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I'm reposting this question with a clarification and deleting the old one.

From the comments on that question I learned that $$\zeta^k(s)=\sum_{n=1}^\infty\frac{d_k(n)}{n^s}$$ where $d_k(n)$ is the number of ways of writing $n$ as a product of $k$ positive integers (I assume that the order of the terms in the products matters in order to make $d_2(n)=d(n)$ but check me on this).

I had previously said I was interested in the dirichlet series of $\zeta^k(s)$ for $k$ being an integer, but I realized that I need $k$ to be any positive real number.

According to this generating function for the coefficients of such a dirichlet series, $k\in\mathbb{R}$ is valid.

What would the coefficients of our dirichlet series be?

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  • $\begingroup$ We recommend against deleting questions except in the extreme case. Questions are not just for your own viewing, their for everyone to look at. And now one less potentially useful question for me to read. Also, you mention things you've previously said, but as you've deleted that question, I imagine most of us are rather confused as to what you are saying. $\endgroup$ – Simply Beautiful Art Jun 19 '17 at 23:51
  • $\begingroup$ Also you have virtually no good reason to look at $\zeta(s)^a$. $\endgroup$ – reuns Jun 19 '17 at 23:56
  • $\begingroup$ I apologize. I'll refrain from doing this again. $\endgroup$ – tyobrien Jun 19 '17 at 23:57
  • $\begingroup$ See my exercice comment below. $\endgroup$ – reuns Jun 20 '17 at 0:10
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Let $$d_k(n) = \# \{ m \in \mathbb{Z}_{\ge 1}^k,\prod_{i=1}^k m_i = n\}, \qquad e_k(n) = \# \{ m \in \mathbb{Z}_{\ge 2}^k,\prod_{i=1}^k m_i = n\}$$ For $k \in \mathbb{Z}_{\ge 0}$ : $$\zeta(s)^k = (\sum_{m=1}^\infty m^{-s})^k = \sum_{m \in \mathbb{Z}_{\ge 1}^k } \prod_{i=1}^k m_i^{-s} =\sum_{n=1}^\infty n^{-s} d_k(n)$$

$$(\zeta(s)-1)^k = (\sum_{m=2}^\infty m^{-s})^k = \sum_{m \in \mathbb{Z}_{\ge 2}^k } \prod_{i=1}^k m_i^{-s} =\sum_{n=1}^\infty n^{-s} e_k(n)$$

For $a \in \mathbb{C}\setminus - \mathbb{N}$ since $\zeta(s)-1 \to 0$ as $\Re(s) \to \infty$ : $$\zeta(s)^a=(1+(\zeta(s)-1))^a = \sum_{k=0}^\infty {a \choose k}(\zeta(s)-1)^k =\sum_{k=0}^\infty {a \choose k} \sum_{n=1}^\infty n^{-s} e_k(n) \\=\sum_{n=1}^\infty n^{-s}\sum_{k=0}^\infty {a \choose k} e_k(n) $$

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    $\begingroup$ From this, can you deduce $(\sum_{n=1}^\infty \chi(n) n^{-s})^a$ where $\chi$ is any completely multiplicative function ? $\endgroup$ – reuns Jun 19 '17 at 23:57
  • $\begingroup$ As soon as I saw the sum in the question I felt the Dirichlet $\chi$ had something to do with it... $\endgroup$ – Klangen Jul 25 '17 at 21:18

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