# 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 $\omega(n)\vert n$?

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No. Consider how the product p/(p-1) grows where the product is taken over the first omega many primes. This puts a sharp upper bound on omega and should imply n is even for the first equation to hold. If omega is greater than 2, an exhaustive search should finish it off. Gerhard "Ask Me About Pi Inverse" Paseman, 2013.05.14 –  Gerhard Paseman May 14 '13 at 17:48
Even if you relax the first equation to n being multiply perfect, I suspect there will still be sharp limits for divisibility, e.g. omega might need to be even and possibly abundant itself in order for it to be a factor of n and significantly greater than 2. Gerhard "Feels That's How It Is" Paseman, 2013.05.14 –  Gerhard Paseman May 14 '13 at 17:56

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.

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Darn. Looks like I spoke too soon. I will see if I can prove that there are no other examples. Gerhard "Drats! Foiled Again By Counterexample!" Paseman, 2013.05.14 –  Gerhard Paseman May 14 '13 at 18:03
@Gerhard: Well, this can happen easily. By the way, integers $n\ge 1$ with $\sigma(n)=mn$ for some $m\ge 2$ are called m-multiple perfect numbers, and there are several of them (see matwbn.icm.edu.pl/ksiazki/mon/mon42/mon4204.pdf). –  Dietrich Burde May 14 '13 at 18:51
And it looks like 30240 is another example. I suspect there are not many more. Gerhard "Ask Me About Premature Speaking" Paseman, 2013.05.14 –  Gerhard Paseman May 14 '13 at 20:33

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

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I just noticed in the above that I did not use the second of the above conditions, only the first. If we use a weaker condition, such as which multiperfect (or other rare class of) numbers n are divisible by the number of there own prime factors, I still think there will be finitely many such, but more machinery will be needed to prove it. This exercise makes me wonder if Zsigmondy's Theorem can be used for studying odd perfects. Gerhard "Odd Perfects? How Perfectly Odd." Paseman, 2013.05.16 –  Gerhard Paseman May 16 '13 at 10:06

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 $w (n) > 16$.

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Interesting. I use arithmetic and not much more than that to get $\omega(n) \lt 5$. You might try a less analytic approach. Gerhard "Doing Much More With Less" Paseman, 2013.05.14 –  Gerhard Paseman May 14 '13 at 21:27
Thanks for the comment. I wrote down the first thing that went through my mind... And I haven't noticed your post below! –  kiskis May 14 '13 at 22:19