%I #10 Apr 08 2016 03:19:46
%S 1,10,-80,820,-8180,81820,-818180,8181820,-81818180,818181820,
%T -8181818180,81818181820,-818181818180,8181818181820,-81818181818180,
%U 818181818181820,-8181818181818180,81818181818181820
%N a(n) = (10/11)*(2+9*(-10)^(n-1)).
%H G. C. Greubel, <a href="/A165750/b165750.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 (-9, 10).
%F a(n) = -9*a(n-1) + 10*a(n-2), a(0)=1, a(1)=10.
%F a(n) = (-10)*a(n-1) + 20 for n>=1, a(0)=1.
%F G.f.: (1+19x)/(1+9x-10x^2).
%F a(n)= Sum_{0<=k<=n} A112555(n,k)*9^(n-k).
%F E.g.f.: (1/11)*(20*exp(x) - 9*exp(-10*x)). - _G. C. Greubel_, Apr 07 2016
%t Table[(10/11)*(2 + 9*(-10)^(n - 1)), {n, 0, 50}] (* or *)
%t LinearRecurrence[{-9,10}, {1,10}, 50] (* _G. C. Greubel_, Apr 07 2016 *)
%o (PARI) x='x+O('x^99); Vec((1+19*x)/(1+9*x-10*x^2)) \\ _Altug Alkan_, Apr 07 2016
%K easy,sign
%O 0,2
%A _Philippe Deléham_, Sep 26 2009