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Smallest Lucas number L(m) > L(n) that is divisible by the n-th Lucas number L(n) = A000204(n).
1

%I #36 Jan 05 2022 16:32:30

%S 3,18,76,322,1364,5778,24476,103682,439204,1860498,7881196,33385282,

%T 141422324,599074578,2537720636,10749957122,45537549124,192900153618,

%U 817138163596,3461452808002,14662949395604,62113250390418,263115950957276,1114577054219522

%N Smallest Lucas number L(m) > L(n) that is divisible by the n-th Lucas number L(n) = A000204(n).

%C Property: a(1) = L(2) and a(n) = L(3*n), for n >=2, where L = A000204 are the Lucas numbers.

%H Harvey P. Dale, <a href="/A245580/b245580.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,1).

%F a(n) = A014448(n), n>1.

%F From _Colin Barker_, Jul 29 2014: (Start)

%F a(n) = (2-sqrt(5))^n+(2+sqrt(5))^n for n>1.

%F a(n) = 4*a(n-1)+a(n-2) for n>3.

%F G.f.: -x*(x^2+6*x+3) / (x^2+4*x-1). (End)

%e a(4) = 322 is the first Lucas number that is divisible by 7, the 4th Lucas number, so a(4) = 322. With the property a(n) = L(3*n), a(4) = A000204(12).

%t Table[k=1;While[Mod[LucasL[k],LucasL[n]] !=0||LucasL[k]==LucasL[n],k++];LucasL[k],{n,0,30}]

%t LinearRecurrence[{4,1},{3,18,76},30] (* _Harvey P. Dale_, Jan 05 2022 *)

%o (PARI) Vec(-x*(x^2+6*x+3)/(x^2+4*x-1) + O(x^100)) \\ _Colin Barker_, Jul 31 2014

%Y Cf. A000204, A237268, A014448 .

%K nonn,easy

%O 1,1

%A _Michel Lagneau_, Jul 26 2014