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Are there any natural numbers $n$ that satisfy the condition $7921\sigma(n) = 15840n$, where $\sigma(n)$ denotes the sum of divisors of $n$?

This question arises from the theory of immaculate groups (or, equivalently, Leinster groups). An immaculate group is a group, such that its order is equal to the sum of all orders of its proper normal subgroups.

It is easy to see, that if $A$ is a non-abelian simple group then $A\times\mathbb{Z}_n$ is immaculate iff $(|A|+1)\sigma(n) = 2|A|n$. Two well known examples of immaculate groups of that form are $A_5\times\mathbb{Z}_{15128}$ and $A_6\times\mathbb{Z}_{366776}$. In terms of immaculate groups this question thus can be reworded as: "Does there exist such $n$, that $M_{11}\times\mathbb{Z}_n$ is immaculate?", where $M_{11}$ stands for Mathieu simple group of order $7920$.

Currently I know only two facts about such $n$-s: if they exist, then $7921|n$, and that such $n$-s, if they exist, are too large to be found by exhaustive search.

Any help will be appreciated.

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    $\begingroup$ This seems like such a random question to ask until you see the entirely reasonable motivation. $\endgroup$ – Arthur Jul 20 '18 at 9:13
  • $\begingroup$ @Arthur Tbh I saw a motivation at the start, this being essentially a question of how close $\sigma(n)$ can be to $2n$. $\endgroup$ – fretty Jul 20 '18 at 9:54
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    $\begingroup$ An explantion why it is hopeless to search such a number with brute force, would be nice. $\endgroup$ – Peter Jul 20 '18 at 11:25
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    $\begingroup$ I figured out a positive integer $n$ with $$7921\sigma(n)=2\cdot 15840\cdot n$$ $$n=563432025678613816032$$ Not sure whether this helps $\endgroup$ – Peter Jul 20 '18 at 12:17
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    $\begingroup$ I have tried hard to find a solution, but I failed. My conjecture is that there is no solution, but a proof will be very difficult considering that it is , for example , an open question whether there is an odd perfect number. $\endgroup$ – Peter Jul 21 '18 at 16:46
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$$\mathbf{\color{green}{Fixed\ version}}$$

$$\mathbf{\color{brown}{Constraints\ of\ the\ task}}$$

The issue equation is $$89^2\sigma(n)=2^53^25^111^1n.\tag1$$ Let WLOG $$n=2^A3^B5^C11^D89^{U+2}m,\quad (A,B,C,D,U)\in\mathbb N\cup {0} ,\quad \gcd(m, 2\cdot3\cdot5\cdot11\cdot89)=1,\tag2$$ then $(1)$ takes the form of $$89^2\cdot\dfrac{2^{A+1}-1}1 \cdot\dfrac{3^{B+1}-1}2 \cdot\dfrac{5^{C+1}-1}4\cdot\dfrac{11^{D+1}-1}{10}\cdot\dfrac{89^{U+3}-1}{88}\cdot\sigma(m) =$$ $$2^{A+5}3^{B+2}5^{C+1}11^{D+1}89^{U+2}m,\tag3$$ or $$\dfrac{2^{A+1}-1}{2^{A+1}}\cdot\dfrac{3^{B+1}-1}{3^{B+1}}\cdot\dfrac{5^{C+1}-1}{5^{C+1}}\cdot\dfrac{11^{D+1}-1}{11^{D+1}}\cdot\dfrac{89^{U+3}-1}{88\cdot89^U}\cdot\sigma(m) = 2^{8}\cdot3\cdot5\cdot{m},\tag4$$ $$\left(1-\dfrac{2^{-A}}2\right)\cdot\left(1-\dfrac{3^{-B}}3\right)\cdot\left(1-\dfrac{5^{-C}}5\right)\cdot\left(1-\dfrac{11^{-D}}{11}\right)\cdot\dfrac{89^3-89^{-U}}{88}\cdot\sigma(m) = 3840m.\tag5$$ Easy to see that $$\dfrac{89^3-89^{-U}}{88} \ge \dfrac{89^3-1}{88} = 8011.$$ At the same time, $\sigma(m)$ is the sum of divisors, so $$\sigma(m)\ge 1+m.$$ Thus, $$8011\cdot\left(1-\dfrac{2^{-A}}2\right)\cdot\left(1-\dfrac{3^{-B}}3\right)\cdot\left(1-\dfrac{5^{-C}}5\right)\cdot\left(1-\dfrac{11^{-D}}{11}\right) \le 3840.\tag6$$ Similarly, \begin{cases} 8011\cdot\left(1-\dfrac{2^{-A}}2\right)\cdot\dfrac23\cdot\dfrac45\cdot\dfrac{10}{11} \le 3840\\ 8011\cdot\left(1-\dfrac{2^{-A}}2\right)\cdot\left(1-\dfrac{3^{-B}}3\right)\cdot\dfrac45\cdot\dfrac{10}{11} \le 3840\\ 8011\cdot\left(1-\dfrac{2^{-A}}2\right)\cdot\left(1-\dfrac{5^{-C}}5\right)\cdot\dfrac23\cdot\dfrac{10}{11} \le3840\\ 8011\cdot\left(1-\dfrac{2^{-A}}2\right)\cdot\left(1-\dfrac{11^{-D}}{11}\right)\cdot\dfrac23\cdot\dfrac45 \le 3840\\ 8011\cdot\left(1-\dfrac{2^{-1}}2\right)\cdot\left(1-\dfrac{5^{-C}}5\right)\cdot\left(1-\dfrac{11^{-D}}{11}\right)\cdot\dfrac23 \le 3840,\tag7\\ \end{cases} or \begin{cases} 8011\cdot\left(1-\dfrac{2^{-A}}2\right) \le 7920\\ 8011\cdot\left(1-\dfrac{2^{-A}}2\right)\cdot\left(1-\dfrac{3^{-B}}3\right) \le 5280\\ 8011\cdot\left(1-\dfrac{2^{-A}}2\right)\cdot\left(1-\dfrac{5^{-C}}5\right) \le 6336\\ 8011\cdot\left(1-\dfrac{2^{-A}}2\right)\cdot\left(1-\dfrac{11^{-D}}{11}\right)\le 7200\\ 8011\cdot\left(1-\dfrac{5^{-C}}5\right)\cdot\left(1-\dfrac{11^{-D}}{11}\right) \le 7680,\quad A=1, \ B=0.\tag8\\ \end{cases} The system $(8)$ has the solution $$\left[ \begin{aligned} &A=0\\ &A=1,\ B=0,\ C=D=1\\ &A=1,\ B=0,\ C>2\\ &A=2,\ B=C=0\\ &5\ge A\ge3,\ B=C=D=0, \end{aligned}\right.\tag9\\$$ or $$\left[ \begin{align} &2 \not|\ n\hspace{40pt}\\ &((n=2m)\vee(n=8m)\vee(n=16m)\vee(n=32m)\vee(n=110m))\\ &\quad\wedge(\gcd(2\cdot3\cdot5\cdot11,m)=1)\\ &(n=50m)\wedge(\gcd(2\cdot3, m)=1)\\ &(n=4m)\wedge(\gcd(2\cdot3\cdot5,m)=1)\\ \end{align}\right..\tag{10} $$

