# Existence of infinitely many n such that sigma(varphi(n)) divides n?

Problem: Do there exist infinitely many positive integers $n$ such that $$\sigma ( \varphi (n)) | n$$, where $\varphi$ is the Euler function and $\sigma$ is the sum of divisors.

Source: Own.

Attempt: Call $n$ "lucky" if $\sigma ( \varphi (n)) | n$. By programming, some small lucky numbers are found: $1,2,3,6,15,28,30,255,510,744,2418$.

Now let's consider Fermat number $${F_m} = {2^{{2^m}}} + 1.$$ It's well-known that $F_m$ is prime when $m \leqslant 4$. Let $$n = {2^{{2^k}}} - 1 = \prod\limits_{m = 0}^{k - 1} {{F_m}}$$ and it's easy to check $n$ is lucky when $k \leqslant 5$. Yet I haven't found any other lucky numbers with certain modality.

Something else to mention (seem useless): $2n$ is lucky as well if $n$ is a lucky odd number. I have also thought of induction, but the prime divisor of $\varphi( n )$ is not easy to determine.

• $25$ is not a lucky number (in your list), since $42\nmid 25$. Aug 25, 2018 at 11:02
• @DietrichBurde It is a typo, correct is "$255$" Aug 25, 2018 at 11:18
A partial answer can be found for the positive integers $n$ with $\sigma(\phi(n))=n$, see the sequence A018784 at OEIS. If $p$ is a known Fermat prime, then $p-2$ is in the sequence. It seems to be open, whether or not there are infinitely many such $n$, although likely. The case $\sigma(\phi(n))\mid n$ can be reduced to the equality, if we know that proper divisibility arises only for $2n$ if $n$ is in the list for equality.