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%I #13 Mar 17 2024 08:31:07
%S 1,1,8,99,1704,37625,1014348,32300359,1186399952,49376357109,
%T 2296400723220,118031059900523,6643848377509368,406471060412884753,
%U 26856124898028246044,1905791887135240982415,144563460111417997403040,11673024609379676114380877,999663240630210837032231460
%N a(n) = n! * [x^n] 1/(2 - exp(x))^n.
%H Vaclav Kotesovec, <a href="/A305919/b305919.txt">Table of n, a(n) for n = 0..350</a>
%H N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>
%F a(n) = [x^n] Sum_{k>=0} binomial(n+k-1,k)*k!*x^k/Product_{j=1..k} (1 - j*x).
%F a(n) = Sum_{k=0..n} Stirling2(n,k)*binomial(n+k-1,k)*k!.
%F a(n) ~ n! * c * ((1 + r)*(1 + 2*r))^n / sqrt(n), where r = (-1 + 1/(-1 + LambertW(2*exp(1))))/2 = 0.833964643008471735434624869020826957702396269585... is the root of the equation (2 + 1/r) * (1 + r*LambertW(-exp(-1/r)/r)) = 1 and c = 1/sqrt(2*Pi*(1 + LambertW(2*exp(1)))) = 0.258877607195571655640738032164006... Equivalently, a(n) ~ LambertW(2*exp(1))^n * n^n / (sqrt(1 + LambertW(2*exp(1))) * 2^n * exp(n) * (LambertW(2*exp(1)) - 1)^(2*n)). - _Vaclav Kotesovec_, Dec 15 2019, updated Mar 17 2024
%t Table[n! SeriesCoefficient[1/(2 - Exp[x])^n, {x, 0, n}], {n, 0, 18}]
%t Table[SeriesCoefficient[Sum[Binomial[n + k - 1, k] k! x^k/Product[1 - j x, {j, 1, k}], {k, 0, n}], {x, 0, n}], {n, 0, 18}]
%t Table[Sum[StirlingS2[n, k] Binomial[n + k - 1, k] k!, {k, 0, n}], {n, 0, 18}]
%Y Cf. A000670, A005649, A053492, A226513, A226515.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Jun 14 2018
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