1):
            r = a - k*b - (k+1)*c
            if k < n : T[k+2] = u;
            a,b,c = T[k-1],a,b
            u = r
        T[1] = u; R.append(u)
    return R
A014182_list(27)  # Peter Luschny, Nov 01 2012

E.g.f.: exp(1-x-exp(-x)).
a(n) = Sum_{k=0..n} (-1)^(n-k)*Stirling2(n+1,k+1). - Paul D. Hanna, Aug 12 2006
A000587(n+1) = -a(n). - Michael Somos, May 12 2012
G.f.: 1/x/(U(0)-x) -1/x  where U(k)= 1 - x + x*(k+1)/(1 - x/U(k+1)); (continued fraction). - Sergei N. Gladkovskii, Oct 12 2012
G.f.: 1/(U(0) - x) where U(k) = 1 + x*(k+1)/(1 - x/U(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 12 2012
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1+k*x+x)/(1-x/(x-1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 17 2013
G.f.: G(0)/(1+x)-1 where G(k) = 1 + 1/(1 + k*x - x*(1+k*x)*(1+k*x+x)/(x*(1+k*x+x) + (1+k*x+2*x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 09 2013
G.f.: S-1 where S = Sum_{k>=0} (2 + x*k)*x^k/Product_{i=0..k} (1+x+x*i). - Sergei N. Gladkovskii, Feb 09 2013
G.f.: G(0)*x^2/(1+x)/(1+2*x) + 2/(1+x) - 1 where G(k) = 1 + 2/(x + k*x - x^3*(k+1)*(k+2)/(x^2*(k+2) + 2*(1+k*x+3*x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 09 2013
G.f.: 1/(x*Q(0)) -1/x, where Q(k) = 1 - x/(1 + (k+1)*x/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Sep 27 2013
G.f.: G(0)/(1-x)/x - 1/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
G.f.:  (1 - Sum_{k>0} k * x^k / ((1 + x) * (1 + 2*x) + ... (1 + k*x))) / (1 - x). - Michael Somos, Nov 07 2014
a(n) = exp(1) * (-1)^n * Sum_{k>=0} (-1)^k * (k + 1)^n / k!. - Ilya Gutkovskiy, Dec 20 2019

A189233 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals upwards, where the e.g.f. of column k is exp(k*(e^x-1)).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 5, 6, 3, 1, 0, 15, 22, 12, 4, 1, 0, 52, 94, 57, 20, 5, 1, 0, 203, 454, 309, 116, 30, 6, 1, 0, 877, 2430, 1866, 756, 205, 42, 7, 1, 0, 4140, 14214, 12351, 5428, 1555, 330, 56, 8, 1, 0, 21147, 89918, 88563, 42356, 12880, 2850, 497, 72, 9, 1

Offset: 0

Peter Luschny, Apr 18 2011

A(n,k) is the n-th moment of a Poisson distribution with mean = k. - Geoffrey Critzer, Dec 23 2018

			Square array begins:
       A000007 A000110 A001861 A027710 A078944 A144180 A144223 A144263
A000012   1,    1,    1,    1,    1,     1,     1,     1, ...
A001477   0,    1,    2,    3,    4,     5,     6,     7, ...
A002378   0,    2,    6,   12,   20,    30,    42,    56, ...
A033445   0,    5,   22,   57,  116,   205,   330,   497, ...
          0,   15,   94,  309,  756,  1555,  2850,  4809, ...
          0,   52,  454, 1866, 5428, 12880, 26682, 50134, ...

Vincenzo Librandi, Table of n, a(n) for n = 0..5150
E. T. Bell, Exponential numbers, Amer. Math. Monthly, 41 (1934), 411-419.
Peter Luschny, Set partitions and Bell numbers

Columns: A000007, A000110, A001861, A027710, A078944, A144180, A144223, A144263.
Rows: A000012, A001477, A002378, A033445.
Main diagonal gives A242817.
Cf. A000587, A144150.

