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.

Showing 1-10 of 12 results. Next

A143395 Triangle read by rows: T(n,k) = number of forests of k labeled rooted trees of height at most 1, with n labels, where any root may contain >= 1 labels, n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 7, 9, 1, 0, 15, 55, 18, 1, 0, 31, 285, 205, 30, 1, 0, 63, 1351, 1890, 545, 45, 1, 0, 127, 6069, 15421, 7770, 1190, 63, 1, 0, 255, 26335, 116298, 95781, 24150, 2282, 84, 1, 0, 511, 111645, 830845, 1071630, 416451, 62370, 3990, 108, 1
Offset: 0

Views

Author

Alois P. Heinz, Aug 12 2008

Keywords

Comments

This is the Sheffer triangle (1,exp(x)*(exp(x)-1)) (Jobotinsky type). See the e.g.f. given by V. Jovovic below, and the W. Lang link under A006232 (second part) for general Sheffer remarks and the conversion to the umbral notation of S. Roman's book. - Wolfdieter Lang, Oct 08 2011
From Peter Bala, Jan 07 2015: (Start)
T(n,k) counts the ways a set of size n can be partitioned into k nonempty blocks and then a nonempty subset chosen from each block. An example is given below.
This triangle is the particular case a = 1, b = 1, c = 0 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. A008277 is the case a = 1, b = 0, c = 0.
Define a polynomial sequence x_(n) by putting x_(0) = 1, x_(1) = x and for n >= 2 setting x_(n) = x*(x - (n+1))*(x - (n+2))*...*(x - (2*n-1)), that is, x_(n) = (-1)^(n+1)*n!*(x/(2*n - x))*binomial(2*n - x,n) for n >= 0. Then this table is the triangle of connection constants for expressing the monomial polynomials x^n in terms of the basis polynomials x_(k), that is, x^n = sum {k = 0..n} T(n,k)*x_(k), n = 0,1,2,.... Examples are given below.
Matrix factorization: Let M be the infinite lower unit triangular array with (n,k)-th entry (2^(n+1-k)-1)*binomial(n,k). For k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. It follows from the recurrence equation given in the Formula section that the infinite product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to the present triangle (but with the first row and column omitted). See the Example section. (End)
The Bell transform of 2^(n+1)-1. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

Examples

			T(3,2) = 9: {1}{2}<-3, {1}{3}<-2, {1}{2,3}, {2}{1}<-3, {2}{3}<-1, {2}{1,3}, {3}{1}<-2, {3}{2}<-1, {3}{1,2}.
Triangle begins:
  1;
  0,   1;
  0,   3,    1;
  0,   7,    9,     1;
  0,  15,   55,    18,    1;
  0,  31,  285,   205,   30,    1;
  0,  63, 1351,  1890,  545,   45,  1;
  0, 127, 6069, 15421, 7770, 1190, 63,  1;
  ...
From _Peter Bala_, Jan 07 2015: (Start)
T(4,2) = 55: There are 7 partitions of the set {1,2,3,4} into 2 blocks. For the 3 set partitions of the type {a,b}{c,d} we can choose a nonempty subset from each block in one of 3*3 ways giving 3*3*3 = 27 possibilities in all. The remaining 4 set partitions of {1,2,3,4} into 2 blocks are of the form {a,b,c}{d} and we can choose a nonempty subset from each block in 7*1 ways giving 4*7*1 = 28 possible choices. Thus in total T(4,2) = 27 + 28 = 55.
Recurrence equation example:
T(4,2) = sum {j = 1..3} (2^(4-j) - 1)*binomial(3,j)*T(j,1) = 7*3*1 + 3*3*3 + 1*1*7 = 55.
Connection constants:
Row 3 = [0, 7, 9, 1]. Hence x^3 = 7*x + 9*x*(x - 3) + x*(x - 4)*(x - 5); Row 4 = [0, 15, 55, 18, 1]. Hence x^4 = 15*x + 55*x*(x - 3) + 18*x*(x - 4)*(x - 5) + x*(x - 5)*(x - 6)*(x - 7).
With the array M(k) as defined in the Comments section, the infinite product M(0)*M(1)*M(2)*... begins
/ 1        \/1           \/1       \       / 1        \
| 3  1     ||0  1        ||0 1      |      | 3  1      |
| 7  6  1  ||0  3  1     ||0 0 1    |... = | 7  9  1   |
|15 21 9 1 ||0  7  6  1  ||0 0 3 1  |      |15 55 18 1 |
|...       ||0 15 21  9 1||0 0 7 6 1|      |...        |
|...       ||...         ||...      |      |           |
(End)

