# An identity which involves Euler's totient function

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?

• Have you tried playing around with a bivariate generating function such as $\sum_{n \geq 1} \frac{\phi(n) y^n}{1-x^n}$? Nov 15, 2012 at 1:25
• It looks fairly hopeless. Unless $2$ is somehow extra-special, I see no way to get any decent answer (if you want to know why, I can post later). The bright side of this is that you can now introduce this constant and name it after yourself. :) Nov 15, 2012 at 1:33
• Sinai - No, I haven't. I guess I'm not sure how... fedja - Thanks, but this constant is not that important. It's just a part of a technical argument, nothing more. Nov 15, 2012 at 12:51
• Plouffe’s inverter appears to be down. Too bad. Nov 15, 2012 at 14:45