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Number of primitive prime factors in Fibonacci(n).
14

%I #16 Dec 08 2017 11:25:43

%S 0,0,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,1,1,2,1,1,1,

%T 1,1,3,1,1,1,2,1,1,2,1,2,1,1,2,2,1,1,2,1,2,1,2,2,2,1,2,1,1,2,1,1,3,2,

%U 3,2,2,1,2,1,1,1,2,2,2,2,3,1,1,2,2,2,2,3,2,2,2,2,1,1,3,2,4,1,2,2,2,2,3,2,1

%N Number of primitive prime factors in Fibonacci(n).

%C A prime factor of Fibonacci(n) is called primitive if it does not divide Fibonacci(r) for any r < n. It can be shown that there is at least one primitive prime factor for n > 12. When n is prime, all the prime factors of Fibonacci(n) are primitive; see A080345 for a count of these.

%H T. D. Noe, <a href="/A086597/b086597.txt">Table of n, a(n) for n = 1..1000</a>

%H R. D. Carmichael, <a href="https://doi.org/10.2307%2F1967797">On the numerical factors of the arithmetic forms α^n ± β^n</a>, Annals of Math., 15 (1/4) (1913), 30-70.

%H Blair Kelly, <a href="http://mersennus.net/fibonacci//">Fibonacci and Lucas Factorizations</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonacciNumber.html">Fibonacci Number</a>

%F a(n) = Sum{d|n} mu(n/d) A022307(d), inverse Mobius transform of A022307.

%F a(n) = 0 if and only if n = 1, 2, 6, or 12, by Carmichael's theorem. - _Jonathan Sondow_, Dec 07 2017

%e a(19) = 2 because Fibonacci(19) = 37*113 and neither 37 nor 113 divide a smaller Fibonacci number.

%t pLst={}; Join[{0, 0}, Table[f=Transpose[FactorInteger[Fibonacci[n]]][[1]]; f=Complement[f, pLst]; cnt=Length[f]; pLst=Union[pLst, f]; cnt, {n, 3, 150}]]

%o (PARI) a(n)=my(t=fibonacci(n),g); fordiv(n,d, if(d==n, break); g=fibonacci(d); while((g=gcd(g, t))>1, t /= g)); omega(t) \\ _Charles R Greathouse IV_, Oct 06 2016

%Y Cf. A022307 (number of distinct prime factors), A038575 (number of prime factors, counting multiplicity), A061446 (primitive part of Fibonacci(n)), A080345.

%K nonn

%O 1,19

%A _T. D. Noe_, Jul 24 2003