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A083584 a(n) = (8*4^n - 5)/3.

%I A083584 #47 May 22 2022 01:52:02
%S A083584 1,9,41,169,681,2729,10921,43689,174761,699049,2796201,11184809,
%T A083584 44739241,178956969,715827881,2863311529,11453246121,45812984489,
%U A083584 183251937961,733007751849,2932031007401,11728124029609,46912496118441
%N A083584 a(n) = (8*4^n - 5)/3.
%C A083584 a(n) = A007583(n+1) - 2 = A020988(n) - 1 = A039301(n+2) - 3. - _Ralf Stephan_, Jun 14 2003
%C A083584 Sum of n-th row of triangle of powers of 4: 1; 4 1 4; 16 4 1 4 16; 64 16 4 1 4 16 64; .... - _Philippe Deléham_, Feb 24 2014
%H A083584 Vincenzo Librandi, <a href="/A083584/b083584.txt">Table of n, a(n) for n = 0..170</a>
%H A083584 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,-4).
%F A083584 a(n) = (8*4^n - 5)/3.
%F A083584 G.f.: (1+4*x)/((1-x)*(1-4*x)).
%F A083584 E.g.f.: (8*exp(4*x) - exp(x))/3.
%F A083584 a(0)=1, a(1)=9, a(n) = 5*a(n-1) - 4*a(n-2). - _Harvey P. Dale_, Oct 23 2011
%F A083584 a(n) = 4*a(n-1) + 5, a(0) = 1. - _Philippe Deléham_, Feb 24 2014
%F A083584 a(n+1) = 2^(2^n+1) + a(n), a(1)=1. - _Ben Paul Thurston_, Dec 27 2015
%e A083584 a(0) = 1;
%e A083584 a(1) = 4 + 1 + 4 = 9;
%e A083584 a(2) = 16 + 4 + 1 + 4 + 16 = 41;
%e A083584 a(3) = 64 + 16 + 4 + 1 + 4 + 16 + 64 = 169; etc. - _Philippe Deléham_, Feb 24 2014
%t A083584 (8 4^Range[0,30]-5)/3 (* or *) LinearRecurrence[{5,-4},{1,9},30] (* _Harvey P. Dale_, Oct 23 2011 *)
%o A083584 (Magma) [(8*4^n-5)/3: n in [0..40] ]; // _Vincenzo Librandi_, Apr 28 2011
%o A083584 (PARI) a(n)=(8*4^n-5)/3 \\ _Charles R Greathouse IV_, Oct 07 2015
%o A083584 (Python) print([8*4**n//3 - 1 for n in range(50)]) # _Karl V. Keller, Jr._, May 21 2022
%Y A083584 Cf. A083855.
%K A083584 easy,nonn
%O A083584 0,2
%A A083584 _Paul Barry_, May 01 2003