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A223486
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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.
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4
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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, 18, -1, 7, -1, 15, -1, -1, -1, -1, -1, -1, 9, -1, -1, 10, -1, 22, 15, -1, 12, 8, -1, 28, -1, -1, -1, -1, 18, -1, -1, -1, 21, 29, -1, -1, 15, -1, -1, -1, -1, 34, -1
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OFFSET
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1,3
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COMMENTS
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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
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REFERENCES
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A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 25.
S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
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LINKS
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EXAMPLE
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a(9) = 6 because L_6 = 18 is the first number in the Lucas sequence (A000032) that 9 divides.
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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EXTENSIONS
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Edited. Added "k >= 0" in the name and added cross references. - Wolfdieter Lang, Jan 20 2015
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STATUS
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approved
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