A126390 -id:A126390 - cp's OEIS frontend

A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

1, 1, 3, 5, 15, 33, 91, 221, 583, 1465, 3795, 9653, 24831, 63441, 162763, 416525, 1067575, 2733673, 7003971, 17938661, 45954543, 117709185, 301527355, 772364093, 1978473511

Offset: 0

Clark Kimberling, Aug 02 2011

nonn

Suppose that p=p(0)*x^n+p(1)*x^(n-1)+...+p(n-1)*x+p(n) is a polynomial of positive degree and that Q is a sequence of polynomials: q(k,x)=t(k,0)*x^k+t(k,1)*x^(k-1)+...+t(k,k-1)*x+t(k,k), for k=0,1,2,... The Q-downstep of p is the polynomial given by D(p)=p(0)*q(n-1,x)+p(1)*q(n-2,x)+...+p(n-1)*q(0,x)+p(n).

Since degree(D(p))

Example: let p(x)=2*x^3+3*x^2+4*x+5 and q(k,x)=(x+1)^k.

D(p)=2(x+1)^2+3(x+1)+4(1)+5=2x^2+7x+14

D(D(p))=2(x+1)+7(1)+14=2x+23

D(D(D(p)))=2(1)+23=25;

the Q-residue of p is 25.

We may regard the sequence Q of polynomials as the triangular array formed by coefficients:

t(0,0)

t(1,0)....t(1,1)

t(2,0)....t(2,1)....t(2,2)

t(3,0)....t(3,1)....t(3,2)....t(3,3)

and regard p as the vector (p(0),p(1),...,p(n)). If P is a sequence of polynomials [or triangular array having (row n)=(p(0),p(1),...,p(n))], then the Q-residues of the polynomials form a numerical sequence.

Following are examples in which Q is the triangle given by t(i,j)=1 for 0<=i<=j:

Q.....P...................Q-residue of P

1.....1...................A000079, 2^n

1....(x+1)^n..............A007051, (1+3^n)/2

1....(x+2)^n..............A034478, (1+5^n)/2

1....(x+3)^n..............A034494, (1+7^n)/2

1....(2x+1)^n.............A007582

1....(3x+1)^n.............A081186

1....(2x+3)^n.............A081342

1....(3x+2)^n.............A081336

1.....A040310.............A193649

1....(x+1)^n+(x-1)^n)/2...A122983

1....(x+2)(x+1)^(n-1).....A057198

1....(1,2,3,4,...,n)......A002064

1....(1,1,2,3,4,...,n)....A048495

1....(n,n+1,...,2n).......A087323

1....(n+1,n+2,...,2n+1)...A099035

1....p(n,k)=(2^(n-k))*3^k.A085350

1....p(n,k)=(3^(n-k))*2^k.A090040

1....A008288 (Delannoy)...A193653

1....A054142..............A101265

1....cyclotomic...........A193650

1....(x+1)(x+2)...(x+n)...A193651

1....A114525..............A193662

More examples:

Q...........P.............Q-residue of P

(x+1)^n...(x+1)^n.........A000110, Bell numbers

(x+1)^n...(x+2)^n.........A126390

(x+2)^n...(x+1)^n.........A028361

(x+2)^n...(x+2)^n.........A126443

(x+1)^n.....1.............A005001

(x+2)^n.....1.............A193660

A094727.....1.............A193657

(k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)

(k+1).....(x+1)^n.........A112091

(x+1)^n...(k+1)...........A029761

(k+1)......A049310........A193663

(In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)

A051162...(x+1)^n.........A193658

A094727...(x+1)^n.........A193659

A049310...(x+1)^n.........A193664

A075362....A075362........A193665

Changing the notation slightly leads to the Mathematica program below and the following formulation for the Q-downstep of p: first, write t(n,k) as q(n,k). Define r(k)=Sum{q(k-1,i)*r(k-1-i) : i=0,1,...,k-1} Then row n of D(p) is given by v(n)=Sum{p(n,k)*r(n-k) : k=0,1,...,n}.

