Find $f(x)$ such that $\tau(x) \leq f(x)$ for $\forall x \in \mathbb{N}$, $\not\exists g(x)$ such that $\tau(x) \leq g(x)\leq f(x).$($f(x), g(x)$ is not related with $\tau(x)$.(For instance, $f(x)$ can't be $\tau(x)$.))
$\tau(x)$ is the number of factors of $x$.
First, I tried to arrange the value of $\tau(x)$.
\begin{align}
&\tau(1)=1. \\
&\tau(2)=2. \\
&\tau(3)=2. \\
&\tau(4)=3. \\
&\tau(5)=2. \\
&\tau(6)=4. \\
&\tau(7)=2. \\
&\tau(8)=4. \\
&\tau(9)=3. \\
&\tau(10)=4. \\
&\vdots
\end{align}
The Graph of $\tau(x)$ looks like:
Note that the graph of $\tau(x)$ is only the points. I drew the lines which are connecting the points so that we can easily assume the graph of $f(x)$.
First, I tried $\displaystyle f(x)=\log_{\sqrt{2}}x+1$.
Just for a try, I just substituted all of the $\log$ functions, since I could assume the upper bound of $\tau(x)$.
Please leave the answer if you have found any functions which satisfy the condition.
