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%I #6 Dec 19 2019 16:47:16
%S 1,0,3,-14,155,-1834,27867,-492246,10068459,-232990178,6025718963,
%T -172182404734,5387697769467,-183214963001082,6728091949444491,
%U -265348057242998822,11185888456798395563,-501937946696294628946,23886968118494957119011,-1201674025637823778926414
%N a(n) = Sum_{k>=0} (k - n)^n / 2^(k + 1).
%C The n-th term of the n-th inverse binomial transform of A000670.
%F a(n) = n! * [x^n] exp(-n*x) / (2 - exp(x)).
%F a(n) = Sum_{k=0..n} binomial(n,k) * (-n)^(n - k) * A000670(k).
%F a(n) ~ (-1)^n * n^n / (2 - exp(-1)). - _Vaclav Kotesovec_, Dec 19 2019
%t Table[Sum[(k - n)^n/2^(k + 1), {k, 0, Infinity}], {n, 0, 19}]
%t Table[HurwitzLerchPhi[1/2, -n, -n]/2, {n, 0, 19}]
%t Table[n! SeriesCoefficient[Exp[-n x]/(2 - Exp[x]), {x, 0, n}], {n, 0, 19}]
%Y Cf. A000670, A052841, A290219, A292916.
%K sign
%O 0,3
%A _Ilya Gutkovskiy_, Dec 19 2019
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