Integers n such that sigma(n)=omega(n)n and omega(n) divides n

Question

Are there other integers $n$ than even perfect numbers such that $\sigma(n)=\omega(n)n$ and $\omega(n)\vert n$?
Thanks in advance.

Answer 1

For $n=120$ we have $\omega(120)=3$ and $\sigma(120)=360=3\cdot 120=\omega(120)\cdot 120$ with $\omega(120)\mid 120$. This is not an even perfect number.

Answer 2

Humbled by Dietrich Burde's example, here is my motivation for saying that there won't be many such.

Consider $\sigma(n)/n$. This is bounded above by a number I will call $P(n)$ and define as $P(n) = \prod_{0 \lt i \leq \omega(n)} \frac{p_i}{p_i - 1}$, which involves the smallest primes $p_i$. Note that when $\omega(n) \gt 4, P(n) \lt \omega(n).$ So any hope of the first equation having a solution implies that $n$ has at most 4 distinct prime factors. The case when $\omega(n)=2$ nicely captures the even perfect numbers, so let us move on to $\omega(n)=4$. Using the finer inequality $\sigma(n)/n \lt \prod_{p |n, p \text{ prime}}\frac{p}{p-1}$ gives $n = 2^a3^b5^cp^d$ for some prime p and positive integers $a,b,c$ and $d$, and further $15p/4(p-1) \gt 4$, so $p \lt 16$.

Now that we have a limit on p, we can use Zsigmondy's theorem and multiplicativity of $\sigma(p^k) = \frac{p^{k+1} - 1}{p - 1}$ to limit the exponents $a,b,c$ and $d$ to at most $6$. So there will be at most finitely many cases to check. More on the finitely many cases later.

For $\omega(n)=3$, a similar analysis implies $n=2^a3^bp^c$, although there seem to be more primes $p$ to check. However, appealing to Zsigmondy again gives bounds on the exponents, and again there will be finitely many cases.

Having exhausted myself before the exhaustive search, I will report back later with additional findings.

EDIT 2013.05.16 So I was right, but in a somewhat surprising fashion. There are finitely many cases to check, it can be done by hand, Zsigmondy's theorem can help, and there are examples, two for $\omega(n)=3,$ and one for $\omega(n)=4$.

In the 4 case, as noted above $p$ is one of $7, 11,$ or $13$, and then one computes $\sigma(p^d)$ and notes those whose prime factors fall in the set of primes at most 5. Zsigmondy's theorem says we can restrict our attention to $d \lt 5$. This leaves $p^d$ being one of $7, 11,$ or $343$. Then one computes $\sigma(q^k)$ for $q=2,3,5$ to ensure its prime power factors do not lie outside of $2^x,3^y,5^z,$ or one of the three choices for $p^d$. Zsigmondy tells us we can stop bumping up $k$ once we've seen all the small primes as factors for each $q$, which is at most 9 when $q=2$, and smaller for the other choices of $q$.

After doing this, we rule out 11 as a candidate for $p^d$, and find that $3^b=3^3$ and $5^c=5^1$, leaving $2^a$ to determine. Since $p=7$, this leads quickly to $a=5$ and $d=1$, so $30240$ is the unique example with 4 distinct prime factors.

In the 3 case, Zsigmondy tells us that if $\sigma(2^a3^bp^c) =2^a3^bp^c3 $ for some prime p, then $c$ is at most 2. This is because $\sigma(p^c)$ will be divisible by some prime larger than $3$ when $p$ is a prime larger than 3 and $c$ is larger than 2. But $\sigma(p^2)= p^2+p+1$ is odd and congruent to 3 mod 9, and cannot be of the form $2^x3^y$ for large enough $x$ and $y$. So as a result, $c=1$, and we now must have $\sigma(2^a)$ being a power of 3 or $\sigma(3^b)$ being a power of 2, each of which relates to a simple case of the Catalan conjecture (or use Zsigmondy yet again to bound either $a$ or $b$). It develops that $b=1$ and then $p \lt 9$ and one quickly finds the examples 120 and 672. END EDIT 2013.05.16

Gerhard "Off To Find More Coffee" Paseman, 2013.05.14

Answer 3

I will prove that if $w (n) > 16$ then there are no integers with the property that $\sigma (n) = w (n) n$ and $w (n) |n$. I assumed the Riemann Hypothesis to simplify some explicit bounds, but with additional pains this could be obviously removed. Combined with some result of Pomerance I think this could show that odd perfect numbers cannot arise in this form (I think it's Pomerance who showed that odd perfect numbers need to have many prime factors...)

Write $n = w (n) m$. Get $\sigma (w (n) m) = w (n)^2 m$. We note that for any integers $m, n$ we have $\sigma (mn) \leqslant \sigma (m) \sigma (n)$. Therefore we get $$ w (n)^2 m \leqslant \sigma (w (n)) \sigma (m) $$ This leads to the inequality $$ w (n)^2 / \sigma (w (n)) \leqslant \sigma (m) / m $$ Now $\sigma (p^{\alpha}) / p^{\alpha} = (1 + p + \ldots + p^{\alpha}) / p^{\alpha} = 1 + 1 / p + 1 / p^2 + \ldots$ . Therefore we notice that $$ \frac{\sigma (m)}{m} \leqslant \prod_{p \leqslant w (m)} \left( 1 - \frac{1}{p} \right)^{- 1} \leqslant \prod_{p \leqslant w (n)} \left( 1 - \frac{1}{p} \right)^{- 1} \leqslant 2.2 \log w (n) $$ for $w (n) > 14$. Therefore we get the inequality $w (n)^2 \leqslant 2.2 \sigma (w (n)) \log w (n)$. We have $\sigma (w (n)) \leqslant 2.6 w (n) \log\log w (n)$ for $w (n) > 7$. Therefore we end up with $w (n) \leqslant 5.72 \log w (n) \log\log w (n)$. This inequality fails as soon as $w (n) > 16$.