A088529 - OEIS

A088529

Numerator of Bigomega(n)/Omega(n).

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

Antti Karttunen, Table of n, a(n) for n = 2..10000

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.

Index entries for sequences computed from exponents in factorization of n.

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