A154022 -id:A154022 - cp's OEIS frontend

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

0, 4, 64, 1020, 16256, 259076, 4128960, 65804284, 1048739584, 16714029060, 266375725376, 4245297576956, 67658385505920, 1078288870517764, 17184963542778304, 273881127813935100, 4364913081480183296

Offset: 1

Vincenzo Librandi, Jan 04 2009

If a(n)=x and a(n+1)=y, then 16=(x^2+y^2)/(xy+1).

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".

Vincenzo Librandi, Table of n, a(n) for n = 1..800
Index entries for linear recurrences with constant coefficients, signature (16,-1).

Cf. A065100, A154022-A154027.

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

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 *)
LinearRecurrence[{16, -1}, {0, 4}, 20] (* T. D. Noe, Apr 17 2011 *)

From R. J. Mathar, Jan 05 2011: (Start)
G.f.: 4*x^2/(1 -16*x +x^2).
a(n) = 4*A077412(n-2). (End)

375725376 replaced by 266375725376 - R. J. Mathar, Jan 07 2009

Edited by N. J. A. Sloane, Jun 23 2010 at the suggestion of Joerg Arndt.

cp's OEIS Frontend

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

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