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a(n) = exp(-1/3) * Sum_{k>=0} (3*k - 1)^n / (3^k * k!).
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%I #7 Jun 26 2022 12:44:02

%S 1,0,3,9,54,351,2673,22842,216513,2248965,25351704,307699965,

%T 3995419365,55207193328,808078734999,12480510487509,202697232446070,

%U 3451417004044323,61450890989472837,1141331486235356178,22066085726516137149,443236553318792110113,9233934519951699602400

%N a(n) = exp(-1/3) * Sum_{k>=0} (3*k - 1)^n / (3^k * k!).

%F G.f. A(x) satisfies: A(x) = (1 - 3*x + x*A(x/(1 - 3*x))) / (1 - 2*x - 3*x^2).

%F G.f.: (1/(1 + x)) * Sum_{k>=0} (x/(1 + x))^k / Product_{j=1..k} (1 - 3*j*x/(1 + x)).

%F E.g.f.: exp((exp(3*x) - 1) / 3 - x).

%F a(0) = 1; a(n) = Sum_{k=1..n-1} binomial(n-1,k) * 3^k * a(n-k-1).

%F a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * A004212(k).

%F a(n) ~ 3^(n - 1/3) * n^(n - 1/3) * exp(n/LambertW(3*n) - n - 1/3) / (sqrt(1 + LambertW(3*n)) * LambertW(3*n)^(n - 1/3)). - _Vaclav Kotesovec_, Jun 26 2022

%t nmax = 22; CoefficientList[Series[Exp[(Exp[3 x] - 1)/3 - x], {x, 0, nmax}], x] Range[0, nmax]!

%t a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k] 3^k a[n - k - 1], {k, 1, n - 1}]; Table[a[n], {n, 0, 22}]

%t Table[Sum[(-1)^(n - k) Binomial[n, k] 3^k BellB[k, 1/3], {k, 0, n}], {n, 0, 22}]

%Y Cf. A000296, A003575, A004212, A337038, A337040, A337041, A337042, A337043.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Aug 12 2020