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A088529 Numerator of Bigomega(n)/Omega(n). 30
1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 3, 1, 1, 1, 4, 1, 3, 1, 3, 1, 1, 1, 2, 2, 1, 3, 3, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 5, 2, 3, 1, 3, 1, 2, 1, 2, 1, 1, 1, 4, 1, 1, 3, 6, 1, 1, 1, 3, 1, 1, 1, 5, 1, 1, 3, 3, 1, 1, 1, 5, 4, 1, 1, 4, 1, 1, 1, 2, 1, 4, 1, 3, 1, 1, 1, 3, 1, 3, 3, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET
2,3
REFERENCES
H. Z. Cao, On the average of exponents, Northeast. Math. J., Vol. 10 (1994), pp. 291-296.
LINKS
R. L. Duncan, On the factorization of integrers, Proc. Amer. Math. Soc. 25 (1970), 191-192.
Steven Finch, Errata and Addenda to Mathematical Constants, arXiv:2001.00578 [math.HO], 2020-2022.
FORMULA
Let B = number of prime divisors of n with multiplicity, O = number of distinct prime divisors of n. Then a(n) = numerator of B/O.
a(n) = A136565(n) = A181591(n) for n: 2 <= n < 24. - Reinhard Zumkeller, Nov 01 2010
Sum_{k=2..n} a(k)/A088530(k) ~ n + O(n/log(log(n))) (Duncan, 1970). - Amiram Eldar, Oct 14 2022
Sum_{k=2..n} a(k)/A088530(k) = n + c_1 * n/log(log(n)) + c_2 * n/log(log(n))^2 + O(n/log(log(n))^3), where c_1 = A136141 and c_2 = A272531 (Cao, 1994; Finch, 2020). - Amiram Eldar, Dec 15 2022
EXAMPLE
bigomega(24) / omega(24) = 4/2 = 2, so a(24) = 2.
MATHEMATICA
Table[Numerator[PrimeOmega[n]/PrimeNu[n]], {n, 2, 100}] (* Michael De Vlieger, Jul 12 2017 *)
PROG
(PARI) for(x=2, 100, y=bigomega(x)/omega(x); print1(numerator(y)", "))
(Python)
from sympy import primefactors, Integer
def bigomega(n): return 0 if n==1 else bigomega(Integer(n)/primefactors(n)[0]) + 1
def omega(n): return Integer(len(primefactors(n)))
def a(n): return (bigomega(n)/omega(n)).numerator()
print([a(n) for n in range(2, 51)]) # Indranil Ghosh, Jul 13 2017
CROSSREFS
Cf. A001221, A001222, A070012, A070013, A070014, A088530 (gives the denominator).
Sequence in context: A293515 A326622 A292777 * A267116 A136565 A181591
KEYWORD
nonn,frac
AUTHOR
Cino Hilliard, Nov 16 2003
STATUS
approved

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