The function $f(n) ={\ln(\tau(n)) \above 1.5pt \ln(2)}$ and the number $2^{\tau(n)}+1$

Question

Suppose that $n\in\mathbb{N}$. Let $\tau(n)$ counts the number of divisors of $n$. Let $f(n):\mathbb{N}\rightarrow\mathbb{R^{+}}$ be defined by the map $$f(n) ={\ln(\tau(n)) \above 1.5pt \ln(2)}$$ I would like to prove the following claim.

If $f(n)\in\mathbb{N}$ then $2^{\tau(n)}+1$ is prime.

In particular we have the following:

• If $f(n)=0$ then $2^{\tau(n)}+1=3$; so $\tau(n)=1$
• If $f(n)=1$ then $2^{\tau(n)}+1=5$; so $\tau(n)=2$
• If $f(n)=2$ then $2^{\tau(n)}+1=17$; so $\tau(n)=4$
• If $f(n)=3$ then $2^{\tau(n)}+1=257$; so $\tau(n)=8$
• If $f(n)=4$ then $2^{\tau(n)}+1=65537$; so $\tau(n)=16$

Analytically is there any reason to be believe the integral values of $f(n)$ are bounded and achieves a maximum at $4$?

Note I can easily show the opposite of the claim, namely : If $2^{\tau(n)}+1$ is prime then $f(n)\in \mathbb{N}$

Answer 1

This is false. For any $n$ wth $\tau(n)=2^5$ you have $f(n)=5 \in \mathbb{N}$ but $2^{2^{5}}+1=641 \cdot 6700417$ is not prime.

In fact, notice that $f(n) \in \mathbb{N}$ if any only if $\tau(n)$ is a power of $2$. Hence your claim is identical to the conjecture that every Fermat number $F_k=2^{2^{k}}+1$ is prime. This is known to be false, failing at $k=5,6,7,...,31$ (and probably many more times).

Answer 2

This is false. Consider $n=2^{31}$, $\tau(n)=32$. Then

$$2^{32}+1 = 641\cdot 6700417$$

More generally, we have that

$$\tau\left(2^{2^n-1}\right) = 2^n$$

for all positive integer $n$. So, your conjecture basically says that

$$2^{2^n}+1$$

is prime for all positive integer $n$. After seeing the first few values, this seems like a reasonable guess to make. Fermat said the same thing (these are Fermat numbers). However, all known Fermat numbers with $n>4$ are composite, and it is conjectured that all of them are with the exception of the first few.