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%I #18 Jan 22 2015 15:46:30
%S 0,0,2,3,-1,6,4,-1,6,-1,5,-1,-1,12,-1,-1,-1,6,9,-1,-1,15,12,-1,-1,-1,
%T 18,-1,7,-1,15,-1,-1,-1,-1,-1,-1,9,-1,-1,10,-1,22,15,-1,12,8,-1,28,-1,
%U -1,-1,-1,18,-1,-1,-1,21,29,-1,-1,15,-1,-1,-1,-1,34,-1
%N Lucas entry points: a(n) = least k such that n divides Lucas number L_k (=A000032(k), for k >= 0), or -1 if there is no such k.
%C If one takes L_k, for k >= 1, that is A000204, then a(1) = 1 and a(2) = 3 followed by the given numbers. This fits then with A106291(n) = A253808(n)*a(n), n >= 1 (where in A253808 a negative entry at position n indicates, as in the present sequence, that the Lucas numbers are not divisible by n. For odd primes not dividing any Lucas numbers see A053028. No power 2^m, m >= 3 divides any Lucas number, see, e.g., Vajda, p. 81). - _Wolfdieter Lang_, Jan 20 2015
%D A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 25.
%D S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
%H T. D. Noe, <a href="/A223486/b223486.txt">Table of n, a(n) for n = 1..10000</a>
%e a(9) = 6 because L_6 = 18 is the first number in the Lucas sequence (A000032) that 9 divides.
%t test[n_] := Module[{a, b, t, cnt = 1}, {a, b} = {2, 1}; While[cnt++; t = b; b = Mod[a + b, n]; a = t; ! (b == 0 || {a, b} == {2, 1})]; If[b == 0, cnt, -1]]; Join[{0, 0}, Table[test[i], {i, Range[3, 100]}]] (* _T. D. Noe_, Mar 22 2013 *)
%Y Cf. A000032, A000204, A001177, A194363, A053028 (primes not dividing any Lucas numbers), A106291, A253808.
%K sign
%O 1,3
%A _Casey Mongoven_, Mar 20 2013
%E Edited. Added "k >= 0" in the name and added cross references. - _Wolfdieter Lang_, Jan 20 2015
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