It is well known that $$ \sum_{n=1}^\infty \frac{\phi(n)x^n}{1-x^n}=\frac x{(1-x)^2}, $$ where $\phi$ is Euler's totient function and $|x|<1$ - see [Hardy and Wright, Theorem 309]. For $x=\frac12$ this immediately yields $$ \sum_{n=1}^\infty \frac{\phi(n)}{2^n-1}=2. $$ What I need for my research is the analytic value for $$ \sum_{n=1}^\infty \frac{\phi(n)}{(2^n-1)^2}. $$ Numerically it is $1.1659457\dots$, which doesn't look like something familiar to me (or to Google, for that matter).

Any ideas?