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Dec 14, 2012 at 1:02 answer added Michael Lugo timeline score: 12
Dec 13, 2012 at 23:47 comment added George Lowther To explain my previous comment. Let $S$ be the set of numbers of the form $4^r(2s+1)$. This has density $(1/2)(1+1/4+1/4^2+\cdots)=2/3$. Write $f(n)=\lvert A\cap[1,n]\rvert$. The property that $\Omega(2n)=\Omega(n)+1$ gives $$\lvert A\cap[1,n/2]\cap S\rvert+\lvert A\cap[1,n]\cap S^c\rvert=\lvert S\cap[1,n/2]\rvert.$$ So, $f(n)=\lvert S\cap[1,n/2]\rvert+\lvert A\cap(n/2,n]\cap S\rvert$ giving the inequality $$1/3\le\liminf_{n\to\infty}f(n)/n\le\limsup_{n\to\infty}f(n)/n\le2/3.$$
Dec 13, 2012 at 23:37 vote accept Joel Moreira
Dec 13, 2012 at 22:59 answer added Peter Humphries timeline score: 18
Dec 13, 2012 at 22:47 comment added George Lowther It will indeed have density 1/2. Just looking at power of 2 factors shows that it is in the range [1/3,2/3]. Then looking at power of 3 factors will get a tighter range about 1/2, and so on.
Dec 13, 2012 at 22:42 comment added Gerry Myerson The number of prime factors of $n$, counted with multiplicities, is often denoted $\Omega(n)$. You are asking about $\sum_1^n(-1)^{\Omega(m)}$.
Dec 13, 2012 at 22:29 history asked Joel Moreira CC BY-SA 3.0