What are the coefficients $a_n$ if $\sum_{n\geq1}\frac{a_n}{z^n-1}=e^{-z}$

Question

Is it possible to represent the $e^{-z}$ as a $$\sum_{n\geq1}\frac{a_n}{z^n-1}=e^{-z}$$

If so what are the coefficients $a_n$?

Answer 1

$$\sum_{n=0}^\infty \dfrac{(-z)^n}{n!} = \sum_{n=1}^\infty -a_n\sum_{r=0}^\infty z^{nr}$$

Comparing the coefficient of $z^n$:

$$\forall n \in \Bbb N: \dfrac{(-1)^n}{n!} = - \sum_{d|n,\ d \in \Bbb N} a_d$$

Consider $n=1$:

$$-1 = -a_1$$

Hence $a_1 = 1$.

Consider the primes:

$$\forall p \in \Bbb P: \dfrac{(-1)^p}{p!} = - (a_1 + a_p)$$

Also:

$$\dfrac{(-1)^{p^2}}{(p^2)!} = -(a_1+a_p+a_{p^2})$$

Observe that generally $a_{p^n} = \dfrac{(-1)^{p^{n-1}}}{(p^{n-1})!} - \dfrac{(-1)^{p^n}}{(p^n)!}$.

Consider $n=pq$:

$$\dfrac{(-1)^{pq}}{(pq)!} = -(a_1+a_p+a_q+q_{pq})$$

Hence $a_{pq} = -\dfrac{(-1)^{pq}}{(pq)!} + \dfrac{(-1)^p}{p!} + \dfrac{(-1)^q}{q!} + 1$.

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And at this point I'm quite tired of repeating the process, but the main idea is to factorize a number and to build up the coefficient.

I'm sure there's a method much faster.