# Cf. also the Maple prog. of Alois P. Heinz in A144223 and A144180.
expnums := proc(k,n) option remember; local j;
`if`(n = 0, 1, (1+add(binomial(n-1,j-1)*expnums(k,n-j), j = 1..n-1))*k) end:
A189233_array := (k, n) -> expnums(k,n):
seq(print(seq(A189233_array(k,n), k = 0..7)), n = 0..5);
A189233_egf := k -> exp(k*(exp(x)-1));
T := (n,k) -> n!*coeff(series(A189233_egf(k), x, n+1), x, n):
seq(lprint(seq(T(n,k), k = 0..7)), n = 0..5):
# alternative Maple program:
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(d-k, k), k=0..d), d=0..12);  # Alois P. Heinz, Sep 25 2017

max = 9; Clear[col]; col[k_] := col[k] = CoefficientList[ Series[ Exp[k*(Exp[x]-1)], {x, 0, max}], x]*Range[0, max]!; a[0, ] = 1; a[n?Positive, 0] = 0; a[n_, k_] := col[k][[n+1]]; Table[ a[n-k, k], {n, 0, max}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 26 2013 *)
Table[Table[BellB[n, k], {k, 0, 5}], {n, 0, 5}] // Grid  (* Geoffrey Critzer, Dec 23 2018 *)

A(n,k):=if k=0 and n=0 then 1 else if k=0 then 0 else  sum(stirling2(n,i)*k^i,i,0,n); /* Vladimir Kruchinin, Apr 12 2019 */

E.g.f. of column k: exp(k*(e^x-1)).
A(n,1) = A000110(n), A(n, -1) = A000587(n).
A(n,k) = BellPolynomial(n, k). - Geoffrey Critzer, Dec 23 2018
A(n,k) = Sum_{i=0..n} Stirling2(n,i)*k^i. - Vladimir Kruchinin, Apr 12 2019

A121867 Let A(0) = 1, B(0) = 0; A(n+1) = Sum_{k=0..n} binomial(n,k)B(k), B(n+1) = Sum_{k=0..n} -binomial(n,k)A(k); entry gives A sequence (cf. A121868).

Original entry on oeis.org

1, 0, -1, -3, -6, -5, 33, 266, 1309, 4905, 11516, -22935, -556875, -4932512, -32889885, -174282151, -612400262, 907955295, 45283256165, 573855673458, 5397236838345, 41604258561397, 250231901787780, 756793798761989, -8425656230853383, -213091420659985440, -2990113204010882473

Offset: 0

N. J. A. Sloane, Sep 05 2006

Stirling transform of A056594.

			From _Peter Bala_, Aug 28 2008: (Start)
E_2(k) as linear combination of E_2(i), i = 0..1.
============================
..E_2(k)..|...E_2(0)..E_2(1)
============================
..E_2(2)..|....-1.......1...
..E_2(3)..|....-3.......0...
..E_2(4)..|....-6......-5...
..E_2(5)..|....-5.....-23...
..E_2(6)..|....33.....-74...
..E_2(7)..|...266....-161...
..E_2(8)..|..1309......57...
..E_2(9)..|..4905....3466...
...
(End)

Alois P. Heinz, Table of n, a(n) for n = 0..220
A. Fekete and G. Martin, Problem 10791: Squared Series Yielding Integers, Amer. Math. Monthly, 108 (No. 2, 2001), 177-178.

Cf. A121868, A024430, A024429, A056594.

Cf. A000587, A143623, A143624, A143628, A143631. - Peter Bala, Aug 28 2008

List([0..30], n-> Sum([0..Int(n/2)], k-> (-1)^k*Stirling2(n,2*k)) ); # G. C. Greubel, Oct 09 2019

[(&+[(-1)^k*StirlingSecond(n,2*k): k in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, Oct 09 2019

# Maple code for A024430, A024429, A121867, A121868.
M:=30; a:=array(0..100); b:=array(0..100); c:=array(0..100); d:=array(0..100); a[0]:=1; b[0]:=0; c[0]:=1; d[0]:=0;
for n from 1 to M do a[n]:=add(binomial(n-1,k)*b[k], k=0..n-1); b[n]:=add(binomial(n-1,k)*a[k], k=0..n-1); c[n]:=add(binomial(n-1,k)*d[k], k=0..n-1); d[n]:=-add(binomial(n-1,k)*c[k], k=0..n-1); od: ta:=[seq(a[n],n=0..M)]; tb:=[seq(b[n],n=0..M)]; tc:=[seq(c[n],n=0..M)]; td:=[seq(d[n],n=0..M)];
# Code based on Stirling transform:
stirtr:= proc(p) proc(n) option remember;
            add(p(k) *Stirling2(n,k), k=0..n) end
         end:
a:= stirtr(n-> (I^n + (-I)^n)/2):
seq(a(n), n=0..30);  # Alois P. Heinz, Jan 29 2011

a[n_] := (BellB[n, -I] + BellB[n, I])/2; Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Mar 06 2013, after Alois P. Heinz *)

a(n) = sum(k=0,n\2, (-1)^k*stirling(n,2*k,2));
vector(30, n, a(n-1)) \\ G. C. Greubel, Oct 09 2019

