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a(n) = Sum_{k>=1} (3*k - 1)^n / 2^k.
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%I #7 Jun 26 2022 04:20:58

%S 1,5,43,557,9643,208685,5419243,164184557,5684837803,221440158125,

%T 9584118542443,456289689634157,23698327407870763,1333388917719691565,

%U 80794290325166308843,5245268489291712773357,363231496206350038884523,26725646191850556128889005,2082075690178933613292014443

%N a(n) = Sum_{k>=1} (3*k - 1)^n / 2^k.

%F E.g.f.: exp(2*x) / (2 - exp(3*x)).

%F a(0) = 1; a(n) = 2^n + Sum_{k=1..n} binomial(n,k) * 3^k * a(n-k).

%F a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n-k) * 3^k * A000670(k).

%F a(n) ~ n! * 3^n / (2^(1/3) * log(2)^(n+1)). - _Vaclav Kotesovec_, Jun 24 2022

%t nmax = 18; CoefficientList[Series[Exp[2 x]/(2 - Exp[3 x]), {x, 0, nmax}], x] Range[0, nmax]!

%t a[0] = 1; a[n_] := a[n] = 2^n + Sum[Binomial[n, k] 3^k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 18}]

%Y Cf. A000629, A000670, A007047, A080253, A151919, A328182, A355219, A355220.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Jun 24 2022