Density of the "multiplicative odd numbers" 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 $\bar d(A)>0$.
 A: Gerry has the right idea here: you are asking about the limiting behaviour of the sum
$$\frac{1}{x} \sum_{n \leq x}{\frac{1 - (-1)^{\Omega(n)}}{2}}.$$
The arithmetic function $\lambda(n) = (-1)^{\Omega(n)}$ is known as Liouville's function. It is well-known (and equivalent to the prime number theorem!) that the summatory function of the Liouville function,
$$L(x) = \sum_{n \leq x}{\lambda(n)},$$
satisfies the asymptotic
$$L(x) = o(x)$$
as $x$ tends to infinity. (In fact, one can probably improve this slightly in the usual way to get better error terms in the prime number theorem.) So it is indeed true that
$$d(A) = \lim_{x \to \infty} \frac{1}{x} \sum_{n \leq x}{\frac{1 - (-1)^{\Omega(n)}}{2}} = \frac{1}{2},$$
and with a little work you could actually say something slightly stronger about the rate at which this converges.
A: 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 $1/2$.
