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Consider the exponentiated Riemann-Zeta function $\zeta(s)^p$. If it is represented as

$$\zeta(s)^p = \sum_{n=1}^\infty\frac{a_n}{n^s}$$

Is there any upper bound we can put on $|a_n|$ in terms of $n$ and $p$.

For example, not that when $p = 2$, we get the divisor function which can be bounded above by $n^{\frac{1}{\log \log n}}$

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For $k \in \Bbb{Z}_{\ge 1}$ let $$\zeta(s)^k = \prod_p \frac{1}{(1-p^{-s})^k}= \sum_{n=1}^\infty a_k(n)n^{-s}$$

  • For $n$ square free then $$a_k(n) = \prod_{p | n} a_k(p) =\prod_{p | n} k = k^{\omega(n)}$$

    If $\omega(n) = m$ then $$n \ge \prod_{p \le \pi^{-1}(m)} p = \exp(\sum_{p \le (1+o(1))m \log m} \log p) = \exp( (1+o(1))m\log m)$$ thus $\omega(n) \le (1+o(1)) \frac{\log n}{\log\log n}$ and $$a_k(n) \le k^{(1+o(1)) \frac{\log n}{\log \log n}} = n^{(1+o(1)) \frac{\log k}{\log \log n}}$$

  • Since $a_k(p^l) = \frac{\prod_{j=0}^{l-1} (k+j)}{l!} \le k^{l+1}$ we have for any $n$ $$a_k(n)\le \sup_l (n^{1/l})^{(1+o(1)) \frac{\log k^{l+1}}{\log \log n}} \le n^{(1+o(1)) \frac{\log k}{\log \log n}}$$

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