$$\mathbf{\color{brown}{Applying\ of\ the\ constraints}}$$ Obtained constraints $(10)$ allow to filter the possible solutions.

At first, becames impossible the case $$n=2^{10}89^3k.$$ Secondly, if $$n=89^2\cdot8011\cdot2003\cdot167k,\quad \gcd(89\cdot167\cdot2003\cdot8011,k) = 1,$$ with the equation $$28\sigma(k)=55k,$$ then \begin{cases} 49\sigma(m)=55m,\text{ if }k=4m,\ \gcd(2\cdot3\cdot5,m)=1 \\ 21\sigma(m)=22m,\text{ if }k=8m,\ \gcd(2\cdot3\cdot5\cdot11,m)=1 \\ 31\cdot7\sigma(m)=4\cdot55m\text{ if }k=16m,\ \gcd(2\cdot3\cdot5\cdot11,m)=1 \\ 21^2\sigma(m)=8\cdot55m\text{ if }k=32m,\ \gcd(2\cdot3\cdot5\cdot11,m)=1 .\tag{11} \end{cases} Taking in account that $$\sigma(7)=2^3,\quad\sigma(49)= 2^2\cdot5^2,$$ the system $(11)$ really can not be satisfied.

So I think that the issue equation $\mathbf{\color{brown}{has\ not\ solutions\ in\ the\ positive\ integer\ numbers.}}$

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    $\begingroup$ $m$ not divisible by 2, 3, 5, 11, and 89? $\endgroup$ – qwr Jul 29 '18 at 4:50
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    $\begingroup$ How did you get from $89^2\cdot\ldots\cdot\frac{89^{U+1}-1}{88}$ to $\frac{89^{U+3}-1}{88}$? $\endgroup$ – jpvee Jul 29 '18 at 6:50
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    $\begingroup$ @YuriNegometyanov: I believe that equation (3) is still wrong. I think the right-hand side should read $2^{11}\cdot3\cdot5\cdot11\cdot m$. $\endgroup$ – jpvee Jul 30 '18 at 11:35
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    $\begingroup$ @YuriNegometyanov: I like the style of your comments - very poetic! But I don't understand what you are trying to say :) $\endgroup$ – jpvee Jul 30 '18 at 12:19
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    $\begingroup$ @YuriNegometyanov: Thanks for the clarification. Then please tell me what you did to the $88$ in the denominator in the equation before (3) - I see that the $8$ has become part of the $2^{11}$ on the right-hand side of (3), but I do not see what became of the remaining $11$ of the $88=8\cdot 11$. Also the same question for the $10$ in the denominator. And finally, the $m$ vanished from the upper equation completely when going to (3). $\endgroup$ – jpvee Jul 30 '18 at 13:14
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This is not an answer, but an extended comment, that should help anyone interested in using brute-force numerical search to find the solution.