[sum((-1)^k*stirling_number2(n,2*k) for k in (0..floor(n/2))) for n in (0..30)] # G. C. Greubel, Oct 09 2019

From Peter Bala, Aug 28 2008: (Start)
This sequence and its companion A121868 are related to the pair of constants cos(1) + sin(1) and cos(1) - sin(1) and may be viewed as generalizations of the Uppuluri-Carpenter numbers (complementary Bell numbers) A000587. Define E_2(k) = Sum_{n >= 0} (-1)^floor(n/2) * n^k/n! for k = 0,1,2,... . Then E_2(0) = cos(1) + sin(1) and E_2(1) = cos(1) - sin(1). It is easy to see that E_2(k+2) = E_2(k+1) - Sum_{i = 0..k} 2^i*binomial(k,i)*E_2(k-i) for k >= 0. Hence E_2(k) is an integral linear combination of E_2(0) and E_2(1) (a Dobinski-type relation). For example, E_2(2) = - E_2(0) + E_2(1), E_2(3) = -3*E_2(0) and E_2(4) = - 6*E_2(0) - 5*E_2(1). More examples are given below.
To find the precise result, show F(k) := Sum_{n >= 0} (-1)^floor((n+1)/2)*n^k/n! satisfies the above recurrence with F(0) = E_2(1) and F(1) = -E_2(0) and then use the identity Sum_{i = 0..k} binomial(k,i)*E_2(i) = -F(k+1) to obtain E_2(k) = A121867(k) * E_2(0) - A121868(k) * E_2(1). For similar results see A143628. The decimal expansions of E_2(0) and E_2(1) are given in A143623 and A143624 respectively. (End)
E.g.f.: A(x) = cos(exp(x)-1).
a(n) = Sum_{k=0..floor(n/2)} stirling2(n,2*k)*(-1)^(k). - Vladimir Kruchinin, Jan 29 2011

A143628 Define E(n) = Sum_{k >= 0} (-1)^floor(k/3)*k^n/k! for n = 0,1,2,... . Then E(n) is an integral linear combination of E(0), E(1) and E(2). This sequence lists the coefficients of E(0).

Original entry on oeis.org

1, 0, 0, -1, -6, -25, -89, -280, -700, -380, 13452, 149831, 1214852, 8700263, 57515640, 351296151, 1909757620, 8017484274, 5703377941, -428273438434, -7295220035921, -89868583754993, -970185398785810, -9657428906237364

Offset: 0

Peter Bala, Sep 05 2008

This sequence and its companion sequences A143629 and A143630 may be viewed as generalizations of the Uppuluri-Carpenter numbers (complementary Bell numbers) A000587. Define E(n) = Sum_{k >= 0} (-1)^floor(k/3)*k^n/k! = 0^n/0! + 1^n/1! + 2^n/2! - 3^n/3! - 4^n/4! - 5^n/5! + + + - - - ... for n = 0,1,2,... . It is easy to see that E(n+3) = 3*E(n+2) - 2*E(n+1) - Sum_{i = 0..n} 3^i* binomial(n,i)*E(n-i) for n >= 0. Thus E(n) is an integral linear combination of E(0), E(1) and E(2). Some examples are given below.

This sequence lists the coefficients of E(0). See A143629 and A143630 for the sequence of coefficients of E(1) and E(2) respectively. The functions F(n) := Sum_{k >= 0} (-1)^floor((k+1)/3)*k^n/k! and G(n) = Sum_{k >= 0} (-1)^floor((k+2)/3)*k^n/k! both satisfy the above recurrence as well as the identities E(n+1) = Sum_{i = 0..n} binomial(n,i)*F(i), F(n+1) = Sum_{i = 0..n} binomial(n,i)*G(i) and G(n+1) = - Sum_{i = 0..n} binomial(n,i)*E(i). This leads to the precise result for E(n) as a linear combination of E(0), E(1) and E(2), namely, E(n) = A143628(n)*E(0) + A143629(n)*E(1) + A143630(n)*E(2). Compare with A121867 and A143815.

			E(n) as linear combination of E(i),
i = 0..2.
====================================
..E(n)..|.....E(0).....E(1)....E(2).
====================================
..E(3)..|......-1......-2........3..
..E(4)..|......-6......-7........7..
..E(5)..|.....-25.....-23.......14..
..E(6)..|.....-89.....-80.......16..
..E(7)..|....-280....-271......-77..
..E(8)..|....-700....-750.....-922..
..E(9)..|....-380....-647....-6660..
..E(10).|...13452...13039...-41264..
...
a(5) = -25 because E(5) = -25*E(0) - 23*E(1) + 14*E(2).
a(6) = -89 because E(6) = -89*E(0) - 80*E(1) + 16*E(2).

