|
|
A038575
|
|
Number of prime factors of n-th Fibonacci number, counted with multiplicity.
|
|
18
|
|
|
0, 0, 0, 1, 1, 1, 3, 1, 2, 2, 2, 1, 6, 1, 2, 3, 3, 1, 5, 2, 4, 3, 2, 1, 9, 3, 2, 4, 4, 1, 7, 2, 4, 3, 2, 3, 10, 3, 3, 3, 6, 2, 7, 1, 5, 5, 3, 1, 12, 3, 6, 3, 4, 2, 8, 4, 7, 5, 3, 2, 12, 2, 3, 5, 6, 3, 7, 3, 5, 5, 7, 2, 14, 2, 4, 6, 5, 4, 8, 2, 9, 7, 3, 1, 13, 4, 3, 4, 9, 2, 12, 5, 6, 4, 2, 6, 16, 4, 5, 6, 10, 2, 8
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,7
|
|
COMMENTS
|
|
|
LINKS
|
Douglas Lind, Problem H-145, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 6, No. 6 (1968), p. 351; Factor Analysis, Solution to Problem H-145 by the proposer, ibid., Vol. 8, No. 4 (1970), pp. 386-387.
|
|
FORMULA
|
|
|
EXAMPLE
|
a(12) = 6 because Fibonacci(12) = 144 = 2^4 * 3^2 has 6 prime factors.
|
|
MAPLE
|
with(numtheory):with(combinat):a:=proc(n) if n=0 then 0 else bigomega(fibonacci(n)) fi end: seq(a(n), n=0..102); # Zerinvary Lajos, Apr 11 2008
|
|
MATHEMATICA
|
Join[{0, 0}, Table[Plus@@(Transpose[FactorInteger[Fibonacci[n]]][[2]]), {n, 3, 102}]]
Join[{0}, PrimeOmega[Fibonacci[Range[110]]]] (* Harvey P. Dale, Apr 14 2018 *)
|
|
PROG
|
(Haskell)
a038575 n = if n == 0 then 0 else a001222 $ a000045 n
(Python)
from sympy import primeomega, fibonacci
def a(n): return 0 if n == 0 else primeomega(fibonacci(n))
|
|
CROSSREFS
|
Cf. A022307 (number of distinct prime factors), A086597 (number of primitive prime factors).
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|