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Lucas(k)/(3k) for k = 2*3^n, where Lucas(k) is k-th Lucas number (A000032).
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%I #5 Mar 31 2012 13:20:25

%S 1,107,1190741689,14769352340699478579719327005523,

%T 253650450218391594062880777243777017638488805917392303113120204411172926964476779033181303378188721

%N Lucas(k)/(3k) for k = 2*3^n, where Lucas(k) is k-th Lucas number (A000032).

%C a(n) divides a(n+1). a(n+1)/a(n) = {107, 11128427, 12403489755282666163307, 17174107866559209832245996776509546318861182768126017871860347845227, ...}. a(n+1)/a(n) is prime for n = {1, 2, 4}.

%F a(n) = ( Fibonacci[ 2*3^n - 1 ] + Fibonacci[ 2*3^n + 1 ] ) / ( 2*3^(n+1) ). a(n) = A000032[ 2*3^n ] / ( 2*3^(n+1) ).

%t Table[ ( Fibonacci[ 2*3^n - 1 ] + Fibonacci[ 2*3^n + 1 ] ) / ( 2*3^(n+1) ), {n,1,5} ]

%Y Cf. A000032, A016089 = numbers n such that n divides n-th Lucas number. Cf. A128935 = Fibonacci(5^n) / 5^n.

%K nonn

%O 1,2

%A _Alexander Adamchuk_, May 13 2007