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A154021 a(n+2) = 16*a(n+1) - a(n), with a(1)=0, a(2)=4.

%I A154021 #36 Dec 31 2023 11:35:55
%S A154021 0,4,64,1020,16256,259076,4128960,65804284,1048739584,16714029060,
%T A154021 266375725376,4245297576956,67658385505920,1078288870517764,
%U A154021 17184963542778304,273881127813935100,4364913081480183296
%N A154021 a(n+2) = 16*a(n+1) - a(n), with a(1)=0, a(2)=4.
%C A154021 If a(n)=x and a(n+1)=y, then 16=(x^2+y^2)/(xy+1).
%C A154021 In general, the sequence a(1)=0, a(2)=U; a(n+2)=U^2*a(n+1)-a(n) has the property that "If a(n)=x and a(n+1)=y then (x^2+y^2)/(xy+1)=U^2".
%H A154021 Vincenzo Librandi, <a href="/A154021/b154021.txt">Table of n, a(n) for n = 1..800</a>
%H A154021 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (16,-1).
%F A154021 From _R. J. Mathar_, Jan 05 2011: (Start)
%F A154021 G.f.: 4*x^2/(1 -16*x +x^2).
%F A154021 a(n) = 4*A077412(n-2). (End)
%t A154021 Nest[Append[#,16Last[#]-#[[-2]]]&,{0,4},20]  (* or *) Rest[CoefficientList[Series[4x^2/(1-16x+x^2), {x,0,22}], x]]  (* _Harvey P. Dale_, Apr 17 2011 *)
%t A154021 LinearRecurrence[{16, -1}, {0, 4}, 20] (* _T. D. Noe_, Apr 17 2011 *)
%o A154021 (Magma) I:=[0,4]; [n le 2 select I[n] else 16*Self(n-1)-Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Feb 25 2012
%Y A154021 Cf. A065100, A154022-A154027.
%K A154021 nonn,easy
%O A154021 1,2
%A A154021 _Vincenzo Librandi_, Jan 04 2009
%E A154021 375725376 replaced by 266375725376 - _R. J. Mathar_, Jan 07 2009
%E A154021 Edited by _N. J. A. Sloane_, Jun 23 2010 at the suggestion of _Joerg Arndt_.