TITLE: Density of the "multiplicative odd numbers"
QUESTION [10 upvotes]: I am interested in the set $A$ of all positive integer numbers such that when factored into primes, the sum of the exponents is odd (I think of $A$ as the multiplicative odd numbers).
I want to know if it has positive upper density, more precisely
$$\bar d(A):=\limsup_{n\to\infty}\frac{|A\cap[1,n]|}n$$
I think I read somewhere that it has density $1/2$ (and the $\lim$ exist, not just the $\limsup$), but I would be happy with a proof that $\bar d(A)>0$.

REPLY [12 votes]: Your sequence is http://oeis.org/A026424.
Define the zeta density of the set of integers $A$ as 
$$ d(A) = \lim_{x \to 1+} \frac1{\zeta(x)}  \sum_{k \in A} k^{-x}. $$
Then from results given at the OEIS, this is
$$ d(A) = \lim_{x \to 1+} {1 \over \zeta(x)} \left[ {\zeta(x)^2 - \zeta(2x) \over 2\zeta(x)} \right]. $$
After some simplifcation this is
$$ \lim_{x \to 1} \left( {1 \over 2} - {\zeta(2x) \over 2 \zeta(x)^2} \right) $$
and recalling that $\lim_{x \to 1} \zeta(x)$ is infinity while $\zeta(2) = \pi^2/6$, this is $1/2$.   
Now, if a set has a natural density, then it has a zeta density, and the two densities are equal; see for example Chapter 2 of Diaconis' PhD dissertation.  So we can conclude that if your set has a natural density, then that natural density is $1/2$.