login
a(n) = (11*9^n+1)/2.
1

%I #14 Oct 23 2024 10:15:38

%S 6,50,446,4010,36086,324770,2922926,26306330,236756966,2130812690,

%T 19177314206,172595827850,1553362450646,13980262055810,

%U 125822358502286,1132401226520570,10191611038685126,91724499348166130,825520494133495166

%N a(n) = (11*9^n+1)/2.

%H Vincenzo Librandi, <a href="/A199680/b199680.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10,-9).

%F a(n) = 9*a(n-1)-4.

%F a(n) = 10*a(n-1)-9*a(n-2).

%F G.f.: 2*(3-5*x)/((1-x)*(1-9*x)).

%t (11*9^Range[0, 20] + 1)/2 (* _Wesley Ivan Hurt_, Apr 26 2023 *)

%t LinearRecurrence[{10,-9},{6,50},20] (* _Harvey P. Dale_, Oct 23 2024 *)

%o (Magma) [(11*9^n+1)/2: n in [0..30]];

%K nonn,easy

%O 0,1

%A _Vincenzo Librandi_, Nov 09 2011