Links

Crossrefs

Columns k=0-9: A000007, A000225, A016269(n-2), A028025(n-3), A143399, A143400, A143401, A143402, A143403, A143404.
Diagonal: A000012.
T(2*n,n) gives A383869.
See also A048993, A008277, A007318, A143405 for row sums.

Programs

  • Magma
    [[(&+[Binomial(n,j)*StirlingSecond(j,k)*k^(n-j): j in [k..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Mar 07 2019
  • Maple
    T:= (n, k)-> add(binomial(n,t)*Stirling2(t,k)*k^(n-t), t=k..n):
seq(seq(T(n, k), k=0..n), n=0..11);
  • Mathematica
    t[0, 0]=1; t[n_, k_]:= SeriesCoefficient[Exp[y*Exp[x]*(Exp[x]-1)], {x, 0, n}, {y, 0, k}]*n!; Table[t[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Jean-François Alcover, Dec 05 2013, after Vladeta Jovovic *)
Table[If[n==k==0, 1, If[k==0, 0, Sum[Binomial[n, j]*StirlingS2[j, k]* k^(n-j), {j,k,n}]]], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 07 2019 *)
  • PARI
    {T(n,k) = sum(j=k, n, binomial(n,j)*stirling(j,k,2)*k^(n-j))};
for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Mar 07 2019
  • Sage
    # uses[bell_matrix from A264428]
bell_matrix(lambda n: 2^(n+1)-1, 10) # Peter Luschny, Jan 18 2016

Formula

G.f. for column k: x^k/Product_{t=k..2*k} (1-t*x).
T(n,k) = Sum_{t=k..n} C(n,t) * Stirling2(t,k) * k^(n-t).
E.g.f.: exp(y*exp(x)*(exp(x)-1)). - Vladeta Jovovic, Dec 08 2008
T(n,k) = Sum_{m=0..k} Stirling2(n,k+m)*(k+m)!/(m!*(k-m)!). - Vladimir Kruchinin, Apr 06 2011
Let P be Pascal's triangle A007318. The first column of the array exp(t*(P^2-P)) gives the row generating polynomials of this triangle.
The row polynomials R(n,t) satisfy the recurrence R(n+1,t) = t*(Sum_{k = 0..n} (2^(k+1)-1)*C(n,k)*R(n-k,t)) with R(0,t) = 1. For example, the row 4 polynomial R(4,t) = 15*t + 55*t^2 + 18*t^3 + t^4 = t*((7*t + 9*t^2 + t^3) + 3*3*(3*t+t^2) + 7*3*t + 15*1). - Peter Bala, Oct 12 2011
From Peter Bala, Jan 07 2015: (Start)
T(n,k) = (1/k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(j + k)^n.
Recurrence equation: T(n+1,k+1) = Sum_{j = k..n} (2^(n-j+1) - 1)*binomial(n,j)*T(j,k) with T(0,0) = 1 and T(n,0) = 0 for n >= 1. This leads to the matrix factorization noted in the Comments section.
The inverse array is a signed version of A038455. (End)

A055882 a(n) = 2^n*Bell(n). E.g.f.: exp(exp(2*x)-1).

Original entry on oeis.org

1, 2, 8, 40, 240, 1664, 12992, 112256, 1059840, 10827264, 118758400, 1389711360, 17258893312, 226463227904, 3127694491648, 45316785602560, 686826595745792, 10861264214949888, 178802342273744896, 3058036745204924416, 54236710945813430272, 995874184692762673152
Offset: 0

Views

Author

Christian G. Bower, Jun 09 2000

Keywords

Comments

a(n) is the number of set partitions of {1,2,...,n} with a (possibly empty) subset of designated elements in each block. - Geoffrey Critzer, Sep 16 2012

Links

Crossrefs

Cf. A000079, A000110, A055883, A143405.