Seiichi Manyama, Table of n, a(n) for n = 0..578
Eric Weisstein's World of Mathematics, Bell Polynomial.

A000587, A121867, A143629, A143630, A143631, A143815, A143816, A143817, A143818.

# Compare with A143815
#
M:=24: a:=array(0..100): b:=array(0..100): c:=array(0..100):
a[0]:=1: b[0]:=0: c[0]:=0:
for n from 1 to M do
a[n]:= -add(binomial(n-1,k)*c[k], k=0..n-1);
b[n]:= add(binomial(n-1,k)*a[k], k=0..n-1);
c[n]:= add(binomial(n-1,k)*b[k], k=0..n-1);
end do:
A143628:=[seq(a[n], n=0..M)];

m = 23; a[0] = 1; b[0] = 0; c[0] = 0; For[n = 1, n <= m, n++, a[n] = -Sum[ Binomial[n-1, k]*c[k], {k, 0, n-1}]; b[n] = Sum[ Binomial[n-1, k]*a[k], {k, 0, n-1}]; c[n] = Sum[ Binomial[n-1, k]*b[k], {k, 0, n-1}]]; Table[a[n], {n, 0, m}] (* Jean-François Alcover, Mar 06 2013, after Maple *)

Bell_poly(n, x) = exp(-x)*suminf(k=0, k^n*x^k/k!);
a(n) = my(w=(-1+sqrt(3)*I)/2); round(Bell_poly(n, -1)+Bell_poly(n, -w)+Bell_poly(n, -w^2))/3; \\ Seiichi Manyama, Oct 15 2022

Define three sequences A(n), B(n) and C(n) by the relations: A(n+1) = - Sum_{i = 0..n} binomial(n,i)*C(i), B(n+1) = Sum_{i = 0..n} binomial(n,i)*A(i), C(n+1) = Sum_{i = 0..n} binomial(n,i)*B(i), with initial conditions A(0) = 1, B(0) = C(0) = 0. Then a(n) = A(n). The other sequences are B(n) = A143630 and C(n) = A143629. Compare with A143815. Also a(n) = A143629(n) + A000587(n).
From Seiichi Manyama, Oct 15 2022: (Start)
a(n) = Sum_{k = 0..floor(n/3)} (-1)^k * Stirling2(n,3*k).
a(n) = ( Bell_n(-1) + Bell_n(-w) + Bell_n(-w^2) )/3, where Bell_n(x) is n-th Bell polynomial and w = exp(2*Pi*i/3). (End)

A121868 Let A(0) = 1, B(0) = 0; A(n+1) = Sum_{k=0..n} binomial(n,k)B(k), B(n+1) = Sum_{k=0..n} -binomial(n,k)A(k); entry gives B sequence (cf. A121867).

Original entry on oeis.org

0, -1, -1, 0, 5, 23, 74, 161, -57, -3466, -27361, -155397, -687688, -1888525, 4974059, 134695952, 1400820897, 11055147275, 70658948426, 327448854237, 223871274083, -19116044475298, -314203665206509, -3562429698724513, -33024521386113840, -250403183401213513

Offset: 0

N. J. A. Sloane, Sep 05 2006

sign

Stirling transform of (I^(n+1)+(-I)^(n+1))/2 = (0,-1,0,1,..) repeated.

			From _Peter Bala_, Aug 28 2008: (Start)
E_2(k) as a linear combination of E_2(i), i = 0..1.
============================
..E_2(k)..|...E_2(0)..E_2(1)
============================
..E_2(2)..|....-1.......1...
..E_2(3)..|....-3.......0...
..E_2(4)..|....-6......-5...
..E_2(5)..|....-5.....-23...
..E_2(6)..|....33.....-74...
..E_2(7)..|...266....-161...
..E_2(8)..|..1309......57...
..E_2(9)..|..4905....3466...
(End)

Alois P. Heinz, Table of n, a(n) for n = 0..250
A. Fekete and G. Martin, Problem 10791: Squared Series Yielding Integers, Amer. Math. Monthly, 108 (No. 2, 2001), 177-178.
V. V. Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.

Cf. A121867, A024430, A024429.

Cf. A000587, A143623, A143624, A143628, A143631.

