Terence Tao gives at his blog the following formula for something called the singular series:
$$\large\mathfrak{S}(h)=2\Pi_{2}\prod\limits_{p|h;p>2}\frac{p-2}{p-1}$$
where $\Pi_{2}=0.66016...$ is the twin prime constant.
I can't understand how to compute the product: $$\prod\limits_{p|h;p>2}\frac{p-2}{p-1}$$
Could you possibly compute numerically a few examples and maybe explain what $$p|h;p>2$$ means?
Usually $p|h$ means $p$ divides $h$, but since $h$ can be a small integer, I don't understand how it works.
I believe it just means to use all of the odd prime numbers, $p$, which are factors of $h$ in the product of $\frac{p-2}{p-1}$. Thus, for example, if $h = 90 = 2 \times 3^2 \times 5$, then the odd prime factors $p$ are $3$ and $5$, thus giving that
$$\prod\limits_{p|h;p>2}\frac{p-2}{p-1} = \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) = \frac{3}{8} \tag{1}\label{eq1}$$