			First five rows of Q, coefficients of Fibonacci polynomials (A049310):
1
1...0
1...0...1
1...0...2...0
1...0...3...0...1
To obtain a(4)=15, downstep four times:
D(x^4+3*x^2+1)=(x^3+x^2+x+1)+3(x+1)+1: (1,1,4,5) [coefficients]
DD(x^4+3*x^2+1)=D(1,1,4,5)=(1,2,11)
DDD(x^4+3*x^2+1)=D(1,2,11)=(1,14)
DDDD(x^4+3*x^2+1)=D(1,14)=15.

Cf. A192872 (polynomial reduction), A193091 (polynomial augmentation), A193722 (the upstep operation and fusion of polynomial sequences or triangular arrays).

q[n_, k_] := 1;
r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}];
f[n_, x_] := Fibonacci[n + 1, x];
p[n_, k_] := Coefficient[f[n, x], x, k]; (* A049310 *)
v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
Table[v[n], {n, 0, 24}]    (* A193649 *)
TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
Table[r[k], {k, 0, 8}]  (* 2^k *)
TableForm[Table[p[n, k], {n, 0, 6}, {k, 0, n}]]

Conjecture: G.f.: -(1+x)*(2*x-1) / ( (x-1)*(4*x^2+x-1) ). - R. J. Mathar, Feb 19 2015

A126617 a(n) = Sum_{i=0..n} (-2)^(n-i)B(i)binomial(n,i) where B(n) = Bell numbers A000110(n).

Original entry on oeis.org

1, -1, 2, -3, 7, -10, 31, -21, 204, 307, 2811, 12100, 74053, 432211, 2768858, 18473441, 129941283, 956187814, 7351696139, 58897405759, 490681196604, 4242903803727, 38014084430983, 352341755256348, 3373662303816313, 33326335433122711, 339232538387804530

Offset: 0

N. J. A. Sloane, Aug 04 2007

sign

a(n) is positive starting at n=8. - Karol A. Penson and Olivier Gérard, Oct 22 2007

Hankel transform is A000178. - Paul Barry, Apr 23 2009

			G.f.: 1 - 1*x + 2*x^2 - 3*x^3 + 7*x^4 - 10*x^5 + 31*x^6 - 21*x^7 + 204*x^8 + 307*x^9 + 2811*x^10 + 12100*x^11 + 74053*x^12 + 432211*x^13 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000110, A000296, A005493, A124311, A126390, A153732.

Table[ Sum[ (-2)^(n - k) Binomial[n, k] BellB[k], {k, 0, n}], {n, 0, 50}] (* Karol A. Penson and Olivier Gérard, Oct 22 2007 *)
With[{nn=30},CoefficientList[Series[Exp[Exp[x]-2x-1],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Jan 19 2025 *)

E.g.f.: exp(exp(x)-2*x-1). - Vladeta Jovovic, Aug 04 2007
a(n) = e^(-1) * Sum_{k>=0} (k-2)^n / k!. This is a Dobinski-type formula. - Karol A. Penson and Olivier Gérard, Oct 22 2007
G.f.: 1/(1+x-x^2/(1-2x^2/(1-x-3x^2/(1-2x-4x^2/(1-3x-5x^2/(1-.... (continued fraction). - Paul Barry, Apr 23 2009
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>=1, a(n)=(-1)^(n)charpoly(A,2). - Milan Janjic, Jul 08 2010
G.f.: -1/U(0)  where U(k) = x*k - 1 - x - x^2*(k+1)/U(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 28 2012
G.f.: 1/G(0) where G(k) = 1 + 2*x/(1 + 1/(1 - 2*x*(k+1)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 23 2012
G.f.: G(0)/(1+3*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-x-1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 19 2012
From Sergei N. Gladkovskii, Feb 13 2013: (Start)
Conjecture: if the e.g.f. is E(x)= exp( exp(x) -1 + p*x) then
g.f.: (x+1-p*x)/x/(G(0)-x) - 1/x where G(k) = 2*x + 1 - p*x - x*k + x*(x*k - x - 1 + p*x)/G(k+1); (continued fraction).
So, for this sequence (p=-2), g.f.: (3*x+1)/x/( G(0)-x ) - 1/x where G(k) = 4*x + 1 - x*k + x*(x*k - 3*x - 1)/G(k+1);
(End)
G.f.: 1/Q(0), where Q(k) = 1 + 2*x - x/(1 - x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 22 2013
a(0) = 1; a(n) = -2 * a(n-1) + Sum_{k=0..n-1} binomial(n-1,k) * a(k). - Ilya Gutkovskiy, Jul 30 2021
a(n) ~ n^(n-2) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n-2)). - Vaclav Kotesovec, Jun 27 2022

More terms from Karol A. Penson and Olivier Gérard, Oct 22 2007

A154537 Triangle T(n,m) read by rows: let p(n,x) = exp(-x) * Sum_{m >= 0} (2m + 1)^n x^m/m!; then T(n,m) = [x^m] p(n,x).

