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a(n) = (5*11^n+1)/2.
1

%I #15 Nov 22 2023 21:49:02

%S 3,28,303,3328,36603,402628,4428903,48717928,535897203,5894869228,

%T 64843561503,713279176528,7846070941803,86306780359828,

%U 949374583958103,10443120423539128,114874324658930403,1263617571248234428

%N a(n) = (5*11^n+1)/2.

%H Vincenzo Librandi, <a href="/A199754/b199754.txt">Table of n, a(n) for n = 0..900</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (12,-11).

%F a(n) = 11*a(n-1)-5.

%F a(n) = 12*a(n-1)-11*a(n-2).

%F G.f.: (3-8*x)/((1-x)*(1-11*x)).

%t (5 11^Range[0,20]+1)/2 (* or *) LinearRecurrence[{12,-11},{3,28},20] (* _Harvey P. Dale_, Apr 12 2022 *)

%o (Magma) [(5*11^n+1)/2: n in [0..30]];

%K nonn,easy

%O 0,1

%A _Vincenzo Librandi_, Nov 10 2011