I'm looking for a good approximation of the following sum: $$\sum_{n\leq x}\omega\left(n\right)$$ where $n$ is odd.
On the wikipedia page for the prime omega function, there is a section called: Example I: A modified summatory function. In that section there is that formula:
$$ S_{odd}\left(x\right)=\frac{x}{2}\log\log x+\frac{\left(2B_{1}-1\right)x}{4}+\left\{ \frac{x}{4}\right\} -\left[x\equiv2,3\mod4\right]_{\delta}+O\left(\frac{x}{\log x}\right) $$
If I read correctly, I think this is exactly what I need, but there are some parts I don't understand. When I use only the first 2 terms, I already get a good approximation, but I would like if someone could explain these: $$\left\{ \frac{x}{4}\right\} $$ $$\left[x\equiv2,3\mod4\right]_{\delta}$$
$\left\{ \dfrac{x}{4}\right\} =\dfrac{x}{4}\ -\bigg\lfloor\dfrac{x}{4} \bigg\rfloor$ is the fractional part of $\dfrac x4$,
and $\left[x\equiv2,3\mod4\right]_{\delta}$ is the Iverson bracket, which means it is $1$ if $x\equiv 2, 3\bmod 4$, and it is $0$ otherwise.