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a(n) = 3*a(n-1) + 5 with a(0) = 1.
6

%I #19 Mar 13 2024 10:51:17

%S 1,8,29,92,281,848,2549,7652,22961,68888,206669,620012,1860041,

%T 5580128,16740389,50221172,150663521,451990568,1355971709,4067915132,

%U 12203745401,36611236208,109833708629,329501125892,988503377681,2965510133048,8896530399149

%N a(n) = 3*a(n-1) + 5 with a(0) = 1.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-3).

%F a(n) = (7/2)*3^n-(5/2). - _Emeric Deutsch_, Apr 01 2006

%F a(n) = 4*a(n-1)-3*a(n-2). G.f.: (4*x+1) / ((x-1)*(3*x-1)). - _Colin Barker_, Jul 18 2013

%e The second term is 8 since a(1) = 3*a(0) + 5 = 3*1 + 5 = 8.

%p a:=n->(7*3^n-5)/2: seq(a(n),n=0..27);

%p a[0]:=1: for n from 1 to 27 do a[n]:=3*a[n-1]+5 od: seq(a[n],n=0..27);

%t a[0] := 1; a[n_] := 3a[n - 1] + 5; Table[a[n], {n, 0, 30}]

%t NestList[3#+5&,1,30] (* or *) LinearRecurrence[{4,-3},{1,8},30] (* _Harvey P. Dale_, Jul 20 2018 *)

%K nonn,easy

%O 0,2

%A _Parthasarathy Nambi_, Mar 29 2006

%E More terms from _Emeric Deutsch_ and _Stefan Steinerberger_, Apr 01 2006

%E More terms from _Colin Barker_, Jul 18 2013