Original entry on oeis.org

1, 1, 2, 1, 8, 4, 1, 26, 36, 8, 1, 80, 232, 128, 16, 1, 242, 1320, 1360, 400, 32, 1, 728, 7084, 12160, 6320, 1152, 64, 1, 2186, 36876, 99288, 81200, 25312, 3136, 128, 1, 6560, 188752, 768768, 929376, 440832, 91392, 8192, 256, 1, 19682, 956880, 5758880, 9901920, 6707904, 2069760, 305664, 20736, 512

Offset: 0

Roger L. Bagula, Jan 11 2009

Row sums are A126390.
These numbers are related to Stirling numbers of the second kind as MacMahon numbers A060187 are related to Eulerian numbers.
Let p and q denote operators acting on a function f(x) by pf(x) = x*f(x) and qf(x) = d/dx(f(x)). Let A be the anticommutator operator qp + pq. Then A^n = Sum_{k = 0..n} T(n,k) p^k q^k. For example, A^3(f) = f + 26*x*df/dx + 36*x^2*d^2(f)/dx^2 + 8*x^3*d^3(f)/dx^3. - Peter Bala, Jul 24 2014
From Peter Bala, May 21 2023: (Start)
Compare the definition of the polynomial p(n,x) with Dobiński's formula for the Bell polynomials (row polynomials of A008277 for n >= 1): Bell(n,x) = exp(-x) * Sum_{m >= 0} m^n * x^m/m!.
Boyadzhiev has shown that Bell(n,x) = d/dx( exp(-x) * Sum_{m >= 0} (1^n + 2^n + ... + (m-1)^n) * x^m/m! ). The corresponding result for this table is that the n-th row polynomial p(n,x) = d/dx( exp(-x) * Sum_{m >= 0} (1^n + 3^n + ... + (2*m-1)^n) * x^m/m! ). (End)

			Triangle begins:
  {1},
  {1, 2},
  {1, 8, 4},
  {1, 26, 36, 8},
  {1, 80, 232, 128, 16},
  {1, 242, 1320, 1360, 400, 32},
  {1, 728, 7084, 12160, 6320, 1152, 64},
  {1, 2186, 36876, 99288, 81200, 25312, 3136, 128},
  {1, 6560, 188752, 768768, 929376, 440832, 91392, 8192, 256},
  {1, 19682, 956880, 5758880, 9901920, 6707904, 2069760, 305664, 20736, 512},
  ...
Boas-Buck recurrence for column m = 2, and n = 4: T(4,2) = (1/2)*[4*3*T(3, 2) + 2*6*(-2)^2*Bernoulli(2)*T(2,2)] = (1/2)*(12*36 + 12*4*(1/6)*4) = 232. - _Wolfdieter Lang_, Aug 11 2017

Khristo N. Boyadzhiev, New identities with Stirling, hyperharmonic, and derangement numbers, Bernoulli and Euler polynomials, powers, and factorials, arXiv:2011.03101v3 [math.NT], 2020-2021.
Paweł Hitczenko, A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality, arXiv:2403.03422 [math.CO], 2024. See p. 9.
Wolfdieter Lang, On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
Eric Weisstein's World of Mathematics, Dobiński's formula

Cf. A008277, A000110, A039755, A048993, A060187.

p[x_, n_] = Sum[(2*m + 1)^n*x^m/m!, {m, 0, Infinity}]/(Exp[x]);
Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}]
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}];
Flatten[%]

