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A159964 a(n) = 2^n*(1-n).

%I A159964 #38 Jan 13 2021 06:18:05
%S A159964 1,0,-4,-16,-48,-128,-320,-768,-1792,-4096,-9216,-20480,-45056,-98304,
%T A159964 -212992,-458752,-983040,-2097152,-4456448,-9437184,-19922944,
%U A159964 -41943040,-88080384,-184549376,-385875968,-805306368,-1677721600,-3489660928
%N A159964 a(n) = 2^n*(1-n).
%C A159964 Hankel transform of A124791. Binomial transform of -A060747.
%C A159964 {1} U A159964 is a composition of generating functions of A165747 and A000012, with H=G(F(x)) with F(x) for A000012 and G(x) for A165747. - _Oboifeng Dira_, Aug 29 2019
%H A159964 Oboifeng Dira, <a href="http://www.seams-bull-math.ynu.edu.cn/downloadfile.jsp?filemenu=_201706&amp;filename=07_41(6).pdf">A Note on Composition and Recursion</a>, Southeast Asian Bulletin of Mathematics (2017), Vol. 41, Issue 6, 849-853.
%H A159964 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4).
%F A159964 G.f.: (1-4x)/(1-2x)^2.
%F A159964 a(n) = -A058922(n). - _Jeffrey R. Goodwin_, Nov 11 2011
%F A159964 E.g.f.: U(0) where U(k)=  1 - 2*x/(2 - 4/(2 - (k+1)/U(k+1))) ; (continued fraction, 3-step). - _Sergei N. Gladkovskii_, Oct 18 2012
%F A159964 a(n) = Sum_{k=0..n} (1-2k) * C(n,k). - _Wesley Ivan Hurt_, Sep 23 2017
%F A159964 From _Amiram Eldar_, Jan 13 2021: (Start)
%F A159964 Sum_{n>=2} 1/a(n) = -log(2)/2.
%F A159964 Sum_{n>=2} (-1)^n/a(n) = -log(3/2)/2. (End)
%t A159964 LinearRecurrence[{4,-4},{1,0},30] (* _Harvey P. Dale_, May 02 2016 *)
%Y A159964 Cf. A058922, A060747, A124791.
%K A159964 easy,sign
%O A159964 0,3
%A A159964 _Paul Barry_, Apr 28 2009