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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?

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Have you tried playing around with a bivariate generating function such as $\sum_{n \geq 1} \frac{\phi(n) y^n}{1-x^n} $? –  Sinai Robins Nov 15 '12 at 1:25
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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. :) –  fedja Nov 15 '12 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. –  Nikita Sidorov Nov 15 '12 at 12:51
    
Plouffe’s inverter appears to be down. Too bad. –  Emil Jeřábek Nov 15 '12 at 14:45

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