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a(n) = exp(-1/4) * Sum_{k>=0} (4*k + 3)^n / (4^k * k!).
4

%I #12 Jun 27 2023 11:35:26

%S 1,4,20,128,1008,9280,96704,1120768,14274816,197833728,2958521344,

%T 47415508992,809838505984,14670950907904,280760761434112,

%U 5655835404271616,119561580162646016,2645030742360588288,61087848487323959296,1469652941137655103488,36758243982057508175872,954111239026567129595904

%N a(n) = exp(-1/4) * Sum_{k>=0} (4*k + 3)^n / (4^k * k!).

%H Michael De Vlieger, <a href="/A355167/b355167.txt">Table of n, a(n) for n = 0..464</a>

%H Adam Buck, Jennifer Elder, Azia A. Figueroa, Pamela E. Harris, Kimberly Harry, and Anthony Simpson, <a href="https://arxiv.org/abs/2306.13034">Flattened Stirling Permutations</a>, arXiv:2306.13034 [math.CO], 2023. See p. 14.

%F E.g.f.: exp(3*x + (exp(4*x) - 1) / 4).

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

%F a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * A004213(k).

%F a(n) ~ 2^(2*n + 3/2) * n^(n + 3/4) * exp(n/LambertW(4*n) - n - 1/4) / (sqrt(1 + LambertW(4*n)) * LambertW(4*n)^(n + 3/4)). - _Vaclav Kotesovec_, Jun 27 2022

%t nmax = 21; CoefficientList[Series[Exp[3 x + (Exp[4 x] - 1)/4], {x, 0, nmax}], x] Range[0, nmax]!

%t a[0] = 1; a[n_] := a[n] = 3 a[n - 1] + Sum[Binomial[n - 1, k - 1] 4^(k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 21}]

%t Table[Sum[Binomial[n, k] 3^(n - k) 4^k BellB[k, 1/4], {k, 0, n}], {n, 0, 21}]

%Y Cf. A003576, A004213, A285064, A355164, A355165.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Jun 22 2022