A088530 Denominator of bigomega(n)/omega(n).

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1

Offset

2,11

Comments

a(n) is the denominator of A022559(n)/A000720(n). - Robert Israel, Jan 08 2024

Links

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

R. L. Duncan, On the factorization of integers, Proc. Amer. Math. Soc. 25 (1970), 191-192.

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) = denominator of B/O.

Example

bigomega(24) / omega(24) = 4/2 = 2/1, so a(24) = 1.

Maple

N:= 100:
W:= ListTools:-PartialSums(map(numtheory:-bigomega, [$1..N])):
seq(denom(W[i]/numtheory:-pi(i)), i=2..N); # Robert Israel, Jan 08 2024

Mathematica

Table[Denominator[PrimeOmega[n]/PrimeNu[n]], {n, 2, 100}] (* Harvey P. Dale, Mar 22 2012 *)

Prog

(PARI) for(x=2, 100, y=bigomega(x)/omega(x); print1(denominator(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)).denominator()
print([a(n) for n in range(2, 51)]) # Indranil Ghosh, Jul 13 2017

Crossrefs

Cf. A001221, A001222, A000720, A022559, A070012, A070013, A070014, A088529 (gives the numerator).

Sequence in context: A330749 A294883 A348940 * A058060 A338160 A336137

Adjacent sequences: A088527 A088528 A088529 * A088531 A088532 A088533

Keyword

nonn,frac,look

Author

Cino Hilliard, Nov 16 2003

Status

approved