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%I #47 Dec 16 2023 15:53:32
%S 1,16,121,841,5776,39601,271441,1860496,12752041,87403801,599074576,
%T 4106118241,28143753121,192900153616,1322157322201,9062201101801,
%U 62113250390416,425730551631121,2918000611027441,20000273725560976
%N a(n) = Lucas(4*n+2)-2, or Lucas(2*n+1)^2.
%C Conjecture: a(n) = Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k). - _Alex Ratushnyak_, May 06 2012
%D Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.
%H Vincenzo Librandi, <a href="/A081071/b081071.txt">Table of n, a(n) for n = 0..1000</a>
%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 G.f.: -(1+8*x+x^2)/((x-1)*(x^2-7*x+1)). - _Colin Barker_, Jun 26 2012
%F From _Peter Bala_, Nov 19 2019: (Start)
%F Sum_{n >= 1} 1/(a(n) + 5) = (3*sqrt(5) - 5)/30.
%F Sum_{n >= 1} 1/(a(n) - 5) = (15 - 4*sqrt(5) )/60.
%F Sum_{n >= 1} (-1)^(n+1)/(a(n) - 5) = 1/12.
%F Sum_{n >= 1} (-1)^(n+1)/(a(n) - 25/a(n)) = (5 + 2*sqrt(5))/120. (End)
%F Sum_{n>=0} 1/a(n) = (1/sqrt(5)) * Sum_{n>=1} n/F(2*n), where F(n) is the n-th Fibonacci number (A000045). - _Amiram Eldar_, Oct 05 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 0 to 40 do printf(`%d,`,luc(4*n+2)-2) od: # _James A. Sellers_, Mar 05 2003
%t CoefficientList[Series[-(1+8*x+x^2)/((x-1)*(x^2-7*x+1)),{x,0,40}],x] (* or *) LinearRecurrence[{8,-8,1},{1,16,121},50] (* _Vincenzo Librandi_, Jun 26 2012 *)
%t LucasL[4*Range[0,20]+2]-2 (* _Harvey P. Dale_, Nov 25 2012 *)
%o (Magma) I:=[1, 16, 121]; [n le 3 select I[n] else 8*Self(n-1)-8*Self(n-2)+Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Jun 26 2012
%o (PARI) x='x+O('x^30); Vec((1+8*x+x^2)/((1-x)*(x^2-7*x+1))) \\ _G. C. Greubel_, Dec 21 2017
%Y Cf. A000032 (Lucas numbers), A000045, A002878 is Lucas(2n+1), A081069.
%K nonn,easy
%O 0,2
%A _R. K. Guy_, Mar 04 2003
%E More terms from _James A. Sellers_, Mar 05 2003
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