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%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
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