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%I #14 Feb 26 2022 04:24:18
%S 1,-4,-24,-32,80,384,448,-1024,-4608,-5120,11264,49152,53248,-114688,
%T -491520,-524288,1114112,4718592,4980736,-10485760,-44040192,
%U -46137344,96468992,402653184,419430400,-872415232,-3623878656,-3758096384
%N a(n) = sin((2*n+5)*Pi/6)*(n+1)*2^(n+1).
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-12,16,-16).
%F Sum_{k>=0} 1/a(k) = log(2), from log((1+x)/(1-x+x^2)) at x=1/2.
%F G.f.: (1-8*x+4*x^2)/(1-2*x+4*x^2)^2.
%F Sum_{n>=0} (-1)^n/a(n) = log(7/2). - _Amiram Eldar_, Feb 26 2022
%t Table[Sin[(2*n + 5)*Pi/6]*(n + 1)*2^(n + 1), {n, 0, 27}] (* _Amiram Eldar_, Feb 26 2022 *)
%o (PARI) a(n)=[1,-1,-2,-1,1,2][n%6+1]*(n+1)*2^n
%Y Cf. A002162.
%K sign
%O 0,2
%A _Jaume Oliver Lafont_, Apr 28 2010
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