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
Examples
a(4) = A060719(3) + 1 = 29 + 1 = 30.
Links
- G. C. Greubel, Table of n, a(n) for n = 0..575
Crossrefs
Programs
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Magma
[Bell(n)*(2-0^n): n in [0..50]]; // Vincenzo Librandi, Apr 06 2011
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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
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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 *)
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PARI
x='x+O('x^50); Vec(serlaplace(2*exp(exp(x) - 1) -1)) \\ G. C. Greubel, Jul 24 2017
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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+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
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