List([0..30], n-> Sum([0..Int(n/2)], k-> (-1)^(k+1)* Stirling2(n,2*k+1)) ); # G. C. Greubel, Oct 09 2019

[(&+[(-1)^(k+1)*StirlingSecond(n,2*k+1): k in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, Oct 09 2019

# Maple code for A024430, A024429, A121867, A121868.
M:=30; a:=array(0..100); b:=array(0..100); c:=array(0..100); d:=array(0..100); a[0]:=1; b[0]:=0; c[0]:=1; d[0]:=0;
for n from 1 to M do a[n]:=add(binomial(n-1,k)*b[k], k=0..n-1); b[n]:=add(binomial(n-1,k)*a[k], k=0..n-1); c[n]:=add(binomial(n-1,k)*d[k], k=0..n-1); d[n]:=-add(binomial(n-1,k)*c[k], k=0..n-1); od: ta:=[seq(a[n],n=0..M)]; tb:=[seq(b[n],n=0..M)]; tc:=[seq(c[n],n=0..M)]; td:=[seq(d[n],n=0..M)];
# Code based on Stirling transform:
stirtr:= proc(p) proc(n) option remember;
            add(p(k) *Stirling2(n, k), k=0..n) end
         end:
a:= stirtr(n-> (I^(n+1) + (-I)^(n+1))/2):
seq(a(n), n=0..30);  # Alois P. Heinz, Jan 29 2011

stirtr[p_] := Module[{f}, f[n_] := f[n] = Sum[p[k]*StirlingS2[n, k], {k, 0, n}]; f]; a = stirtr[(I^(#+1)+(-I)^(#+1))/2&]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Mar 11 2014, after Alois P. Heinz *)
Table[Im[BellB[n, -I]], {n, 0, 25}] (* Vladimir Reshetnikov, Oct 22 2015 *)

a(n) = sum(k=0,n\2, (-1)^(k+1)*stirling(n,2*k+1,2));
vector(30, n, a(n-1)) \\ G. C. Greubel, Oct 09 2019

[sum((-1)^(k+1)*stirling_number2(n,2*k+1) for k in (0..floor(n/2))) for n in (0..30)] # G. C. Greubel, Oct 09 2019

From Peter Bala, Aug 28 2008: (Start)
This sequence and its companion A121867 are related to the pair of constants cos(1) + sin(1) and cos(1) - sin(1) and may be viewed as generalizations of the Uppuluri-Carpenter numbers (complementary Bell numbers) A000587.
Define E_2(k) = Sum_{n >= 0} (-1)^floor(n/2) *n^k/n! for k = 0,1,2,... . Then E_2(0) = cos(1) + sin(1) and E_2(1) = cos(1) - sin(1). Furthermore, E_2(k) is an integral linear combination of E_2(0) and E_2(1) (a Dobinski-type relation). For example, E_2(2) = - E_2(0) + E_2(1), E_2(3) = -3*E_2(0) and E_2(4) = - 6*E_2(0) - 5*E_2(1). More examples are given below. The precise result is E_2(k) = A121867(k) * E_2(0) - A121868(k) * E_2(1).
For similar results see A143628. The decimal expansions of E_2(0) and E_2(1) are given in A143623 and A143624 respectively. (End)
From Vladimir Kruchinin, Jan 26 2011: (Start)
E.g.f.: A(x) = -sin(exp(x)-1).
a(n) = Sum_{k = 0..floor(n/2)} Stirling2(n,2*k+1)*(-1)^(k+1). (End)

A143630 Define E(n) = Sum_{k >= 0} (-1)^floor(k/3)*k^n/k! for n = 0,1,2,.... Then E(n) is an integral linear combination of E(0), E(1) and E(2). This sequence lists the coefficients of E(2).

Original entry on oeis.org

0, 0, 1, 3, 7, 14, 16, -77, -922, -6660, -41264, -233828, -1218392, -5607225, -19220589, 4397930, 1016675382, 14251497833, 151695504253, 1432992328055, 12527186450276, 102042171190168, 760272520469199, 4849866087637364

Offset: 0

Peter Bala, Sep 05 2008

This sequence and its companion sequences A143628 and A143629 may be viewed as generalizations of the Uppuluri-Carpenter numbers (complementary Bell numbers) A000587. Define E(n) = Sum_{k >= 0} (-1)^floor(k/3)*k^n/k! = 0^n/0! + 1^n/1! + 2^n/2! - 3^n/3! - 4^n/4! - 5^n/5! + + + - - - ... for n = 0,1,2,.... It is easy to see that E(n+3) = 3*E(n+2) - 2*E(n+1) - Sum_{i = 0..n} 3^i*binomial(n,i)*E(n-i) for n >= 0. Thus E(n) is an integral linear combination of E(0), E(1) and E(2). Some examples are given below. This sequence lists the coefficients of E(2). The precise result is E(n) = A143628(n)*E(0) + A143629(n)*E(1) + A143630(n)*E(2). Compare with A121867 and A143815.

