# 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.)

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The function $d(n)=\prod_k (d_k+1)$ is the number of positive divisors of $n$. Also sometimes written $\tau(n)$. So your $c_n=\frac{d(|n|)}{|n|}$ – Thomas Andrews Nov 5 '12 at 21:48
Thanks, I know this. I am interested in the function given by the series. – Limanac Nov 5 '12 at 21:54
Ah, when you said "I have met the function on the unit circle," I assumed that meant you have a known function, and were trying to understand the coefficients. – Thomas Andrews Nov 5 '12 at 21:58
Sorry if my words admit ambiguity. – Limanac Nov 5 '12 at 22:01
your function is closely related to the logarithm of the modular form $\Delta$ (the discriminant) – user8268 Nov 5 '12 at 22:15

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.

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To get your formula, one can use the relation $f(z)=\sum_{n=1}^\infty \sum_{k=1}^\infty\frac{z^{kn}}{kn}$. – Limanac Nov 8 '12 at 17:56
For my question, one should take the real part. Thanks! – Limanac Nov 8 '12 at 17:57