A126390 a(n) = Sum_{i=0..n} 2^i*B(i)*binomial(n,i) where B(n) = Bell numbers A000110(n).
1, 3, 13, 71, 457, 3355, 27509, 248127, 2434129, 25741939, 291397789, 3510328695, 44782460313, 602513988107, 8518757813637, 126179029108463, 1952609274344353, 31492811964616163, 528249539951292461, 9197240228562763687, 165923214676585626729
Offset: 0
Keywords
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Toufik Mansour and Mark Shattuck, A recurrence related to the Bell numbers, INTEGERS 11 (2011), #A67.
Programs
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Maple
with(combstruct):seq(count(([S, {N=Union(Z, S, P), S=Set(Union(Z, P), card>=0), P=Set(Union(Z, Z), card>=1)}, labeled], size=n)), n=0..20); # Zerinvary Lajos, Mar 18 2008 -
Mathematica
Table[ Sum[ 2^k Binomial[n, k] BellB[k], {k, 0, n}], {n, 0, 30}] (* Karol A. Penson and Olivier Gérard, Oct 22 2007 *) -
PARI
x='x+O('x^66); Vec(serlaplace((exp(exp(2*x)-1+x)))) \\ Joerg Arndt, May 13 2013
Formula
E.g.f.: exp(exp(2*x)-1+x). - Vladeta Jovovic, Aug 04 2007
a(n) = e^(-1)* 2^n * Sum_{k>=0} (k + 1/2)^n / k!. This is a Dobinski-type formula. - Karol A. Penson and Olivier Gérard, Oct 22 2007
G.f.: 1/Q(0), where Q(k)= 1 - (2*k+3)*x - 4*(k+1)*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: 1/Q(0), where Q(k)= 1 - x - 2*x/(1 - 2*x*(2*k+1)/(1 - x - 2*x/(1 - 2*x*(2*k+2)/Q(k+1)))); (continued fraction). - Sergei N. Gladkovskii, May 13 2013
a(0) = 1; a(n) = a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 2^k * a(n-k). - Ilya Gutkovskiy, Jun 21 2022
From Vaclav Kotesovec, Jun 22 2022: (Start)
a(n) ~ Bell(n) * (2 + LambertW(n)/n)^n.
a(n) ~ Bell(n) * 2^n * sqrt(n) * log(n)^(-1/2 + 1/(2*log(n)) - 1/(2*log(n)^2)) * exp(log(log(n))^2/(4*log(n)^2)). (End)
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 27 2022