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A086600
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Number of primitive prime factors in Lucas(n).
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4
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0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 2, 2, 1, 2, 2, 3, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 2, 2, 3, 2, 1, 1, 2, 2, 3, 1, 2, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 3, 2, 2, 2, 1, 2, 2, 2, 1, 1, 2, 2, 3, 3, 1, 2, 2, 3, 2, 3, 2, 3, 3, 2
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OFFSET
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1,22
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COMMENTS
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A prime factor of Lucas(n) is called primitive if it does not divide Lucas(r) for any r < n. It can be shown that there is at least one primitive prime factor for n > 6. When n is prime, all the prime factors of Lucas(n) are primitive.
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..1000 (using Blair Kelly's data)
Eric Weisstein's World of Mathematics, Lucas Number
Blair Kelly, Fibonacci and Lucas Factorizations
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FORMULA
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a(n) = Sum{d|n and n/d odd} mu(n/d) A086600(d) -1 if 6|n and n/6 is a power of 2.
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EXAMPLE
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a(22) = 2 because Lucas(22) = 3*43*307 and neither 43 nor 307 divide a smaller Lucas number.
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MATHEMATICA
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Lucas[n_] := Fibonacci[n+1] + Fibonacci[n-1]; pLst={}; Join[{0}, Table[f=Transpose[FactorInteger[Lucas[n]]][[1]]; f=Complement[f, pLst]; cnt=Length[f]; pLst=Union[pLst, f]; cnt, {n, 2, 150}]]
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CROSSREFS
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Cf. A000204 (Lucas numbers), A086598 (number of distinct prime factors), A086599 (number of prime factors, counting multiplicity).
Sequence in context: A137454 A030613 A092984 * A218450 A025912 A029441
Adjacent sequences: A086597 A086598 A086599 * A086601 A086602 A086603
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KEYWORD
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hard,nonn
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AUTHOR
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T. D. Noe, Jul 24 2003
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STATUS
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approved
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