Programs

  • Magma
    [2^n*Bell(n): n in [0..20]]; // Vincenzo Librandi, Sep 19 2014
  • Maple
    seq(add(binomial(n, k)*(bell(n)), k=0..n), n=0..18); # Zerinvary Lajos, Dec 01 2006
# second Maple program:
a:= proc(n) option remember; `if`(n=0, 1, add(
      a(n-j) *binomial(n-1, j-1)*2^j, j=1..n))
    end:
seq(a(n), n=0..23);  # Alois P. Heinz, Oct 04 2019
  • Mathematica
    nn=20;a=Exp[2x]-1;Range[0,nn]!CoefficientList[Series[Exp[a],{x,0,nn}],x]  (* Geoffrey Critzer, Sep 16 2012 *)
Table[2^n BellB[n], {n, 0, 20}] (* Vincenzo Librandi, Sep 19 2014 *)
  • Python
    # Python 3.2 or higher required
from itertools import accumulate
A055882_list, blist, b, n2 = [1,2], [1], 1, 4
for _ in range(2, 201):
    blist = list(accumulate([b]+blist))
    b = blist[-1]
    A055882_list.append(b*n2)
    n2 *= 2 # Chai Wah Wu, Sep 19 2014

Formula

a(n) = exp(-1)*2^n*Sum_{k>=0} k^n/k!. - Benoit Cloitre, May 20 2002
G.f.: 1/(1-2*x/(1-2*x/(1-2*x/(1-4*x/(1-2*x/(1-6*x/(1-2*x/(1-8*x/(1-... (continued fraction). - Paul Barry, Oct 11 2009
G.f.: 1/(U(0) - 2*x) where U(k) = 1 + 2*x - 2*x*(k+1)/(1 - 2*x/U(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Oct 12 2012
G.f.: G(0)/(1+2*x) where G(k) = 1 - 4*x*(k+1)/((2*k+1)*(4*x*k-1) - 2*x*(2*k+1)*(2*k+3)*(4*x*k-1)/(2*x*(2*k+3) - 2*(k+1)*(4*x*k+2*x-1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Dec 22 2012
G.f.: G(0)/2 where G(k) = 1 - (2*x*k + 1)/(2*x*k - 1 - 2*x*(2*x*k - 1)/(2*x + (2*x*k + 1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 30 2013
G.f.: 1/Q(0), where Q(k) = 1 - 2*(k+1)*x - 4*(k+1)*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
a(n) = Sum_{k=0..n} binomial(n,k) * A004211(k) * A004211(n-k). - Vaclav Kotesovec, Apr 17 2020

A143396 Triangle T(n,k) = number of forests of labeled rooted trees of height at most 1, with n labels, k of which are used for root nodes and any root may contain >= 1 labels, n >= 0, 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 0, 2, 2, 0, 3, 9, 5, 0, 4, 30, 40, 15, 0, 5, 90, 220, 185, 52, 0, 6, 255, 1040, 1485, 906, 203, 0, 7, 693, 4550, 9905, 9891, 4718, 877, 0, 8, 1820, 19040, 59850, 87416, 66808, 26104, 4140, 0, 9, 4644, 77448, 341082, 686826, 750120, 463212, 153063, 21147
Offset: 0

Views

Author

Alois P. Heinz, Aug 12 2008

Keywords

Examples

			T(3,2) = 9: {1,2}<-3, {1,3}<-2, {2,3}<-1, {1}<-3{2}, {1}{2}<-3, {1}<-2{3}, {1}{3}<-2, {2}<-1{3}, {2}{3}<-1.
Triangle begins:
  1;
  0,  1;
  0,  2,  2;
  0,  3,  9,   5;
  0,  4, 30,  40,  15;
  0,  5, 90, 220, 185,  52;
  ...

