# Any counter example for this: ${\phi(2^n-1)} \bmod \tau(2^n-1)=0$ for every integer $n \geq 1$? [closed]

I asked this question here In S.E but i don't received any resposnes for it, I would like to know if it is appropriate for M.O.

I'm always interesting for properties of the following series : $\sum_{n=1}^{\infty }\frac{\tau(2^n-1)}{\phi(2^n-1)}$ such that: $\tau(N)$ is the number of divisors of $N$ and $\phi$ is Euler totient function, In this question i w'd be interest to know a little bit about the reciprocal of each term of the titled series at $N=2^n-1$ for every $n\geq 1$.

Note:01 According to my calculations that i run in wolfram alpha from $n=1$ to $100$ I have got this result as shown here such that it works which letting me to ask the following question.

My question here is: Is ${\tau(2^n-1)}$ always divides $\phi(2^n-1)$ for every integer $n\geq 1$ and any counterexample for it?

Thank you for any help !!!

## closed as off-topic by Myshkin, Alexey Ustinov, Franz Lemmermeyer, Wolfgang, András BátkaiNov 10 '16 at 7:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Myshkin, Alexey Ustinov, Franz Lemmermeyer, Wolfgang
If this question can be reworded to fit the rules in the help center, please edit the question.

• The smallest counterexample occurs at $n=253$. – Emanuele Tron Nov 9 '16 at 23:33
• If odd $m$ is squarefree, say, $m=p_1p_2\cdots p_r$, then $\tau(m)=2^r$, and $$\phi(m)=2^r{p_1-1\over2}{p_2-1\over2}\cdots{p_r-1\over2}$$ so counterexamples can only occur when $2^n-1$ is not squarefree. – Gerry Myerson Nov 10 '16 at 0:04
• If $k$ is a square-free odd integer define $\omega(k)$ to be the l.c.m. of all the $\omega(p)$'s with $p|k$; thus $\omega(k)$ is the order of $2 \pmod{k^2}$. Note that $k^2$ divides $2^n-1$ if and only if $\omega(k)$ divides $n$. Now note that $\omega(3)=6$, $\omega(5)=20$, $\omega(7)=21$, $\omega(11)=110$, $\omega(13)=156$ and so on. Thus if $n$ is divisible by $6$, or $20$ or $21$, or $110$, or $156$ etc then $2^n-1$ is not squarefree , for example $n=12$ is divisible by $6$ and here :${2}^{12}-1$ is not counter example , how this ? – zeraoulia rafik Nov 10 '16 at 21:48
• Crossosted at MSE. – Dietrich Burde Nov 15 '16 at 22:24