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E.g.f. satisfies: A(x) = exp( x * (1 + 1/A(x))/2 ).
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%I #26 Nov 03 2023 08:12:20

%S 1,1,0,1,-4,26,-212,2108,-24720,334072,-5112544,87396728,-1650607040,

%T 34132685120,-767025716736,18612106195456,-485013257865472,

%U 13509071081429888,-400505695457942528,12592502771190979712,-418524228123134068224

%N E.g.f. satisfies: A(x) = exp( x * (1 + 1/A(x))/2 ).

%H Seiichi Manyama, <a href="/A349719/b349719.txt">Table of n, a(n) for n = 0..411</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LambertW-Function.html">Lambert W-Function</a>.

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

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

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

%F a(n) ~ -(-1)^n * sqrt(1 + LambertW(exp(-1))) * n^(n-1) / (2^n * exp(n) * LambertW(exp(-1))^(n-1)). - _Vaclav Kotesovec_, Dec 05 2021

%t a[n_] := (1/2^n) * Sum[If[k == n == 1, 1, (-k + 1)^(n - 1)] * Binomial[n, k], {k, 0, n}]; Array[a, 21, 0] (* _Amiram Eldar_, Nov 27 2021 *)

%o (PARI) a(n) = sum(k=0, n, (-k+1)^(n-1)*binomial(n, k))/2^n;

%o (PARI) my(N=40, x='x+O('x^N)); Vec(serlaplace((x/2)/lambertw(x/2*exp(-x/2))))

%o (PARI) my(N=40, x='x+O('x^N)); Vec(2*sum(k=0, N, (-k+1)^(k-1)*x^k/(2-(-k+1)*x)^(k+1)))

%Y Cf. A007889, A202617, A283828, A349714, A349715, A349716, A349720, A349721.

%K sign

%O 0,5

%A _Seiichi Manyama_, Nov 27 2021