From Peter Bala, Oct 28 2011: (Start)
T(n,k) = 1/k!*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n.
Recurrence relation: T(n,k) = 2*T(n-1,k-1) + (2*k+1)*T(n-1,k).
T(n,k) = (2^k)*A039755(n,k).
E.g.f.: exp(x + y*(exp(2*x) - 1)) = 1 + (1 + 2*y)*x + (1 + 8*y + 4*y^2)*x^2/2! + .... (End)
T(n, k) = Sum_{m=0..n} binomial(n, m)*2^m*Stirling2(m, k), 0 <= k <= n, where Stirling2 is A048993. - Wolfdieter Lang, Apr 13 2017
Boas-Buck recurrence for column sequence m: T(n,k) = (1/(n - k))*[n*(1 + m)*T(n-1,k) + k*Sum_{p=m..n-2} binomial(n,p)*(-2)^(n-p)*Bernoulli(n-p)*T(p,k)], for n > m >= 0, with input T(m,m) = 2^m. See a comment in A282629, also for references, and an example below. - Wolfdieter Lang, Aug 11 2017

Edited by N. J. A. Sloane, Jan 12 2009

A124311 a(n) = Sum_{i=0..n} (-2)^ibinomial(n,i)B(i) where B(n) = Bell numbers A000110(n).

Original entry on oeis.org

1, -1, 5, -21, 121, -793, 5917, -49101, 447153, -4421105, 47062773, -535732805, 6484924585, -83079996041, 1121947980173, -15915567647101, 236442490569825, -3668776058118881, 59316847871113445, -997182232031471477, 17397298225094055897, -314449131128077197561

Offset: 0

N. J. A. Sloane, Aug 04 2007

sign

The sequence has strictly alternating signs. The variant Dobinski-type formula e^(-1)* (2)^n * Sum_{k >= 0} ( (k-1/2)^n / k! ) is strictly positive. - Karol A. Penson and Olivier Gérard, Oct 22 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000110, A000296, A005493, A126390, A126617.

A124311:= func< n | (&+[(-2)^k*Binomial(n,k)*Bell(k): k in [0..n]]) >;
[A124311(n): n in [0..30]]; // G. C. Greubel, Aug 25 2023

Table[ Sum[ (-2)^(k) Binomial[n, k] BellB[k], {k, 0, n}], {n, 0, 50}] (* Karol A. Penson and Olivier Gérard, Oct 22 2007 *)
With[{nn=30},CoefficientList[Series[Exp[Exp[-2x]-1+x],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Mar 04 2016 *)

def A124311_list(n):  # n>=1
    T = [0]*(n+1); R = [1]
    for m in (1..n-1):
        a,b,c = 1,0,0
        for k in range(m,-1,-1):
            r = a + 2*(k*(b+c)+c)
            if k < m : T[k+2] = u;
            a,b,c = T[k-1],a,b
            u = r
        T[1] = u;
        R.append((-1)^m*sum(T))
    return R
A124311_list(22)  # Peter Luschny, Nov 02 2012

def A124311(n): return sum( (-2)^k*binomial(n,k)*bell_number(k) for k in range(n+1) )
[A124311(n) for n in range(31)] # G. C. Greubel, Aug 25 2023

E.g.f.: exp(exp(-2*x) - 1 + x). - Vladeta Jovovic, Aug 04 2007
G.f.: 1/U(0) where U(k)=  1 + x*(2*k+1) - 4*x^2*(k+1)/U(k+1) ; (continued fraction, 1-step). - Sergei N. Gladkovskii, Oct 11 2012
a(n) ~ (-2)^n * n^(n - 1/2) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n - 1/2)). - Vaclav Kotesovec, Jun 26 2022
a(0) = 1; a(n) = a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * (-2)^k * a(n-k). - Ilya Gutkovskiy, Nov 29 2023

A284859 Row sums of the Sheffer triangle (exp(x), exp(3*x)-1) given in A282629.

Original entry on oeis.org

1, 4, 25, 199, 1876, 20257, 245017, 3266914, 47450923, 743935375, 12497579698, 223619318215, 4240423494685, 84855613320004, 1785410320771933, 39373503608087299, 907548770965519660, 21810536356271794549, 545305573054110017125, 14155835044848094831018

Offset: 0

Wolfdieter Lang, Apr 05 2017

See A282629 for details. These are generalized Bell numbers (A000110) because A282629 is a generalized Stirling2 triangle.

Cf. A000110, A126390, A282629, A284864.

