|
%I #4 Sep 11 2017 20:04:42
%S 1,3,9,28,84,255,772,2340,7088,21476,65064,197126,597228,1809420,
%T 5481980,16608712,50319264,152451844,461882016,1399359936,4239628608,
%U 12844765920,38915675520,117902483528,357208128400,1082230358448,3278823900048,9933824239744
%N a(n) = (1/2)*A291730(n)..
%H Clark Kimberling, <a href="/A291731/b291731.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2, 2, 2, 4, 0, 2)
%F G.f.: -(((1 + x^2) (1 + x + x^3))/(-1 + 2 x + 2 x^2 + 2 x^3 + 4 x^4 + 2 x^6)).
%F a(n) = 4*a(n-1) + 2*a(n-2) - 4*a(n-3) - 2*a(n-4) for n >= 5.
%t z = 60; s = x + x^3; p = 1 - 2 s - 2 s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A154272 *)
%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291730 *)
%t u / 2 (*A291731)
%Y Cf. A154272, A291728, A291730.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Sep 11 2017
|
