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a(n) = (1/5)*A291389(n).
2

%I #4 Sep 06 2017 21:13:31

%S 2,17,130,940,6550,44475,296250,1944375,12612500,81035000,516537500,

%T 3270615625,20591031250,128998328125,804673593750,5000444062500,

%U 30969644531250,191231146484375,1177627753906250,7234317013671875,44342955390625000,271252632343750000

%N a(n) = (1/5)*A291389(n).

%H Clark Kimberling, <a href="/A291390/b291390.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 (10, -15, -50, -25)

%F G.f.: -(((1 + x) (-2 + 5 x + 5 x^2))/(-1 + 5 x + 5 x^2)^2).

%F a(n) = 10*a(n-1) - 15*a(n-2) - 50*a(n-3) - 25*a(n-4) for n >= 5.

%t z = 60; s = x + x^2; p = (1 - 5 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] (* A291389 *)

%t u / 5 (* A291390 *)

%Y Cf. A019590, A291382, A291389.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Sep 06 2017