Links

Crossrefs

Columns k=0-10 give: A000007, A000027, A273652, A273653, A273654, A273655, A273656, A273657, A273658, A273659, A273660.
Diagonal gives A000110.
Row sums give A143405.
T(2n,n) gives A273661.
Cf. A048993, A008277, A007318.

Programs

  • Maple
    T:= (n, k)-> binomial(n, k)*add(Stirling2(k, t)*t^(n-k), t=0..k):
seq(seq(T(n, k), k=0..n), n=0..11);
  • Mathematica
    T[n_, k_] := T[n, k] = Binomial[n, k]*Sum[StirlingS2[k, t]*If[n == k, 1, t^(n - k)], {t, 0, k}];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 27 2016, translated from Maple, updated Jan 01 2021 *)

Formula

T(n,k) = C(n,k) * Sum_{t=0..k} Stirling2(k,t) * t^(n-k).
E.g.f.: exp(exp(x)*(exp(x*y)-1)). - Vladeta Jovovic, Dec 08 2008

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

Views

Author

Vaclav Kotesovec, Jun 27 2022

Keywords

Comments

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

Links

Crossrefs

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

Programs

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

Formula

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

A355378 Expansion of e.g.f. exp(exp(3*x) - exp(x)).

Original entry on oeis.org

1, 2, 12, 82, 688, 6754, 75096, 928386, 12591392, 185384130, 2938319144, 49799613538, 897495547184, 17118975292514, 344206910941624, 7270287035936706, 160826794265399360, 3716047107259486082, 89472755268582494792, 2240097688067896960674, 58207872357772581544272
Offset: 0

Views

Author

Vaclav Kotesovec, Jun 30 2022

Keywords

Links

Crossrefs

Cf. A143405, A355291, A355379, A355381.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[Exp[3*x] - Exp[x]], {x, 0, nmax}], x] * Range[0, nmax]!
Table[Sum[Binomial[n,k] * 3^k * BellB[k] * BellB[n-k, -1], {k, 0, n}], {n, 0, 20}]
  • PARI
    my(x='x+O('x^25)); Vec(serlaplace(exp(exp(3*x) - exp(x)))) \\ Michel Marcus, Jun 30 2022

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * 3^k * Bell(k) * Bell(n-k, -1).
a(0) = 1; a(n) = Sum_{k=1..n} (3^k - 1) * binomial(n-1,k-1) * a(n-k). - Seiichi Manyama, Jun 30 2022
a(n) ~ exp(exp(3*z) - exp(z) - n) * (n/z)^(n + 1/2) / sqrt(3*(1 + 3*z)*exp(3*z) - (1 + z)*exp(z)), where z = LambertW(n)/3 - 1/(1 + 3/LambertW(n) - 9 * n^(2/3) * (1 + LambertW(n)) / LambertW(n)^(5/3)). - Vaclav Kotesovec, Jul 03 2022
a(n) ~ (3*n/LambertW(n))^n * exp(n/LambertW(n) - (n/LambertW(n))^(1/3) - n) / sqrt(1 + LambertW(n)). - Vaclav Kotesovec, Jul 10 2022

A355381 Expansion of e.g.f. exp(exp(3*x) - exp(2*x)).

Original entry on oeis.org

1, 1, 6, 35, 247, 2102, 20547, 224541, 2707292, 35638329, 507464939, 7757439428, 126538995293, 2191454313661, 40120212534838, 773554002955047, 15656660861190371, 331700076893737054, 7337160433117899959, 169068422994937678185, 4050093664805130165348
Offset: 0

Views

Author

Vaclav Kotesovec, Jun 30 2022

Keywords

Comments

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

Links

Crossrefs

Cf. A143405, A194006, A355291, A355380, A355378.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[Exp[3*x] - Exp[2*x]], {x, 0, nmax}], x] * Range[0, nmax]!
Table[Sum[Binomial[n,k] * 3^k * 2^(n-k) * BellB[k] * BellB[n-k, -1], {k, 0, n}], {n, 0, 20}]
  • PARI
    my(x='x+O('x^25)); Vec(serlaplace(exp(exp(3*x) - exp(2*x)))) \\ Michel Marcus, Jun 30 2022