			E(n) as linear combination of E(i),
i = 0..2.
====================================
..E(n)..|.....E(0)....E(1).....E(2).
====================================
..E(3)..|......-1......-2........3..
..E(4)..|......-6......-7........7..
..E(5)..|.....-25.....-23.......14..
..E(6)..|.....-89.....-80.......16..
..E(7)..|....-280....-271......-77..
..E(8)..|....-700....-750.....-922..
..E(9)..|....-380....-647....-6660..
..E(10).|...13452...13039...-41264..
...
a(5) = 14 because E(5) = -25*E(0) - 23*E(1) + 14*E(2).
a(6) = 16 because E(6) = -89*E(0) - 80*E(1) + 16*E(2).

Seiichi Manyama, Table of n, a(n) for n = 0..578
Eric Weisstein's World of Mathematics, Bell Polynomial.

A000587, A121867, A143628, A143629, A143631, A143815, A143816, A143817, A143818.

# Compare with A143817
#
M:=24: a:=array(0..100): b:=array(0..100): c:=array(0..100):
a[0]:=1: b[0]:=0: c[0]:=0:
for n from 1 to M do
a[n]:= -add(binomial(n-1,k)*c[k], k=0..n-1);
b[n]:= add(binomial(n-1,k)*a[k], k=0..n-1);
c[n]:= add(binomial(n-1,k)*b[k], k=0..n-1);
end do:
A143630:=[seq(c[n], n=0..M)];

m = 23; a[0] = 1; b[0] = 0; c[0] = 0; For[n = 1, n <= m, n++, a[n] = -Sum[Binomial[n - 1, k]*c[k], {k, 0, n - 1}]; b[n] = Sum[Binomial[n - 1, k]*a[k], {k, 0, n - 1}]; c[n] = Sum[Binomial[n - 1, k]*b[k], {k, 0, n - 1}]]; A143630 = Table[c[n], {n, 0, m}] (* Jean-François Alcover, Mar 06 2013, after Maple *)

Bell_poly(n, x) = exp(-x)*suminf(k=0, k^n*x^k/k!);
a(n) = my(w=(-1+sqrt(3)*I)/2); round(Bell_poly(n, -1)+w*Bell_poly(n, -w)+w^2*Bell_poly(n, -w^2))/3; \\ Seiichi Manyama, Oct 15 2022

Define three sequences A(n), B(n) and C(n) by the relations: A(n+1) = - Sum_{i = 0..n} binomial(n,i)*C(i), B(n+1) = Sum_{i = 0..n} binomial(n,i)*A(i), C(n+1) = Sum_{i = 0..n} binomial(n,i)*B(i), with initial conditions A(0) = 1, B(0) = C(0) = 0. Then a(n) = C(n). The other sequences are A(n) = A143628 and B(n) = A143631. Compare with A143817.
From Seiichi Manyama, Oct 15 2022: (Start)
a(n) = Sum_{k = 0..floor((n-2)/3)} (-1)^k * Stirling2(n,3*k+2).
a(n) = ( Bell_n(-1) + w * Bell_n(-w) + w^2 * Bell_n(-w^2) )/3, where Bell_n(x) is n-th Bell polynomial and w = exp(2*Pi*i/3). (End)

A143629 Define E(n) = Sum_{k>=0} (-1)^floor(k/3)*k^n/k! for n = 0,1,2,... . Then E(n) is an integral linear combination of E(0), E(1) and E(2). This sequence lists the coefficients of E(1).

Original entry on oeis.org

0, 1, 0, -2, -7, -23, -80, -271, -750, -647, 13039, 152011, 1232583, 8750796, 57405464, 349329354, 1899818951, 8008845556, 5981853002, -425732481925, -7285403175563, -89895756043392, -970910901819211, -9663021449412616

Offset: 0

Peter Bala, Sep 05 2008

This sequence and its companion sequences A143628 and A143630 may be viewed as generalizations of the Uppuluri-Carpenter numbers (complementary Bell numbers) A000587. Define E(n) = Sum_{k>=0} (-1)^floor(k/3)*k^n/k! = 0^n/0! + 1^n/1! + 2^n/2! - 3^n/3! - 4^n/4! - 5^n/5! + + + - - - ... for n = 0,1,2,... . It is easy to see that E(n+3) = 3*E(n+2) - 2*E(n+1) - Sum_{i = 0..n} 3^i*binomial(n,i)*E(n-i) for n >= 0. Thus E(n) is an integral linear combination of E(0), E(1) and E(2). This sequence lists the coefficients of E(1). Some examples are given below. The precise result for E(n) as a linear combination of E(0), E(1) and E(2) is E(n) = A143628(n)*E(0) + A143629(n)*E(1) + A143630(n)*E(2). Compare with A121867 and A143815.

