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
Examples
bigomega(24) / omega(24) = 4/2 = 2/1, so a(24) = 1.
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
Crossrefs
Programs
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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
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Mathematica
Table[Denominator[PrimeOmega[n]/PrimeNu[n]],{n,2,100}] (* Harvey P. Dale, Mar 22 2012 *)
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PARI
for(x=2,100,y=bigomega(x)/omega(x);print1(denominator(y)","))
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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
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.
Comments