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * 3^k * 2^(n-k) * Bell(k) * Bell(n-k, -1).
a(0) = 1; a(n) = Sum_{k=1..n} (3^k - 2^k) * binomial(n-1,k-1) * a(n-k). - Seiichi Manyama, Jun 30 2022
a(n) ~ exp(exp(3*z) - exp(2*z) - n) * (n/z)^(n + 1/2) / sqrt(3*(1 + 3*z)*exp(3*z) - 2*(1 + 2*z)*exp(2*z)), where z = LambertW(n)/3 - 1/(2 + 3/LambertW(n) - 9 * n^(1/3) * (1 + LambertW(n)) / (2*LambertW(n)^(4/3))). - Vaclav Kotesovec, Jul 03 2022

A355380 Expansion of e.g.f. exp(exp(3*x) + exp(2*x) - 2).

Original entry on oeis.org

1, 5, 38, 355, 3879, 48050, 661163, 9961745, 162598044, 2851150665, 53350521523, 1059447004560, 22224898346989, 490589320542305, 11356591577861398, 274886065370874775, 6939205217774546339, 182273695066097752170, 4971724931587003394863, 140559648864263508395965
Offset: 0

Views

Author

Vaclav Kotesovec, Jun 30 2022

Keywords

Links

Crossrefs

Cf. A143405, A355291, A355381.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[Exp[3*x] + Exp[2*x] - 2], {x, 0, nmax}], x] * Range[0, nmax]!
Table[Sum[Binomial[n,k] * 3^k * 2^(n-k) * BellB[k] * BellB[n-k], {k, 0, n}], {n, 0, 20}]
  • PARI
    my(x='x+O('x^25)); Vec(serlaplace(exp(exp(3*x) + exp(2*x) - 2))) \\ Michel Marcus, Jun 30 2022

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * 3^k * 2^(n-k) * Bell(k) * Bell(n-k).
a(0) = 1; a(n) = Sum_{k=1..n} (3^k + 2^k) * binomial(n-1,k-1) * a(n-k). - Seiichi Manyama, Jun 30 2022
a(n) ~ exp(exp(3*z) + exp(2*z) - 2 - n) * (n/z)^(n + 1/2) / sqrt(3*(1 + 3*z)*exp(3*z) + 2*(1 + 2*z)*exp(2*z)), where z = LambertW(n)/3 - 1/(2 + 3/LambertW(n) + 9 * n^(1/3) * (1 + LambertW(n)) / (2*LambertW(n)^(4/3))). - Vaclav Kotesovec, Jul 03 2022

A355379 Expansion of e.g.f. exp(exp(3*x) + exp(x) - 2).

Original entry on oeis.org

1, 4, 26, 212, 2046, 22588, 278942, 3792916, 56128254, 895795692, 15307847614, 278435732484, 5364073445278, 108994074306268, 2327475127169182, 52069279762495220, 1217024509006768574, 29647115491635327180, 751085909757123127294, 19750410883486281805028
Offset: 0

Views

Author

Vaclav Kotesovec, Jun 30 2022

Keywords

Links

Crossrefs

Cf. A143405, A355291, A355378.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[Exp[3*x] + Exp[x] - 2], {x, 0, nmax}], x] * Range[0, nmax]!
Table[Sum[Binomial[n,k] * 3^k * BellB[k] * BellB[n-k], {k, 0, n}], {n, 0, 20}]
  • PARI
    my(x='x+O('x^25)); Vec(serlaplace(exp(exp(3*x) + exp(x) - 2))) \\ Michel Marcus, Jun 30 2022

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * 3^k * Bell(k) * Bell(n-k).
a(0) = 1; a(n) = Sum_{k=1..n} (3^k + 1) * binomial(n-1,k-1) * a(n-k). - Seiichi Manyama, Jun 30 2022
a(n) ~ exp(exp(3*z) + exp(z) - 2 - n) * (n/z)^(n + 1/2) / sqrt(3*(1 + 3*z)*exp(3*z) + (1 + z)*exp(z)), where z = LambertW(n)/3 - 1/(1 + 3/LambertW(n) + 9 * n^(2/3) * (1 + LambertW(n)) / LambertW(n)^(5/3)). - Vaclav Kotesovec, Jul 03 2022
a(n) ~ (3*n/LambertW(n))^n * exp(n/LambertW(n) + (n/LambertW(n))^(1/3) - n - 2) / sqrt(1 + LambertW(n)). - Vaclav Kotesovec, Jul 10 2022

