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a(n) = (1/12)*A291391(n).
2

%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