The continued fraction representation above had its origins on another problem I was working on sometime ago.
It's based on a very simple way of looking at the Euler's product representation of $\frac{1}{\zeta(s)}$. Interestingly it applies to every infinite product.
And this is as follows
$$
\frac{1}{\zeta(s)}=\left(1-\frac{1}{2^s}\right)-\left(1-\frac{1}{2^s}\right)\frac{1}{3^s}-\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\frac{1}{5^s}-\cdots
$$
From here its easy to derive the above continued fraction using Euler's continued fraction formula.
And thats it, It's nice and eventually a new thing.
EDIT
Just to make it clear, note that
$$
\begin{align*}
\frac{1}{\zeta(s)}&=\left(1-\frac{1}{2^s}\right)\left[\left(1-\frac{1}{3^s}\right)-\left(1-\frac{1}{3^s}\right)\frac{1}{5^s}-\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{5^s}\right)\frac{1}{7^s}-\cdots\right]\\
&=\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\left[\left(1-\frac{1}{5^s}\right)-\left(1-\frac{1}{5^s}\right)\frac{1}{7^s}-\cdots\right]\\
&\vdots\\
&=\prod_{p\in\mathbb{P}}\left(1-\frac{1}{p^{s}}\right)
\end{align*}
$$
where $\mathbb{P}$ is the set of the prime numbers.
EDIT
To derive the continued faction just put $\frac{1}{\zeta(s)}$ in the form
$$
\frac{1}{\zeta(s)}=1-\frac{1}{2^s}\left(1+\frac{2^s-1}{3^s}\left(1+\frac{3^s-1}{5^s}\left(1+\frac{5^s-1}{7^s}\left(1+\frac{7^s-1}{11^s}\left(1+\ddots\right ) \right ) \right ) \right ) \right)
$$
and then just apply the Euler continued fraction formula.
So we can write this as
$$
\frac{1}{\zeta(s)}=1-\frac{1}{2^s}-\frac{1}{2^s}\frac{2^s-1}{3^s}-\frac{1}{2^s}\frac{2^s-1}{3^s}\frac{3^s-1}{5^s}-\frac{1}{2^s}\frac{2^s-1}{3^s}\frac{3^s-1}{5^s}\frac{5^s-1}{7^s}-\cdots
$$
Now, let $a_1=-\frac{1}{2^s};a_2=\frac{2^s-1}{3^s};a_3=\frac{3^s-1}{5^s};a_4=\frac{5^s-1}{7^s}\cdots$ and we'll get the Euler continued fraction formula.
