Known facts about a function In my work I have met the function on the unit circle whose Fourier coefficients are 
$$
c_n=\frac{1}{|n|}\prod (d_k+1)
$$
if $n=\pm\prod p_k^{d_k}$ is the decomposition of the integer $n$ into the product of prime numbers. The formal series of the function is
$$
\sum_{n\in\mathbb Z}c_nz^n.
$$
This function could have already appeared and I would much appreciate any references about it. (I am interested in properties of this function, but now it is not yet easy to say what I really need.)
 A: Doing a Google search for "sum d(n) x^n" gives http://en.wikipedia.org/wiki/Divisor_function as one of the first answers. Looking in there, I find "A Lambert series involving the divisor function is: 
$\sum_{n=1}^{\infty} q^n \sigma_a(n) = \sum_{n=1}^{\infty} \frac{n^a q^n}{1-q^n}$
where $\sigma_a(n) = \sum_{d|n} d^a$."
Setting $a=0$, 
$\sum_{n=1}^{\infty} q^n d(n) = \sum_{n=1}^{\infty} \frac{q^n}{1-q^n}$.
Your series, calling it $f$, is
$f(z) = \sum_{n=1}^{\infty} z^n d(n)/n$.
Differentiating,
$f'(z) = \sum_{n=1}^{\infty} z^{n-1} d(n)
=(1/z)\sum_{n=1}^{\infty} z^n d(n)
= (1/z)\sum_{n=1}^{\infty} \frac{z^n}{1-z^n}
= \sum_{n=1}^{\infty} \frac{z^{n-1}}{1-z^n}
$.
To get $f(z)$ we integrate this term by term,
using $f(z) = \int_0^z f'(y) dy$.
Letting $x = y^n$,
$\int_0^z \frac{y^{n-1}}{1-y^n} dy
=(1/n) \int_0^{z^n} \frac{dx}{1-x} 
=- \frac{\ln{1-y}}{n}]_0^{z^n}
=- \ln(1-z^n)/n
$
so
$f(z) = -\sum_{n=1}^{\infty} \frac{\ln{1-z^n}}{n}
= - \ln\prod_{n=1}^{\infty} (1-z^n)^{1/n}
$.
I'm not sure where to go from here (assuming no errors, which has $p < .7$ in my estimation), so I'll leave it at this.
