Integers n such that sigma(n)=omega(n)n and omega(n) divides n Are there other integers $n$ than even perfect numbers such that $\sigma(n)=\omega(n)n$ and $\omega(n)\vert n$?
Thanks in advance.
 A: 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.
A: 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
A: 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 $w (n) > 16$. 
