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A081078 a(n) = Lucas(4n) - 3, or Lucas(2n-1)*Lucas(2n+1).

%I A081078 #41 Jul 02 2025 16:02:01
%S A081078 4,44,319,2204,15124,103679,710644,4870844,33385279,228826124,
%T A081078 1568397604,10749957119,73681302244,505019158604,3461452807999,
%U A081078 23725150497404,162614600673844,1114577054219519,7639424778862804,52361396397820124,358890350005878079
%N A081078 a(n) = Lucas(4n) - 3, or Lucas(2n-1)*Lucas(2n+1).
%D A081078 Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.
%H A081078 Amiram Eldar, <a href="/A081078/b081078.txt">Table of n, a(n) for n = 1..1196</a>
%H A081078 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-8,1).
%F A081078 a(n) = 8a(n-1) - 8a(n-2) + a(n-3).
%F A081078 From _R. J. Mathar_, Sep 03 2010: (Start)
%F A081078 G.f.: x*(-4-12*x+x^2) / ( (x-1)*(x^2-7*x+1)).
%F A081078 a(n) = A056854(n)-3. (End)
%F A081078 From _Peter Bala_, Nov 30 2013: (Start)
%F A081078 a(n) = Lucas(2*n)^2 - 5.
%F A081078 Sum_{n>=1} 1/a(n) = (5 - sqrt(5))/10. (End)
%F A081078 Sum_{n>=1} 1/a(n) = A244847. - _Amiram Eldar_, Oct 27 2020
%p A081078 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 Sellers_, Mar 05 2003
%t A081078 LinearRecurrence[{8,-8,1},{4,44,319},30] (* _Harvey P. Dale_, Jun 08 2014 *)
%t A081078 Table[LucasL[4*n] - 3, {n, 1, 20}] (* _Amiram Eldar_, Oct 27 2020 *)
%o A081078 (PARI) Vec(x*(-4-12*x+x^2) / ( (x-1)*(x^2-7*x+1)) + O(x^30)) \\ _Michel Marcus_, Dec 23 2014
%Y A081078 Cf. A000032 (Lucas numbers), A056854 (Lucas(4n)), A244847.
%K A081078 nonn,easy
%O A081078 1,1
%A A081078 _R. K. Guy_, Mar 04 2003