			E(n) as linear combination of E(i),
i = 0..2.
====================================
..E(n)..|.....E(0).....E(1)....E(2).
====================================
..E(3)..|......-1......-2........3..
..E(4)..|......-6......-7........7..
..E(5)..|.....-25.....-23.......14..
..E(6)..|.....-89.....-80.......16..
..E(7)..|....-280....-271......-77..
..E(8)..|....-700....-750.....-922..
..E(9)..|....-380....-647....-6660..
..E(10).|...13452...13039...-41264..
...
a(5) = -23 because E(5) = -25*E(0) - 23*E(1) + 14*E(2).
a(6) = -80 because E(6) = -89*E(0) - 80*E(1) + 16*E(2).

A000587, A121867, A143628, A143630, A143631, A143815, A143816, A143817, A143818.

# Compare with A143818
M:=24: a:=array(0..100): b:=array(0..100): c:=array(0..100):
a[0]:=1: b[0]:=0: c[0]:=0:
for n from 1 to M do
a[n]:= -add(binomial(n-1,k)*c[k], k=0..n-1);
b[n]:= add(binomial(n-1,k)*a[k], k=0..n-1);
c[n]:= add(binomial(n-1,k)*b[k], k=0..n-1);
end do:
A143629:=[seq(b[n]-c[n], n=0..M)];

m = 23; a[0] = 1; b[0] = 0; c[0] = 0; For[n = 1, n <= m, n++, a[n] = -Sum[ Binomial[n - 1, k]*c[k], {k, 0, n - 1}]; b[n] = Sum[ Binomial[n - 1, k]*a[k], {k, 0, n - 1}]; c[n] = Sum[ Binomial[n - 1, k]*b[k], {k, 0, n - 1}] ]; A143629 = Table[b[n] - c[n], {n, 0, m}] (* Jean-François Alcover, Mar 06 2013, after Maple *)

Define three sequences A(n), B(n) and C(n) by the relations: A(n+1) = - Sum_{i = 0..n} binomial(n,i)*C(i), B(n+1) = Sum_{i = 0..n} binomial(n,i)*A(i), C(n+1) = Sum_{i = 0..n} binomial(n,i)*B(i), with initial conditions A(0) = 1, B(0) = C(0) = 0. Then a(n) = B(n) - C(n). The other sequences are A(n) = A143628(n) and C(n) = A143630(n). The values of B(n) are recorded in A143631. Compare with A143818. Also a(n) = A143628(n) - A000587(n).

A153229 a(0) = 0, a(1) = 1, and for n >= 2, a(n) = (n-1) * a(n-2) + (n-2) * a(n-1).

Original entry on oeis.org

0, 1, 0, 2, 4, 20, 100, 620, 4420, 35900, 326980, 3301820, 36614980, 442386620, 5784634180, 81393657020, 1226280710980, 19696509177020, 335990918918980, 6066382786809020, 115578717622022980, 2317323290554617020, 48773618881154822980, 1075227108896452857020

Offset: 0

Shaojun Ying (dolphinysj(AT)gmail.com), Dec 21 2008

Previous name was: Weighted Fibonacci numbers.
From Peter Bala, Aug 18 2013: (Start)
The sequence occurs in the evaluation of the integral I(n) := Integral_{u >= 0} exp(-u)*u^n/(1 + u) du.
The result is I(n) = A153229(n) + (-1)^n*I(0), where I(0) = Integral_{u >= 0} exp(-u)/(1 + u) du = 0.5963473623... is known as Gompertz's constant. See A073003.
Note also that I(n) = n!*Integral_{u >= 0} exp(-u)/(1 + u)^(n+1) du. (End)
((-1)^(n+1))*a(n) = p(n,-1), where the polynomials p are defined at A248664. - Clark Kimberling, Oct 11 2014

			a(20) = 19 * a(18) + 18 * a(19) = 19 * 335990918918980 + 18 * 6066382786809020 = 6383827459460620 + 109194890162562360 = 115578717622022980

