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a(n) = 8^n + 2^(n+1).
1

%I #22 Sep 08 2022 08:44:45

%S 3,12,72,528,4128,32832,262272,2097408,16777728,134218752,1073743872,

%T 8589938688,68719484928,549755830272,4398046543872,35184372154368,

%U 281474976841728,2251799813947392,18014398510006272

%N a(n) = 8^n + 2^(n+1).

%C 4th Fibonacci polynomial evaluated at 2^n.

%H Vincenzo Librandi, <a href="/A020530/b020530.txt">Table of n, a(n) for n = 0..140</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10,-16)

%F G.f. 3*(1-6*x)/((8*x-1)*(2*x-1)). - _R. J. Mathar_, Jun 07 2016

%F a(n) = 3*A103334(n+1). - _R. J. Mathar_, Jun 07 2016

%F E.g.f.: (2 + exp(6*x))*exp(2*x). - _Ilya Gutkovskiy_, Jun 07 2016

%p with(combinat,fibonacci):seq(fibonacci(4,2**i),i=0..24);

%t Table[Fibonacci[4,2^i],{i,0,30}] (* _Vladimir Joseph Stephan Orlovsky_, Nov 03 2009 *)

%t LinearRecurrence[{10,-16},{3,12},30] (* _Harvey P. Dale_, May 04 2018 *)

%o (Magma) [8^n + 2^(n+1): n in [0..30]]; // _Vincenzo Librandi_, Apr 26 2011

%K nonn,easy

%O 0,1

%A _Simon Plouffe_