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: graph of τ(x).

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 $\displaystyle f(x)=\log_{\sqrt{2}}x+1$. graph of τ(x) and = log(✓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.



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