login
Lucas(4n+2) - 3, or 5*Fibonacci(2n)*Fibonacci(2n+2).
0

%I #28 Dec 16 2023 17:45:27

%S 0,15,120,840,5775,39600,271440,1860495,12752040,87403800,599074575,

%T 4106118240,28143753120,192900153615,1322157322200,9062201101800,

%U 62113250390415,425730551631120,2918000611027440,20000273725560975

%N Lucas(4n+2) - 3, or 5*Fibonacci(2n)*Fibonacci(2n+2).

%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.: -15*x/(x-1)/(x^2-7*x+1). a(n) = 15*A092521(n) = 5*A058038(n). - _R. J. Mathar_, Sep 03 2010

%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)-3) od: # _James A. Sellers_, Mar 05 2003

%t LinearRecurrence[{8, -8, 1}, {0, 15, 120}, 20] (* _Jean-François Alcover_, Nov 29 2023 *)

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

%K nonn,easy

%O 0,2

%A _R. K. Guy_, Mar 04 2003