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%I #12 Jul 25 2014 05:26:19
%S 1,1,6,45,508,7225,126306,2606065,62075952,1675774089,50565938050,
%T 1686510607111,61609858744248,2446470026497705,104922088624078194,
%U 4833250468667819325,238004208840601580416,12476420334546637657489,693675026024580055139778
%N a(n) = n! * [x^n] (exp(x)+x^2/2!)^n.
%C In general, if a(n) = n! * [x^n] (exp(x) + x^k/k!)^n, k>=1, then limit n-> infinity (a(n)/n!)^(1/n) = ((1-k*r)/(1-r))^(k-1) / (r*k!), where r is the root of the equation exp((k*r-1)/(1-r)) = r*k! * (1-r)^(k-1) / (1-k*r)^k.
%H Vincenzo Librandi, <a href="/A245493/b245493.txt">Table of n, a(n) for n = 0..200</a>
%F a(n) ~ c * d^n * n^n / exp(n), where d = (1-2*r)/(2*r*(1-r)) = 3.177499696443893762475339445134038..., where r = 0.13317988718414524112... is the root of the equation exp((2*r-1)/(1-r)) = 2*r*(1-r)/(1-2*r)^2, and c = 1.061620103934913384222610538939... .
%t Table[n!*SeriesCoefficient[(E^x + x^2/2)^n, {x, 0, n}], {n, 0, 20}]
%t With[{k=2}, Flatten[{1, Table[Sum[Binomial[n, j]*Binomial[n, k*j]*(n-j)^(n-k*j)*(k*j)!/(k!)^j, {j, 0, n/k}], {n, 1, 20}]}]]
%Y Cf. A245406, A245405, A245496.
%K nonn,easy
%O 0,3
%A _Vaclav Kotesovec_, Jul 24 2014
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