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a(n) = Sum_{k=0..floor(n/4)} binomial(n-k, 3k)*(-3)^(3k)*4^(n-4k).
0

%I #22 Jan 23 2019 02:32:31

%S 1,4,16,64,229,592,-224,-18176,-175655,-1265732,-7914560,-44970752,

%T -236014307,-1145932664,-5086940240,-19929220352,-61944816911,

%U -81359219468,858917862064,10785877546432,84667993188757,555461238134080,3268576565244544,17688312222825472,88631554966652233,408731119650234796

%N a(n) = Sum_{k=0..floor(n/4)} binomial(n-k, 3k)*(-3)^(3k)*4^(n-4k).

%H P. S. Bruckman and G. C. Greubel, <a href="https://www.fq.math.ca/Problems/MayAdvanced2014.pdf">Advanced Problem H-725</a>, Fibonacci Quarterly, 52(2):187-190, 2014.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (12,-48,64,-27).

%F a(n) = (9n + 7 + 3^(3n/2)*(11*sqrt(2)*cos(n*arcsin(sqrt(2/27))) + sin(n*arcsin(sqrt(2/27))))/sqrt(2))/18.

%F G.f.: (1-4x)^2/((1-4x)^3+27x^4) = (1-4*x)^2/((x-1)^2*(1 - 10*x + 27*x^2)).

%p Gser:=series((1-4*x)^2/((1-4*x)^3+27*x^4), x = 0,35): seq(coeff(Gser, x, n), n = 0 .. 30);

%t LinearRecurrence[{12,-48,64,-27},{1,4,16,64},30] (* _Harvey P. Dale_, Nov 21 2015 *)

%K sign,easy

%O 0,2

%A _Emeric Deutsch_, Oct 27 2014