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A086600 Number of primitive prime factors in the n-th Lucas number A000204(n). 6
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.
FORMULA
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.
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}]]
PROG
(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
CROSSREFS
Cf. A000204 (Lucas numbers), A058036, A086598 (number of distinct prime factors), A086599 (number of prime factors, counting multiplicity), A274333.
Sequence in context: A338428 A297382 A258257 * A218450 A025912 A029441
KEYWORD
hard,nonn
AUTHOR
T. D. Noe, Jul 24 2003
STATUS
approved

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Last modified March 28 05:39 EDT 2024. Contains 371235 sequences. (Running on oeis4.)