A186021 a(n) = Bell(n)*(2 - 0^n).

1, 2, 4, 10, 30, 104, 406, 1754, 8280, 42294, 231950, 1357140, 8427194, 55288874, 381798644, 2765917090, 20960284294, 165729739608, 1364153612318, 11665484410114, 103448316470744, 949739632313502, 9013431476894646, 88304011710168692, 891917738589610578, 9277180664459998706

Offset: 0

Author

Paul Barry, Feb 10 2011

Keywords

• nonn
• easy

Comments

a(n) is the number of collections of subsets of {1,2,...,n-1} that are pairwise disjoint. a(n+1) = 2*Sum_{j=0..n} C(n,j)*Bell(j). For example a(3)=10 because we have: {}, {{}}, {{1}}, {{2}}, {{1,2}}, {{},{1}}, {{},{2}}, {{},{1,2}}, {{1},{2}}, {{},{1},{2}}. - Geoffrey Critzer, Aug 28 2014

a(n) is the number of collections of subsets of [n] that are pairwise disjoint and cover [n], with [0] = {}. For disjoint collections of nonempty subsets see A000110. For arbitrary collections of subsets see A000371. For arbitrary collections of nonempty subsets see A003465. - Manfred Boergens, May 02 2024 and Apr 09 2025

Examples

			a(4) = A060719(3) + 1 = 29 + 1 = 30.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..575

Crossrefs

Row sums of A186020 and A256894.

Main diagonal of A271466 (shifted) and A381682.

Cf. A000110, A000371, A003465.

Programs

Magma

[Bell(n)*(2-0^n): n in [0..50]]; // Vincenzo Librandi, Apr 06 2011

Maple

A186021List := proc(m) local A, P, n; A := [1,2]; P := [2];
for n from 1 to m - 2 do P := ListTools:-PartialSums([P[-1], op(P)]);
A := [op(A), P[-1]] od; A end: A186021List(26); # Peter Luschny, Mar 24 2022

Mathematica

Prepend[Table[2 Sum[Binomial[n, j] BellB[j], {j, 0, n}], {n, 0, 25}], 1] (* Geoffrey Critzer, Aug 28 2014 *)
With[{nmax = 50}, CoefficientList[Series[2*Exp[Exp[x] - 1] - 1, {x, 0, nmax}], x]*Range[0, nmax]!] (* G. C. Greubel, Jul 24 2017 *)

PARI

x='x+O('x^50); Vec(serlaplace(2*exp(exp(x) - 1) -1)) \\ G. C. Greubel, Jul 24 2017

Python

from itertools import accumulate
def A186021_list(size):
    if size < 1: return []
    L, accu = [1], [2]
    for _ in range(size-1):
        accu = list(accumulate([accu[-1]] + accu))
        L.append(accu[0])
    return L
print(A186021_list(26)) # Peter Luschny, Apr 25 2016

Formula

E.g.f.: 2*exp(exp(x)-1)-1. - Paul Barry, Apr 06 2011

a(n) = A000110(n)*A040000(n).

a(n+1) = 1 + Sum_{k=0..n} C(n,k)*a(k). - Franklin T. Adams-Watters, Oct 02 2011

From Sergei N. Gladkovskii, Nov 11 2012 to Mar 29 2013: (Start)

Continued fractions:

G.f.: A(x)= 1 + 2*x/(G(0)-x) where G(k)= 1 - x*(k+1)/(1 - x/G(k+1)).

G.f.: G(0)-1 where G(k) = 1-(x*k+1)/(x*k - 1 - x*(x*k - 1)/(x + (x*k + 1)/G(k+1))).

G.f.: (G(0)-2)/x - 1 where G(k) = 1 + 1/(1 - x/(x + (1 - x*k)/G(k+1))).

G.f.: (S-2)/x - 1 where S = 2*Sum_{k>=0} x^k/Product_{n=0..k-1}(1 - n*x).

G.f.: 1/(1-x) - x/(G(0)-x^2+x) where G(k) =x^2 + x - 1 + k*(2*x-x^2) - x^2*k^2 + x*(x*k - 1)*(x*k + 2*x - 1)^2/G(k+1).

E.g.f.: E(0) - 1 where E(k) = 1 + 1/(1 - 1/(1 + (k+1)/x*Bell(k)/Bell(k+1)/E(k+1))). (End)

a(n) = A060719(n-1) + 1, and the inverse binomial transform of A060719. - Gary W. Adamson, May 20 2013

G.f. A(x) satisfies: A(x) = 1 + (x/(1 - x)) * (1 + A(x/(1 - x))). - Ilya Gutkovskiy, Jun 30 2020