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a(n) = exp(-1/2) * Sum_{k>=0} (2*k - 1)^n / (2^k * k!).
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%I #10 Jun 26 2022 12:14:01

%S 1,0,2,4,20,96,552,3536,25104,194816,1637408,14792768,142761280,

%T 1464117760,15886137984,181667507456,2182268117248,27456279388160,

%U 360872502280704,4943580063237120,70437638474568704,1041911242274562048,15972832382065977344,253388070573020401664

%N a(n) = exp(-1/2) * Sum_{k>=0} (2*k - 1)^n / (2^k * k!).

%H Robert Israel, <a href="/A337038/b337038.txt">Table of n, a(n) for n = 0..516</a>

%F G.f. A(x) satisfies: A(x) = (1 - 2*x + x*A(x/(1 - 2*x))) / (1 - x - 2*x^2).

%F G.f.: (1/(1 + x)) * Sum_{k>=0} (x/(1 + x))^k / Product_{j=1..k} (1 - 2*j*x/(1 + x)).

%F E.g.f.: exp((exp(2*x) - 1) / 2 - x).

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

%F a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * A004211(k).

%F a(n) ~ 2^(n - 1/2) * n^(n - 1/2) * exp(n/LambertW(2*n) - n - 1/2) / (sqrt(1 + LambertW(2*n)) * LambertW(2*n)^(n - 1/2)). - _Vaclav Kotesovec_, Jun 26 2022

%p E:= exp((exp(2*x)-1)/2-x):

%p S:= series(E,x,31):

%p seq(coeff(S,x,i)*i!,i=0..30); # _Robert Israel_, Aug 26 2020

%t nmax = 23; CoefficientList[Series[Exp[(Exp[2 x] - 1)/2 - x], {x, 0, nmax}], x] Range[0, nmax]!

%t a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k] 2^k a[n - k - 1], {k, 1, n - 1}]; Table[a[n], {n, 0, 23}]

%t Table[Sum[(-1)^(n - k) Binomial[n, k] 2^k BellB[k, 1/2], {k, 0, n}], {n, 0, 23}]

%Y Cf. A000296, A004211, A007405, A124311, A166922, A217203, A337039, A337040, A337041, A337042, A337043.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Aug 12 2020