OFFSET
1,8
COMMENTS
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
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..222 (terms 1..168 from T. D. Noe)
Blair Kelly, Fibonacci and Lucas Factorizations
FORMULA
EXAMPLE
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
MATHEMATICA
Table[Length[FactorInteger[Fibonacci[Prime[n]]]], {n, 60}]
PrimeNu[Fibonacci[Prime[Range[100]]]] (* Harvey P. Dale, Mar 13 2016 *)
PROG
(PARI) a(n) = omega(fibonacci(prime(n))); \\ Michel Marcus, Oct 22 2015
CROSSREFS
KEYWORD
hard,nonn
AUTHOR
T. D. Noe, Feb 16 2003
STATUS
approved