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
KEYWORD
nonn,easy
AUTHOR
Michel Lagneau, Jul 26 2014
STATUS
approved