Alois P. Heinz, Table of n, a(n) for n = 0..200

Cf. A000045, A000255, A000587, A058006.
First differences of A136580.
Column k=0 of A303697 (for n>0).

unsigned long a(unsigned int n) {
if (n == 0) return 0;
if (n == 1) return 1;
return (n - 1) * a(n - 2) + (n - 2) * a(n - 1); }

t1 := sum(n!*x^n, n=0..100): F := series(t1/(1+x), x, 100): for i from 0 to 40 do printf(`%d, `, i!-coeff(F, x, i)) od: # Zerinvary Lajos, Mar 22 2009
# second Maple program:
a:= proc(n) a(n):= `if`(n<2, n, (n-1)*a(n-2) +(n-2)*a(n-1)) end:
seq(a(n), n=0..25); # Alois P. Heinz, May 24 2013

Join[{a = 0}, Table[b = n! - a; a = b, {n, 0, 100}]] (* Vladimir Joseph Stephan Orlovsky, Jun 28 2011 *)
RecurrenceTable[{a[0]==0,a[1]==1,a[n]==(n-1)a[n-2]+(n-2)a[n-1]},a,{n,30}] (* Harvey P. Dale, May 01 2020 *)

a(n)=if(n,my(t=(-1)^n);-t-sum(i=1,n-1,t*=-i),0); \\ Charles R Greathouse IV, Jun 28 2011

a(0) = 0, a(1) = 1, and for n >= 2, a(n) = (n-1) * a(n-2) + (n-2) * a(n-1).
For n>=1, a(n) = A058006(n-1) * (-1)^(n-1).
G.f.: G(0)*x/(1+x)/2, where G(k)= 1 + 1/(1 - x*(k+1)/(x*(k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
G.f.: 2*x/(1+x)/G(0), where G(k)= 1 + 1/(1 - 1/(1 - 1/(2*x*(k+1)) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 29 2013
G.f.: W(0)*x/(1+sqrt(x))/(1+x), where W(k) = 1 + sqrt(x)/( 1 - sqrt(x)*(k+1)/(sqrt(x)*(k+1) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 17 2013
a(n) ~ (n-1)! * (1 - 1/n + 1/n^3 + 1/n^4 - 2/n^5 - 9/n^6 - 9/n^7 + 50/n^8 + 267/n^9 + 413/n^10), where numerators are Rao Uppuluri-Carpenter numbers, see A000587. - Vaclav Kotesovec, Mar 16 2015
E.g.f.: exp(1)/exp(x)*(Ei(1, 1-x)-Ei(1, 1)). - Alois P. Heinz, Jul 05 2018
a(n) = Sum_{k = 0..n-1} (-1)^(n-k-1) * k!. - Peter Bala, Dec 05 2024

Edited by Max Alekseyev, Jul 05 2010

Better name by Joerg Arndt, Aug 17 2013

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

Jan Fricke, Aug 14 2002

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

			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.

Robert Israel, Table of n, a(n) for n = 0..545

Cf. A074052, A143494.

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

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 *)

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

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

More terms from Vladeta Jovovic, Jan 27 2005

cp's OEIS Frontend

A293037 E.g.f.: exp(1 + x - exp(x)).

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A014182 Expansion of e.g.f. exp(1-x-exp(-x)).

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A189233 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals upwards, where the e.g.f. of column k is exp(k*(e^x-1)).

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A121867 Let A(0) = 1, B(0) = 0; A(n+1) = Sum_{k=0..n} binomial(n,k)*B(k), B(n+1) = Sum_{k=0..n} -binomial(n,k)*A(k); entry gives A sequence (cf. A121868).

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A143628 Define E(n) = Sum_{k >= 0} (-1)^floor(k/3)*k^n/k! for n = 0,1,2,... . Then E(n) is an integral linear combination of E(0), E(1) and E(2). This sequence lists the coefficients of E(0).

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A121868 Let A(0) = 1, B(0) = 0; A(n+1) = Sum_{k=0..n} binomial(n,k)*B(k), B(n+1) = Sum_{k=0..n} -binomial(n,k)*A(k); entry gives B sequence (cf. A121867).

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A121867 Let A(0) = 1, B(0) = 0; A(n+1) = Sum_{k=0..n} binomial(n,k)B(k), B(n+1) = Sum_{k=0..n} -binomial(n,k)A(k); entry gives A sequence (cf. A121868).

A121868 Let A(0) = 1, B(0) = 0; A(n+1) = Sum_{k=0..n} binomial(n,k)B(k), B(n+1) = Sum_{k=0..n} -binomial(n,k)A(k); entry gives B sequence (cf. A121867).

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.