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A088529
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Numerator of Bigomega(n)/Omega(n).
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30
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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
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OFFSET
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2,3
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REFERENCES
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H. Z. Cao, On the average of exponents, Northeast. Math. J., Vol. 10 (1994), pp. 291-296.
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LINKS
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FORMULA
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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.
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
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EXAMPLE
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bigomega(24) / omega(24) = 4/2 = 2, so a(24) = 2.
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MATHEMATICA
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Table[Numerator[PrimeOmega[n]/PrimeNu[n]], {n, 2, 100}] (* Michael De Vlieger, Jul 12 2017 *)
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PROG
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(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()
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CROSSREFS
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KEYWORD
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nonn,frac
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AUTHOR
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STATUS
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approved
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