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a(n) = Lucas(4n+2)+2, or 5*Fibonacci(2n+1)^2.
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%I #37 Jul 19 2024 11:31:25

%S 5,20,125,845,5780,39605,271445,1860500,12752045,87403805,599074580,

%T 4106118245,28143753125,192900153620,1322157322205,9062201101805,

%U 62113250390420,425730551631125,2918000611027445,20000273725560980

%N a(n) = Lucas(4n+2)+2, or 5*Fibonacci(2n+1)^2.

%C a(n) is the square of limit of (G(j+2n-1) + G(j-2n+1))/G(j) as j -> infinity, where G(n) is any sequence of the form G(n+1) = G(n) + G(n-1), with any starting values, including non-integer values. G(n) includes Lucas and Fibonacci. Compare with A005248 for even number offsets from j in any such G(n). - _Richard R. Forberg_, Nov 16 2014

%C a(n) = (t(i+6n+3) + t(i))/(t(i+4n+2) + t(i+2n+1)) + 3, where (t) is any sequence of the form t(n+2) = 2t(n+1) + 2t(n) - t(n-1) or of the form t(n+1) = 3t(n) - t(n-1) without regard to initial values as long as t(i+4n+2) + t(i+2n+1) != 0. - _Klaus Purath_, Jun 23 2024

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

%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 G.f.: -5*(x^2-4*x+1)/((x-1)*(x^2-7*x+1)). - _Colin Barker_, Jun 25 2012

%F a(n) ~ phi^(4n+2). - _Charles R Greathouse IV_, Nov 17 2014

%F a(n) = 5*A081068(n). - _R. J. Mathar_, Feb 13 2016

%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 Table[LucasL[4n+2]+2,{n,0,20}] (* or *)

%t Table[5Fibonacci[2n+1]^2,{n,0,30}] (* _Harvey P. Dale_, Apr 18 2011 *)

%o (PARI) a(n)=5*fibonacci(2*n+1)^2 \\ _Charles R Greathouse IV_, Nov 17 2014

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

%K nonn,easy

%O 0,1

%A _R. K. Guy_, Mar 04 2003