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a(n) = Sum_{k>=1} (4*k - 1)^n / 2^k.
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%I #7 Jun 26 2022 04:21:13

%S 1,7,81,1399,32289,931687,32259441,1303134679,60160827969,

%T 3124574220487,180312309395601,11445969681199159,792626097462398049,

%U 59462922484586318887,4804064349575887075761,415847988794676360818839,38396277196654611908582529,3766800071614388562865514887

%N a(n) = Sum_{k>=1} (4*k - 1)^n / 2^k.

%F E.g.f.: exp(3*x) / (2 - exp(4*x)).

%F a(0) = 1; a(n) = 3^n + Sum_{k=1..n} binomial(n,k) * 4^k * a(n-k).

%F a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * 4^k * A000670(k).

%F a(n) ~ n! * 2^(2*n - 1/4) / log(2)^(n+1). - _Vaclav Kotesovec_, Jun 24 2022

%t nmax = 17; CoefficientList[Series[Exp[3 x]/(2 - Exp[4 x]), {x, 0, nmax}], x] Range[0, nmax]!

%t a[0] = 1; a[n_] := a[n] = 3^n + Sum[Binomial[n, k] 4^k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 17}]

%Y Cf. A000629, A000670, A080253, A259533, A285067, A328183, A355218, A355219.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Jun 24 2022