Known facts about a function

Question

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.)

Answer 1

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.