A143397 Triangle T(n,k)=number of forests of labeled rooted trees of height at most 1, with n labels and k nodes, where any root may contain >= 1 labels, n >= 0, 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 1, 6, 10, 0, 1, 11, 36, 41, 0, 1, 20, 105, 230, 196, 0, 1, 37, 285, 955, 1560, 1057, 0, 1, 70, 756, 3535, 8680, 11277, 6322, 0, 1, 135, 2002, 12453, 41720, 80682, 86800, 41393, 0, 1, 264, 5347, 43008, 186669, 485982, 773724, 708948, 293608
Offset: 0

Views

Author

Alois P. Heinz, Aug 12 2008

Keywords

Examples

			T(3,2) = 6: {1,2}{3}, {1,3}{2}, {2,3}{1}, {1,2}<-3, {1,3}<-2, {2,3}<-1.
Triangle begins:
  1;
  0, 1;
  0, 1,  3;
  0, 1,  6,  10;
  0, 1, 11,  36,   41;
  0, 1, 20, 105,  230,  196;
  0, 1, 37, 285,  955, 1560,  1057;
  0, 1, 70, 756, 3535, 8680, 11277, 6322;
  ...

Links

Crossrefs

Columns k=0-2: A000007, A000012, A006127. Diagonal: A000248. See also A048993, A008277, A007318, A143405 for row sums.

Programs

  • Maple
    T:= (n,k)-> add(binomial(n, k-t)*Stirling2(n-(k-t),t)*t^(k-t), t=0..k):
seq(seq(T(n, k), k=0..n), n=0..11);
  • Mathematica
    T[n_, k_] := Sum[Binomial[n, k-t]*StirlingS2[n - (k-t), t]*t^(k-t), {t, 0, k}]; T[0, 0] = 1; T[_, 0] = 0;
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 31 2016, translated from Maple *)

Formula

T(n,k) = Sum_{t=0..k} C(n,k-t) * Stirling2(n-(k-t),t) * t^(k-t).
E.g.f.: exp(y*exp(x*y)*(exp(x)-1)). - Vladeta Jovovic, Dec 08 2008

A347432 E.g.f.: exp( exp(x) * (exp(x) - 1 - x) ).

Original entry on oeis.org

1, 0, 1, 4, 14, 66, 397, 2626, 18797, 148238, 1281134, 11943790, 118998365, 1262189748, 14203022537, 168835162632, 2111832477426, 27708387132906, 380355066174121, 5449577398256414, 81316095965242989, 1261149374033472626, 20293627142875917978, 338263983223664609198
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 02 2021

Keywords

Comments

Exponential transform of A000295.

Crossrefs

Cf. A000295, A000296, A003725, A055882, A143405, A347434, A347435.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n=0, 1, add(
      a(n-j)*binomial(n-1, j-1)*(2^j-j-1), j=1..n))
    end:
seq(a(n), n=0..23);  # Alois P. Heinz, Sep 02 2021
  • Mathematica
    nmax = 23; CoefficientList[Series[Exp[Exp[x] (Exp[x] - 1 - x)], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k - 1] (2^k - k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 23}]

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * A000295(k) * a(n-k).
a(n) = Sum_{k=0..n} (-1)^k * binomial(n,k) * A003725(k) * A143405(n-k).
a(n) ~ n^(n + 1/2) * (exp(exp(r)*(exp(r) - r - 1) - r/2 - n) / (r^(n + 1/2) * sqrt(2*exp(r)*(1 + 2*r) - (2 + r*(4 + r))))), where r = LambertW(n)/2 + (4 + LambertW(n)) * LambertW(n)^(3/2) / (8 * sqrt(n) * (1 + LambertW(n))). - Vaclav Kotesovec, Jul 07 2022
Showing 1-10 of 12 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.