# Find the functions which is always bigger than $\tau(x)$.

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.