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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.

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.

After doing some computations of the divisibility of $\sigma(n)$ with $n+ \varphi(n)$, 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.

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.

After doing some computations of the divisibility of $\sigma(n)$ with $n+ \varphi(n)$, 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.

After doing some computations of the divisibility of $\sigma(n)$ with $n+ \varphi(n)$, 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.