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After doing some computations of the divisibility of $\sigma(n)$ by $n+ \varphi(n)$, mostly with Peter´s help, we found these solutions:

$n=2, 456, 828, 7584 ,33462 , 1357440, 1596048 ,1964544 ,19800384 ,26211264 ,31451136 ,106805184,156868224 ,316113024 ,365395680 ,449746560 ,502349274 ,503291904 $

This is mostly computational and recreational research, to see some connections between $n$ and $\sigma(n)$ and $\varphi(n)$ and some of their interdependencies in this form of divisibility condition.

Of course, other divisibility conditions between exactly the same variables could be researched, and many of them are surely of no lesser value, but this one is particularly simple.

But there is one interesting observation (at least on the surface): if we do not observe $2$, then all of these numbers are divisible by $6$.

I would like to know does somebody has some ideas of how to prove this, if true?

  • That is, if we have $n \neq 2$ and $(n+\varphi(n)) | \sigma(n)$ is it then necessarily true $6 |n$?

The sequence is not in OEIS.

$\sigma$ is sum-of-divisors and $\varphi$ is totient*.

Update: As observed by Peter and Robert Israel, if $p=5 \cdot 2^{d-1}-1$ is prime then for $n=2^d \cdot 3p$ we have $\dfrac{\sigma(n)}{n+\varphi(n)}=2$, and that would give an infinite number of solutions, if there is an infinite number of primes of that form.

Update 2: Some other solutions found by Peter: $$1557940992, 2026608480, 7511094360, 8024671392, 8052965376$$

These are also divisible by $6$.

And a related paper mentioned by TheSimpliFire in his chatroom.

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  • 5
    $\begingroup$ I don't have a proof, but I can see at least one pattern. If $p = 5 \cdot 2^{d-1} - 1$ is prime, then $n = 2^d\cdot 3p$ works with $\sigma(n) = 2 (n + \varphi(n))$. The primes of this form are sequence A050522. $\endgroup$
    – Robert Israel
    Commented Apr 6, 2020 at 13:49
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    $\begingroup$ I think that these kind of questions are interesting. About your thought in your comment I think that maybe it is interesting to explore systematically what useful and simple congruences one can to prove (I evoke to create a table of useful congruences for number theoretic functions if isn't in the literature). Today I proved as an easy variation of the harmonic divisor numbers that if $n$ is an odd perfect number with $\omega(n):=\kappa$ distinct prime factors in its factorization, then for each $2\leq k\leq \kappa$ integer we've that $2(\sigma(n))^k$ divides $\varphi(n^{k+1})$. $\endgroup$
    – user142929
    Commented Apr 6, 2020 at 13:49
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    $\begingroup$ @RobertIsrael Peter also noticed that. Sometimes we have also that the fraction is equal to 3, but so far none equal to 4. $\endgroup$
    – user153451
    Commented Apr 6, 2020 at 13:57
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    $\begingroup$ oeis.org/A099650 tabulates solutions to $x+\phi(x) = \sigma(x)/2$. The connection with primes of the form $5\times2^n-1$ was noticed by Firoozbakht in 2005. $\endgroup$
    – Gerry Myerson
    Commented Apr 7, 2020 at 0:00
  • 4
    $\begingroup$ $$7511094360$$ is an example with fraction $4$ $\endgroup$
    – Peter
    Commented Apr 7, 2020 at 12:47

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