login
a(n) = 6th Fibonacci polynomial evaluated at 2^n.
1

%I #19 Aug 04 2023 15:44:16

%S 8,70,1292,34840,1065008,33685600,1074790592,34368127360,

%T 1099578737408,35184908961280,1125904201812992,36028831378708480,

%U 1152921779484766208,36893490346442383360,1180591638309597396992,37778932003694650163200,1208925820740529081745408

%N a(n) = 6th Fibonacci polynomial evaluated at 2^n.

%H Colin Barker, <a href="/A020532/b020532.txt">Table of n, a(n) for n = 0..664</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (42,-336,512).

%F From _Colin Barker_, May 03 2015: (Start)

%F a(n) = 2^n*(3+4^(1+n)+16^n).

%F a(n) = 42*a(n-1)-336*a(n-2)+512*a(n-3) for n>2.

%F G.f.: -2*(520*x^2-133*x+4) / ((2*x-1)*(8*x-1)*(32*x-1)).

%F (End)

%p with(combinat,fibonacci):seq(fibonacci(6,2^i),i=0..24);

%t Table[Fibonacci[6,2^i],{i,0,30}] (* _Vladimir Joseph Stephan Orlovsky_, Nov 03 2009 *)

%t LinearRecurrence[{42,-336,512},{8,70,1292},20] (* _Harvey P. Dale_, Aug 04 2023 *)

%o (PARI) Vec(-2*(520*x^2-133*x+4)/((2*x-1)*(8*x-1)*(32*x-1)) + O(x^100)) \\ _Colin Barker_, May 03 2015

%K nonn,easy

%O 0,1

%A _Simon Plouffe_