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a(n) = (1/8)*A290915(n).
3

%I #4 Aug 18 2017 21:34:55

%S 0,1,4,18,84,387,1784,8228,37944,174981,806940,3721270,17160940,

%T 79139079,364956336,1683025992,7761412016,35792386313,165059516916,

%U 761185462362,3510269016516,16187892672139,74651791054120,344262840164460,1587596244438120

%N a(n) = (1/8)*A290915(n).

%H Clark Kimberling, <a href="/A290916/b290916.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, 2, 4, -1)

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

%F a(n) = 4*a(n-1) + 2*a(n-2) + 4*a(n-3) - a(n-4).

%F a(n) = (1/8)*A290915(n) for n >= 0.

%t z = 60; s = x/(1 - x)^2; p = 1 - 8 s^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)

%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290915 *)

%t u/8 (* A290916 *)

%Y Cf. A000027, A290890, A290915.

%K nonn,easy

%O 0,3

%A _Clark Kimberling_, Aug 18 2017