T[n_, m_]:= Sum[Binomial[m, k] (-1)^(k - m) (1 + 3k)^n/m!, {k, 0, m}]; Table[Sum[T[n, m], {m, 0, n}], {n, 0, 20}] (* Indranil Ghosh, Apr 10 2017 *)
Table[Sum[3^k*Binomial[n,k]*BellB[k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jun 22 2022 *)

T(n, m) = sum(k=0, m, binomial(m, k) * (-1)^(k - m) * (1 + 3*k)^n/m!);
a(n) = sum(m=0, n, T(n, m)); \\ Indranil Ghosh, Apr 10 2017

from sympy import binomial, factorial
def T(n, m): return sum([binomial(m, k) * (-1)**(k - m) * (1 + 3*k)**n for k in range(m + 1)])//factorial(m)
def a(n): return sum([T(n, k) for k in range(n + 1)])
print([a(n) for n in range(20)]) # Indranil Ghosh, Apr 10 2017

a(n) = Sum_{m=0..n} A282629(n, m).
E.g.f.: exp(x)*exp(exp(3*x) -1).
a(n) = (1/e)*Sum_{m>=0} (1/m!)*(1+3*m)^n, n >= 0. (Dobiński type formula from the A282629(n,m) sum formula, interchanging summations).
a(0) = 1; a(n) = a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 3^k * a(n-k). - Ilya Gutkovskiy, Jun 21 2022
a(n) ~ Bell(n) * (3 + LambertW(n)/n)^n. - Vaclav Kotesovec, Jun 22 2022
a(n) ~ 3^n * n^(n + 1/3) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n + 1/3)). - Vaclav Kotesovec, Jun 27 2022

A285064 Row sums of Sheffer triangle S2[4,1] = A285061.

Original entry on oeis.org

1, 5, 41, 429, 5329, 75989, 1215481, 21453693, 412820385, 8579772325, 191166679497, 4538638641997, 114238219541617, 3035305413035125, 84819458105387417, 2484842038066995485, 76101249873390595905, 2430497813260105226053, 80769536433102942870377, 2787318255464814752951533

Offset: 0

Wolfdieter Lang, Apr 13 2017

See A285061 for details. These are generalized Bell numbers (A000110) because A285061 is a generalized Stirling2 triangle.

For the alternating row sums of A285061 see A285065.

Cf. A000110, A285061, A285065.

Cf. A126390, A284859.

Table[Sum[Binomial[n, k]*BellB[k]*4^k, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 19 2017 *)

from sympy import binomial, bell
def a(n): return sum([binomial(n, k)*bell(k)*4**k for k in range(n + 1)]) # Indranil Ghosh, Apr 19 2017

a(n) = Sum_{m=0..n} A285061(n, m), n >= 0.
E.g.f.: exp(x)*exp(exp(4*x) - 1).
a(n) = (1/e)*Sum_{m>=0} (1/m!)*(1+4*m)^n, n >= 0. (Dobiński type formula from the A285061(n,m) sum formula, after interchange of summations).
a(n) = Sum_{k=0..n} binomial(n, k)*A000110(k)*4^k, n >= 0. From the Vaclav Kotesovec program. This follows from the S2[4,1] formula in terms of Stirling2. - Wolfdieter Lang, Apr 24 2017
a(0) = 1; a(n) = a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 4^k * a(n-k). - Ilya Gutkovskiy, Jun 21 2022
a(n) ~ Bell(n) * (4 + LambertW(n)/n)^n. - Vaclav Kotesovec, Jun 22 2022
a(n) ~ 4^n * n^(n + 1/4) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n + 1/4)). - Vaclav Kotesovec, Jun 27 2022

A355291 Expansion of e.g.f. exp(exp(x)*(exp(x) + 1) - 2).

