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%I #31 Jul 26 2022 22:04:04
%S 0,1,1,1,1,1,1,2,1,1,2,3,2,1,1,2,2,2,3,2,2,2,1,2,4,2,3,2,2,2,2,1,1,3,
%T 4,2,4,4,2,2,3,3,2,2,4,2,4,4,2,5,3,4,3,2,3,3,4,2,2,3,4,2,4,4,4,3,2,3,
%U 5,4,2,1,7,5,4,3,3,2,2,4,3,4,1,1,5,5,1,3,5,3,2,3,4,3,4,6,1,3,4,3
%N a(n) is the number of prime factors in Fibonacci(prime(n)); that is, in the Fibonacci number whose index is the n-th prime.
%C In all known examples, Fibonacci(prime(n)) is squarefree, in which case a(n) is well-defined, i.e., the number of distinct prime factors equals the total number of prime factors. But if for some n, Fibonacci(prime(n)) has a repeated prime factor, then a(n) is not well-defined. - _Jonathan Sondow_, Oct 22 2015
%H Amiram Eldar, <a href="/A080345/b080345.txt">Table of n, a(n) for n = 1..222</a> (terms 1..168 from T. D. Noe)
%H Blair Kelly, <a href="http://mersennus.net/fibonacci//">Fibonacci and Lucas Factorizations</a>
%F a(n) = A001221(A000045(A000040(n))). - _Michel Marcus_, Oct 22 2015
%e a(12) = 3 because the 12th prime is 37 and Fibonacci(37) = 24157817 = 73 * 149 * 2221 has 3 prime factors. - clarified by _Jonathan Sondow_, Oct 21 2015
%t Table[Length[FactorInteger[Fibonacci[Prime[n]]]], {n, 60}]
%t PrimeNu[Fibonacci[Prime[Range[100]]]] (* _Harvey P. Dale_, Mar 13 2016 *)
%o (PARI) a(n) = omega(fibonacci(prime(n))); \\ _Michel Marcus_, Oct 22 2015
%Y Cf. A001605, A022307.
%K hard,nonn
%O 1,8
%A _T. D. Noe_, Feb 16 2003
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