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a(n) = (11*3^n+1)/2.
4

%I #18 Mar 20 2023 11:32:23

%S 6,17,50,149,446,1337,4010,12029,36086,108257,324770,974309,2922926,

%T 8768777,26306330,78918989,236756966,710270897,2130812690,6392438069,

%U 19177314206,57531942617,172595827850,517787483549,1553362450646,4660087351937

%N a(n) = (11*3^n+1)/2.

%H Vincenzo Librandi, <a href="/A199113/b199113.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 (4,-3).

%F a(n) = 3*a(n-1)-1.

%F a(n) = 4*a(n-1)-3*a(n-2).

%F G.f.: (6-7*x)/((1-x)*(1-3*x)). - _Bruno Berselli_, Nov 04 2011

%t LinearRecurrence[{4,-3},{6,17},30] (* or *) (11 3^Range[0,30]+1)/2 (* _Harvey P. Dale_, Mar 25 2012 *)

%o (Magma) [(11*3^n+1)/2: n in [0..30]];

%K nonn,easy

%O 0,1

%A _Vincenzo Librandi_, Nov 04 2011