Original entry on oeis.org

1, 3, 14, 81, 551, 4266, 36803, 348543, 3583484, 39652659, 468970211, 5894584812, 78366374813, 1097537989671, 16136598952718, 248309032411485, 3988468487017379, 66715970326561170, 1159712730763363991, 20909709414253764819, 390374806223071148084, 7534929383736826736007

Offset: 0

Vaclav Kotesovec, Jun 27 2022

nonn

In general, if m > 0, b > d >= 1 and e.g.f. = exp(m*exp(b*x) + r*exp(d*x) + s) then a(n) ~ exp(m*exp(b*z) + r*exp(d*z) + s - n) * (n/z)^(n + 1/2) / sqrt(m*b*(1 + b*z)*exp(b*z) + r*d*(1 + d*z)*exp(d*z)), where z = LambertW(n/m)/b - 1/(d + b/LambertW(n/m) + b^2 * m^(d/b) * n^(1 - d/b) * (1 + LambertW(n/m)) / (d*r*LambertW(n/m)^(2 - d/b))). - Vaclav Kotesovec, Jul 03 2022

In addition, if b/d >=2 then a(n) ~ c * (b*n/LambertW(n/m))^n * exp(n/LambertW(n/m) + r * (n/(m*LambertW(n/m)))^(d/b) - n + s) / sqrt(1 + LambertW(n/m)), where c = 1 for b/d > 2 and c = exp(-r^2/(8*m)) for b/d = 2. - Vaclav Kotesovec, Jul 10 2022

Seiichi Manyama, Table of n, a(n) for n = 0..500
Vaclav Kotesovec, Asymptotics for a certain group of exponential generating functions, arXiv:2207.10568 [math.CO], Jul 13 2022.

Cf. A001861, A055882, A126390, A143405, A355379.

nmax = 20; CoefficientList[Series[Exp[Exp[2*x] - 2 + Exp[x]], {x, 0, nmax}], x] * Range[0, nmax]!
Table[Sum[Binomial[n, k] * 2^k * BellB[k] * BellB[n-k], {k, 0, n}], {n, 0, 20}]

my(x='x+O('x^30)); Vec(serlaplace(exp(exp(x)*(exp(x) + 1) - 2))) \\ Michel Marcus, Jun 27 2022

a(n) = Sum_{k=0..n} binomial(n,k) * 2^k * Bell(k) * Bell(n-k).
a(0) = 1; a(n) = Sum_{k=1..n} (1 + 2^k) * binomial(n-1,k-1) * a(n-k). - Seiichi Manyama, Jul 01 2022
a(n) ~ exp(exp(2*z) + exp(z) - 2 - n) * (n/z)^(n + 1/2) / sqrt(2*(1 + 2*z)*exp(2*z) + (1 + z)*exp(z)), where z = LambertW(n)/2 - 1/(1 + 2/LambertW(n) + 4 * n^(1/2) * (1 + LambertW(n)) / LambertW(n)^(3/2)). - Vaclav Kotesovec, Jul 03 2022
a(n) ~ 2^n * n^n / (sqrt(1 + LambertW(n)) * LambertW(n)^n * exp(n + 17/8 - n/LambertW(n) - sqrt(n/LambertW(n)))). - Vaclav Kotesovec, Jul 08 2022

A367835 Expansion of e.g.f. 1/(2 - x - exp(2*x)).

Original entry on oeis.org

1, 3, 22, 242, 3544, 64872, 1424976, 36517840, 1069533824, 35240047232, 1290137297152, 51955085596416, 2282489348834304, 108630445541684224, 5567741266098944000, 305752314499878569984, 17909736027185859100672, 1114647522476340562132992

Offset: 0

Seiichi Manyama, Dec 02 2023

nonn

Cf. A006155, A367836, A367837.

Cf. A126390, A216794, A343672, A355110, A367838.

A367835 := proc(n)
    option remember ;
    if n = 0 then
        1 ;
    else
        n*procname(n-1)+add(2^k*binomial(n,k)*procname(n-k),k=1..n) ;
    end if;
end proc:
seq(A367835(n),n=0..70) ; # R. J. Mathar, Dec 04 2023

a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=i*v[i]+sum(j=1, i, 2^j*binomial(i, j)*v[i-j+1])); v;

a(0) = 1; a(n) = n * a(n-1) + Sum_{k=1..n} 2^k * binomial(n,k) * a(n-k).

A307066 a(n) = exp(-1) * Sum_{k>=0} (n*k + 1)^n/k!.

