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

%I #24 Jun 14 2024 12:19:51

%S 1,8,43,218,1093,5468,27343,136718,683593,3417968,17089843,85449218,

%T 427246093,2136230468,10681152343,53405761718,267028808593,

%U 1335144042968,6675720214843,33378601074218,166893005371093

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

%H Vincenzo Librandi, <a href="/A117617/b117617.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 (6, -5).

%F G.f.: (1+2*x)/((1-x)*(1-5*x)). - _Philippe Deléham_, Feb 22 2014

%F a(n) = 6*a(n-1) - 5*a(n-2), a(0) = 1, a(1) = 8. - _Philippe Deléham_, Feb 22 2014

%e If n=1 then a(1) = 5*a(0) + 3 = 5*1 + 3 = 8 which is the second term.

%t a[0] := 1; a[n_] := 5*a[n - 1] + 3; Table[a[n], {n, 0, 35}] (* _Stefan Steinerberger_, Apr 08 2006 *)

%t NestList[5#+3&,1,20] (* _Harvey P. Dale_, Jun 14 2024 *)

%Y Cf. A116952.

%K nonn

%O 0,2

%A _Parthasarathy Nambi_, Apr 07 2006

%E More terms from _Stefan Steinerberger_, Apr 08 2006