login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

Lucas(4n)-2, or 5*Fibonacci(2n)^2.
3

%I #38 Dec 16 2023 17:44:56

%S 0,5,45,320,2205,15125,103680,710645,4870845,33385280,228826125,

%T 1568397605,10749957120,73681302245,505019158605,3461452808000,

%U 23725150497405,162614600673845,1114577054219520,7639424778862805

%N Lucas(4n)-2, or 5*Fibonacci(2n)^2.

%D Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

%H Hang Gu and Robert M. Ziff, <a href="http://arxiv.org/abs/1111.5626">Crossing on hyperbolic lattices</a>, arXiv:1111.5626 [cond-mat.dis-nn], 2011-2012 (see Eq. 4).

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-8,1).

%F a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).

%F a(n) = 5*A049684(n).

%F G.f.: 5*x*(x+1)/((1-x)*(x^2-7*x+1)). - _Colin Barker_, Jun 24 2012

%p luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 40 do printf(`%d,`,luc(4*n)-2) od: # _James A. Sellers_, Mar 05 2003

%t LinearRecurrence[{8, -8, 1}, {0, 5, 45}, 20] (* _Jean-François Alcover_, Nov 24 2017 *)

%o (Magma) [Lucas(4*n)-2: n in [0..30]]; // _Vincenzo Librandi_, Apr 21 2011

%o (PARI) a(n) = 5*fibonacci(2*n)^2; \\ _Michel Marcus_, Nov 24 2017

%Y Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A049684.

%K nonn,easy

%O 0,2

%A _R. K. Guy_, Mar 04 2003