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a(n) = (1/2)A291405(n).
2

%I #4 Sep 07 2017 21:30:28

%S 0,1,2,4,12,26,60,145,336,796,1880,4412,10400,24488,57648,135794,

%T 319744,752908,1773016,4175032,9831472,23151400,54516976,128377380,

%U 302304640,711870368,1676321856,3947423264,9295440640,21889019712,51544533376,121377706184

%N a(n) = (1/2)A291405(n).

%H Clark Kimberling, <a href="/A291406/b291406.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0, 2, 4, 4, 8, 12, 8, 2)

%F G.f.: -((x (1 + x)^2 (1 + x^2 + 2 x^3 + x^4))/(-1 + 2 x^2 + 4 x^3 + 4 x^4 + 8 x^5 + 12 x^6 + 8 x^7 + 2 x^8)).

%F a(n) = 2*a(n-2) + 4*a(n-3) + 4*a(n-4) + 5*a(n-5) + 12*a(n-6) + 8*a(n-7) + 2*a(n-8) for n >= 9.

%t z = 60; s = x + x^2; p = 1 - 2 s^2 - 2 s^4;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A019590 *)

%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291405 *)

%t u / 2 (* A291406 *)

%Y Cf. A019590, A291382, A291405.

%K nonn,easy

%O 0,3

%A _Clark Kimberling_, Sep 07 2017