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a(n) = Lucas(4n) - 3, or Lucas(2n-1)*Lucas(2n+1).
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%I #40 Dec 16 2023 15:48:27

%S 4,44,319,2204,15124,103679,710644,4870844,33385279,228826124,

%T 1568397604,10749957119,73681302244,505019158604,3461452807999,

%U 23725150497404,162614600673844,1114577054219519,7639424778862804,52361396397820124,358890350005878079

%N a(n) = Lucas(4n) - 3, or Lucas(2n-1)*Lucas(2n+1).

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

%H Amiram Eldar, <a href="/A081078/b081078.txt">Table of n, a(n) for n = 1..1196</a>

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

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

%F From _R. J. Mathar_, Sep 03 2010: (Start)

%F G.f.: x*(-4-12*x+x^2) / ( (x-1)*(x^2-7*x+1)).

%F a(n) = A056854(n)-3. (End)

%F From _Peter Bala_, Nov 30 2013: (Start)

%F a(n) = Lucas(2*n)^2 - 5.

%F Sum_{n>=1} 1/a(n) = (5 - sqrt(5))/10. (End)

%F Sum_{n>=1} 1/a(n) = A244847. - _Amiram Eldar_, Oct 27 2020

%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 1 to 40 do printf(`%d,`,luc(4*n)-3) od: # _James A. Sellers_, Mar 05 2003

%t LinearRecurrence[{8,-8,1},{4,44,319},30] (* _Harvey P. Dale_, Jun 08 2014 *)

%t Table[LucasL[4*n] - 3, {n, 1, 20}] (* _Amiram Eldar_, Oct 27 2020 *)

%o (PARI) Vec(x*(-4-12*x+x^2) / ( (x-1)*(x^2-7*x+1)) + O(x^30)) \\ _Michel Marcus_, Dec 23 2014

%Y Cf. A000032 (Lucas numbers), A056854 (Lucas(4n)), A244847.

%K nonn,easy

%O 1,1

%A _R. K. Guy_, Mar 04 2003