%I #15 Jul 16 2014 06:31:34

%S 3,2,2,3,2,2,4,2,2,3,2,2,3,2,2,3,2,2,4,2,2,3,2,2,3,2,2,3,2,2,4,2,2,3,

%T 2,2,3,2,2,3,2,2,4,2,2,3,2,2,3,2,2,3,2,2,4,2,2,3,2,2,3,2,2,3,2,2,4,2,

%U 2,3,2,2,3,2,2,3,2,2,4,2,2,3,2,2,3,2

%N Smallest positive non-divisor of the n-th Lucas number (A000032).

%C This sequence seems to be cyclic with period 12, but the equivalent sequence for the Fibonacci numbers (A152727) is not.

%C Lucas numbers modulo 12 are cyclic with period 24 and no 0 in the cycle (unlike Fibonacci numbers): 2, 1, 3, 4, 7, 11, 6, 5, 11, 4, 3, 7, 10, 5, 3, 8, 11, 7, 6, 1, 7, 8, 3, 11. It follows that this sequence is cyclic with period 12: 3, 2, 2, 3, 2, 2, 4, 2, 2, 3, 2, 2. - _Jens Kruse Andersen_, Jul 15 2014

%H Jens Kruse Andersen, <a href="/A245070/b245070.txt">Table of n, a(n) for n = 0..1000</a>

%F For n >= 12, a(n) = a(n-12). - _Jens Kruse Andersen_, Jul 15 2014

%e a(6) = 4 because lucas(6) = 18, both 2 and 3 divide 18, but 4 does not.

%o (PARI) lucas(n) = if(n==0, 2, 2*fibonacci(n-1)+fibonacci(n));

%o vector(1000, n, m=lucas(n-1); d=2; while(m%d==0, d++); d)

%Y Cf. A000032, A152727.

%K nonn

%O 0,1

%A _Colin Barker_, Jul 12 2014