%I #6 Nov 24 2023 15:45:28
%S 1,10,90,765,6264,49968,390960,3013740,22958640,173225952,1296640224,
%T 9640743120,71270772480,524277204480,3840015361536,28018969060032,
%U 203753553511680,1477232299307520,10681095982072320,77040637862485248,554445497303525376
%N a(n) = (1/12)*A291391(n).
%H Clark Kimberling, <a href="/A291392/b291392.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 (12, -24, -72, -36)
%F G.f.: -(((1 + x) (-1 + 3 x + 3 x^2))/(-1 + 6 x + 6 x^2)^2).
%F a(n) = 12*a(n-1) - 24*a(n-2) - 72*a(n-3) - 36*a(n-4) for n >= 5.
%t z = 60; s = x + x^2; p = (1 - 6 s)^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A019590 *)
%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291391 *)
%t u / 12 (* A291392 *)
%t LinearRecurrence[{12,-24,-72,-36},{1,10,90,765},30] (* _Harvey P. Dale_, Nov 24 2023 *)
%Y Cf. A019590, A291382, A291391.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Sep 06 2017