%I #23 Jul 31 2022 23:10:52
%S 5,29,305,4289,66305,1051649,16789505,268484609,4295163905,
%T 68720263169,1099514773505,17592198627329,281475027042305,
%U 4503599828697089,72057594843234305,1152921507828072449,18446744086594453505,295147905230892433409,4722366483075803643905
%N a(n) = 5th Fibonacci polynomial evaluated at 2^n.
%H Colin Barker, <a href="/A020531/b020531.txt">Table of n, a(n) for n = 0..830</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (21,-84,64).
%F From _Colin Barker_, May 03 2015: (Start)
%F a(n) = 1 + 3*4^n + 16^n.
%F a(n) = 21*a(n-1) - 84*a(n-2) + 64*a(n-3) for n > 2.
%F G.f.: -(116*x^2 - 76*x + 5)/((x - 1)*(4*x - 1)*(16*x - 1)). (End)
%F E.g.f.: exp(x)*(1 + 3*exp(3*x) + exp(15*x)). - _Stefano Spezia_, Jul 31 2022
%p with(combinat,fibonacci):seq(fibonacci(5,2^i),i=0..24);
%t Table[Fibonacci[5,2^i],{i,0,30}] (* _Vladimir Joseph Stephan Orlovsky_, Nov 03 2009 *)
%o (PARI) Vec(-(116*x^2-76*x+5)/((x-1)*(4*x-1)*(16*x-1)) + O(x^100)) \\ _Colin Barker_, May 03 2015
%K nonn,easy
%O 0,1
%A _Simon Plouffe_