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 A245580 Smallest Lucas number L(m) > L(n) that is divisible by the n-th Lucas number L(n) = A000204(n). 1
 3, 18, 76, 322, 1364, 5778, 24476, 103682, 439204, 1860498, 7881196, 33385282, 141422324, 599074578, 2537720636, 10749957122, 45537549124, 192900153618, 817138163596, 3461452808002, 14662949395604, 62113250390418, 263115950957276, 1114577054219522 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Property: a(1) = L(2) and a(n) = L(3*n), for n >=2, where L = A000204 are the Lucas numbers. LINKS Harvey P. Dale, Table of n, a(n) for n = 1..1000 Index entries for linear recurrences with constant coefficients, signature (4,1). FORMULA a(n) = A014448(n), n>1. From Colin Barker, Jul 29 2014: (Start) a(n) = (2-sqrt(5))^n+(2+sqrt(5))^n for n>1. a(n) = 4*a(n-1)+a(n-2) for n>3. G.f.: -x*(x^2+6*x+3) / (x^2+4*x-1). (End) EXAMPLE 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). MATHEMATICA Table[k=1; While[Mod[LucasL[k], LucasL[n]] !=0||LucasL[k]==LucasL[n], k++]; LucasL[k], {n, 0, 30}] LinearRecurrence[{4, 1}, {3, 18, 76}, 30] (* Harvey P. Dale, Jan 05 2022 *) PROG (PARI) Vec(-x*(x^2+6*x+3)/(x^2+4*x-1) + O(x^100)) \\ Colin Barker, Jul 31 2014 CROSSREFS Cf. A000204, A237268, A014448 . Sequence in context: A059393 A301820 A094033 * A043008 A241628 A056319 Adjacent sequences: A245577 A245578 A245579 * A245581 A245582 A245583 KEYWORD nonn,easy AUTHOR Michel Lagneau, Jul 26 2014 STATUS approved

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Last modified November 27 11:20 EST 2022. Contains 358397 sequences. (Running on oeis4.)