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a(n) = (9/5)*(1+4*(-9)^(n-1)).
2

%I #17 Apr 07 2016 17:05:18

%S 1,9,-63,585,-5247,47241,-425151,3826377,-34437375,309936393,

%T -2789427519,25104847689,-225943629183,2033492662665,-18301433963967,

%U 164712905675721,-1482416151081471,13341745359733257

%N a(n) = (9/5)*(1+4*(-9)^(n-1)).

%H G. C. Greubel, <a href="/A165749/b165749.txt">Table of n, a(n) for n = 0..500</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (-8, 9).

%F a(n) = -8*a(n-1) + 9*a(n-2), a(0)=1, a(1)=9.

%F a(n) = (-9)*a(n-1) + 18 for n>=1, with a(0)=1.

%F G.f.: (1+17x)/(1+8x-9x^2).

%F a(n) = Sum_{0<=k<=n} A112555(n,k)*8^(n-k).

%F E.g.f.: (1/5)*(9*exp(x) - 4*exp(-9*x)). - _G. C. Greubel_, Apr 07 2016

%t Table[9/5 (1+4(-9)^(n-1)),{n,0,20}] (* or *) LinearRecurrence[{-8,9},{1,9},20] (* _Harvey P. Dale_, Nov 24 2011 *)

%o (PARI) x='x+O('x^99); Vec((1+17*x)/(1+8*x-9*x^2)) \\ _Altug Alkan_, Apr 07 2016

%o (PARI) a(n) = (9-4*(-9)^n)/5 \\ _Charles R Greathouse IV_, Apr 07 2016

%K easy,sign

%O 0,2

%A _Philippe Deléham_, Sep 26 2009