A question about Dirichlet series i have the following question given 2 options as i) and ii)
Let $f(n)$ be the unique positive real-valued arithmetic function that satisfies $\sum_{d | n} f(d) f(n / d)=1$ for all $n$ .
(i.e., $f$ is the positive "Dirichlet square root" of the function 1 ). Let $F(s)=\sum_{n=1}^{\infty} f(n) n^{-s}$ be the Dirichlet series of $F(s)$.
(i) Express $F(s)$ for $\sigma>1$ in terms of the Riemann zeta function.
(ii) Find an explicit formula for $f\left(p^{k}\right),$ where $p$ is prime and $k \geq 1$
 A: i) This is simply calculating $F(s)^2$ by manipulating the resulting series expression. The result is $F(s)=\sqrt{\zeta(s)}$.
ii) From this it follows $$\lim_{s\rightarrow \infty} F(s)=1=f(1) \, .$$
Finally $$1=\sum_{d|p^k} f(d)f(p^k/d) = f(1)f(p^k) + f(p)f(p^{k-1}) + ... + f(p^{k-1})f(p) + f(p^k)f(1) \\
\Rightarrow \quad f(p^k) = \frac{1}{2} - \frac{1}{2}\sum_{m=1}^{k-1} f(p^m) f(p^{k-m}) \, .$$
So clearly $f(1)=1,f(p)=1/2,f(p^2)=3/8,...$ and any higher power will be of the form $f(p^k)=\frac{a_k}{2^{2k-1}}$ for integers $a_k$.
You can calculate these numbers explicitly by the following procedure. Start with the identity
$$1=\sum_{m=0}^k f(p^m)f(p^{k-m}) \\
\Rightarrow \quad \sum_{k=0}^\infty x^k = \frac{1}{1-x} = \sum_{k=0}^\infty x^k \sum_{m=0}^k f(p^m) f(p^{k-m}) \\
\Rightarrow \quad \frac{1}{1-x} = \left(\sum_{k=0}^\infty f(p^k) x^k\right)^2 \, .$$
Hence, $$f(p^k)=\left[x^k\right] \frac{1}{\sqrt{1-x}}=(-1)^k\binom{-1/2}{k}\,.$$

For small values of $n$ you can obtain $f(n)$ by expanding out $\sqrt{\zeta(s)}$ for large values of $s$ i.e.
$$\sqrt{\zeta(s)} = \sqrt{1+\zeta(s)-1} = \sum_{k=0}^\infty \binom{1/2}{k} (\zeta(s)-1)^k \\
=1+\sum_{m=2}^\infty m^{-s} \sum_{k=1}^\infty \binom{1/2}{k} \sum_{\substack{n_1,...,n_k=2 \\ n_1 \cdots n_k \stackrel{!}{=} m}}^\infty 1 \, .$$
Therefore $$f(m)=\sum_{k=1}^\infty \binom{1/2}{k} \sum_{\substack{n_1,...,n_k=2 \\ n_1 \cdots n_k \stackrel{!}{=} m}}^\infty 1 = \sum_{k=1}^\infty \binom{1/2}{k} F_k(m) $$
where $$F_k(m)=\#\left\{n_1 \cdots n_k = m \big| n_1,...,n_k \in \mathbb{N}\setminus 1 \right\}$$
counts the number of ways to write $m$ as a product of $k$ integers greater than $1$, including multiplicity.
For example, obviously $F_1(12)=1$, $12=2\cdot 6=6\cdot 2=3\cdot 4=4\cdot 3$ so $F_2(12)=4$. Also $12=2\cdot 2\cdot 3=2\cdot 3\cdot 2=3\cdot 2 \cdot 2$, so $F_3(12)=3$. Clearly, $12$ can not be represented as a product of 4 or more numbers and we have $$f(12)=\binom{1/2}{1} + \binom{1/2}{2}\cdot 4 + \binom{1/2}{3}\cdot 3 = \frac{3}{16} \, .$$
Generally $F_1(m)=1$. If $m=p^N$ is a prime power, then
$$F_k(p^N)=\# \left\{n_1 + \dots + n_k = N \big| n_1,...,n_k \in \mathbb{N}\right\} = \sum_{l=k-1}^{N-1} F_{k-1}(p^l)$$
with $F_k(p^k)=1$. This recurrence can be turned into closed form $$F_k(p^N)=\binom{N-1}{k-1}$$
and gives the identity $$f(p^k)=(-1)^k \binom{-1/2}{k}=\sum_{l=1}^k \binom{1/2}{l} \binom{k-1}{l-1} \, .$$
