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 A086600 Number of primitive prime factors in Lucas(n). 4
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,22 COMMENTS 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. LINKS 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 FORMULA 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. EXAMPLE a(22) = 2 because Lucas(22) = 3*43*307 and neither 43 nor 307 divide a smaller Lucas number. MATHEMATICA 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}]] CROSSREFS 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 KEYWORD hard,nonn AUTHOR T. D. Noe, Jul 24 2003 STATUS approved

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