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A086600
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Number of primitive prime factors in the n-th Lucas number A000204(n).
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6
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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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FORMULA
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a(n) = Sum{d|n and n/d odd} mu(n/d) a(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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PROG
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(Magma) lst:=[]; pr:=1; for n in [1..105] do pd:=PrimeDivisors(Lucas(n)); d:=1; t:=0; for c in [1..#pd] do f:=pd[c]; if Gcd(pr, f) eq 1 then t+:=1; else d:=d*f; end if; end for; Append(~lst, t); pr:=pr*Truncate(Lucas(n)/d); end for; lst; // Arkadiusz Wesolowski, Jun 22 2016
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CROSSREFS
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KEYWORD
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hard,nonn
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AUTHOR
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STATUS
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approved
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