Original entry on oeis.org

1, 2, 13, 199, 5329, 216151, 12211597, 909102342, 85761187393, 9957171535975, 1390946372509101, 229587693339867567, 44117901231194922193, 9748599124579281064294, 2451233017637221706477037, 695088863051920283838281851, 220558203335628758134165860609

Offset: 0

Ilya Gutkovskiy, Jun 24 2019

nonn

G. C. Greubel, Table of n, a(n) for n = 0..250

Cf. A000110, A126390, A134980, A284859, A285064, A292914, A307080.

A307066:= func< n | (&+[Binomial(n,k)*n^k*Bell(k): k in [0..n]]) >;
[A307066(n): n in [0..31]]; // G. C. Greubel, Jan 24 2024

Table[Exp[-1] Sum[(n k + 1)^n/k!, {k, 0, Infinity}], {n, 0, 16}]
Table[n! SeriesCoefficient[Exp[Exp[n x] + x - 1], {x, 0, n}], {n, 0, 16}]
Join[{1}, Table[Sum[Binomial[n, k] n^k BellB[k], {k, 0, n}], {n, 1, 16}]]

def A307066(n): return sum(binomial(n,k)*n^k*bell_number(k) for k in range(n+1))
[A307066(n) for n in range(31)] # G. C. Greubel, Jan 24 2024

a(n) = n! * [x^n] exp(exp(n*x) + x - 1).

a(n) = Sum_{k=0..n} binomial(n,k) * n^k * Bell(k).

A346417 E.g.f.: exp(exp(2*(exp(x) - 1)) - 1).

Original entry on oeis.org

1, 2, 10, 66, 538, 5186, 57402, 714594, 9853978, 148774914, 2436823034, 42979319202, 811254807770, 16302732719682, 347248840767162, 7809649226242530, 184831773033020826, 4589793199157616770, 119272846472231229818, 3235960069037751550498, 91466308730323104617050

Offset: 0

Ilya Gutkovskiy, Jul 16 2021

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..449

Cf. A000258, A000898, A001861, A055882, A126390, A136658.

b:= proc(n, t, m) option remember; `if`(n=0, `if`(t=1, 1,
      b(m, 1, 0)*2^m) , m*b(n-1, t, m)+b(n-1, t, m+1))
    end:
a:= n-> b(n, 0$2):
seq(a(n), n=0..20);  # Alois P. Heinz, Aug 06 2021

nmax = 20; CoefficientList[Series[Exp[Exp[2 (Exp[x] - 1)] - 1], {x, 0, nmax}], x] Range[0, nmax]!
Table[Sum[StirlingS2[n, k] 2^k BellB[k], {k, 0, n}], {n, 0, 20}]
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k - 1] BellB[k, 2] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 20}]

my(x='x+O('x^25)); Vec(serlaplace(exp(exp(2*(exp(x) - 1)) - 1))) \\ Michel Marcus, Jul 19 2021

a(n) = Sum_{k=0..n} Stirling2(n,k) * 2^k * Bell(k).

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * A001861(k) * a(n-k).

Showing 1-10 of 15 results. Next

cp's OEIS Frontend

A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

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A126617 a(n) = Sum_{i=0..n} (-2)^(n-i)*B(i)*binomial(n,i) where B(n) = Bell numbers A000110(n).

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A154537 Triangle T(n,m) read by rows: let p(n,x) = exp(-x) * Sum_{m >= 0} (2*m + 1)^n * x^m/m!; then T(n,m) = [x^m] p(n,x).

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A124311 a(n) = Sum_{i=0..n} (-2)^i*binomial(n,i)*B(i) where B(n) = Bell numbers A000110(n).

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A284859 Row sums of the Sheffer triangle (exp(x), exp(3*x)-1) given in A282629.

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A285064 Row sums of Sheffer triangle S2[4,1] = A285061.

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A355291 Expansion of e.g.f. exp(exp(x)*(exp(x) + 1) - 2).

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A126617 a(n) = Sum_{i=0..n} (-2)^(n-i)B(i)binomial(n,i) where B(n) = Bell numbers A000110(n).

A154537 Triangle T(n,m) read by rows: let p(n,x) = exp(-x) * Sum_{m >= 0} (2m + 1)^n x^m/m!; then T(n,m) = [x^m] p(n,x).

A124311 a(n) = Sum_{i=0..n} (-2)^ibinomial(n,i)B(i) where B(n) = Bell numbers A000110(n).