The problem is to find $n \in \mathbb{N}$ for which $$7921 \sigma(n) = 15840 n \tag{1}\label{NA1}$$ where $\sigma(n)$ is the sum of all divisors of $n$, as defined in the Wikipedia divisor function article, and as a sequence in OEIS A000203.

Because $\sigma(n) = n + 1$ if $n$ is a prime, and $$7921 (n + 1) \ne 15840 n, \quad n \in \mathbb{N}$$ we already know there is no prime solution to $\eqref{NA1}$.


Consider the prime factorization of $n$. Let $p_i \in \mathbb{N}$ be nonrepeating primes ($p_i = p_j$ if and only if $i = j$), and $1 \le k_i \in \mathbb{N}$. Then, $$n = \prod_{i=0}^{N-1} p_i^{k_i}$$ and $$\sigma(n) = \prod_{i=0}^{N-1} \frac{p_i^{k_i+1} - 1}{p_i - 1}$$ because $\sigma(p^k) = \sum_{j=0}^{k} p^j = (p^{k+1}-1)/(p-1)$ when $p$ is a prime.

We can now rewrite the problem $\eqref{NA1}$ as $$7921 \prod_{i=0}^{N-1} \frac{p_i^{k_i+1} - 1}{p_i - 1} = 15840 \prod_{i=0}^{N-1} p_i^{k_i} \tag{2}\label{NA2}$$ Rearranging the terms yields $$\prod_{i=0}^{N-1} \frac{ p_i^{k_i + 1} - p_i^{k_i} }{ p_i^{k_i + 1} - 1 } = \frac{7921}{15840} = \frac{n}{\sigma(n)} = \frac{89^2}{2^5 \cdot 3^2 \cdot 5 \cdot 11} \tag{3}\label{NA3}$$ Note the term $$f_i = \frac{ p_i^{k_i + 1} - p_i^{k_i} }{ p_i^{k_i + 1} - 1 } = \frac{p_i^{k_i}}{\sum_{j=0}^{k_i} p^j}, \quad \frac{1}{2} \lt f_i \lt 1 \tag{4}\label{NA4}$$ i.e., $$\begin{array}{ll} f_i = \frac{p_i}{p_i + 1}, & k_i = 1 \\ f_i = \frac{p_i^2}{p_i^2 + p_i + 1}, & k_i = 2 \\ f_i = \frac{p_i^3}{p_i^3 + p_i^2 + p_i + 1 }, & k_i = 3 \\ f_i = \frac{p_i^{k_i}}{p_i^{k_i} + p_i^{k_i-1} + \dots + p_i + 1 } & \\ \end{array}$$

Thus, the numerical search problem is now reduced to find the set of terms $f_i$ based on primes $p_i$ and their positive powers $k_i$, so that the product $$\prod_{i=0}^{N-1} f_i = \frac{7921}{15840}$$ In particular, because $f_i \lt 1$, a particular set can be rejected immediately if the product falls below the target ratio.


For example, if $p_0 = 89$, $k_0 = 2$, to eliminate the prime factor in the numerator. Repeating, that leads to $p_1 = 8011$, $k_1 = 1$; $p_2 = 2003$, $k_2 = 1$; and $p_3 = 167$, $k_3 = 1$, to get us to a result with a composite numerator and a denominator: $$\begin{array}{r|l|l} n & \frac{n}{\sigma(n)} & \frac{15840 n}{7921 \sigma(n)} \\ \hline 89^2 \cdot 8011 \cdot 2003 \cdot 167 & \frac{7921}{8064} = \frac{7921}{2^7 \cdot 3^2 \cdot 7} & \frac{55}{28} = \frac{5 \cdot 11}{2^2 \cdot 7} \\ 89^2 \cdot 8011 \cdot 2003 \cdot 167 \cdot 7 & \frac{7921}{9216} = \frac{7921}{2^{10} \cdot 3^2} & \frac{55}{32} = \frac{5 \cdot 11}{2^5} \\ 89^2 \cdot 8011 \cdot 2003 \cdot 167 \cdot 2 & \frac{7921}{12096} = \frac{7921}{2^6 \cdot 3^3 \cdot 7} & \frac{55}{42} = \frac{5 \cdot 11}{2 \cdot 3 \cdot 7} \\ 89^2 \cdot 8011 \cdot 2003 \cdot 167 \cdot 2 \cdot 7 & \frac{7921}{13824} = \frac{7921}{2^9 \cdot 3^3} & \frac{55}{48} = \frac{5 \cdot 11}{2^4 \cdot 3} \\ 89^2 \cdot 8011 \cdot 2003 \cdot 167 \cdot 2^2 & \frac{7921}{14112} = \frac{7921}{2^5 \cdot 3^2 \cdot 7^2} & \frac{55}{49} = \frac{5 \cdot 11}{7^2} \\ \end{array}$$ If you append $p_5 = 7$, $k_5 = 1$ or $k_5 = 2$ to $n$ in the final row above, the rightmost field drops below 1 (to $55/56$ for $k_5 = 1$, and to $55/57$ for $k_5 = 2$), leading nowhere. Similarly, appending $p_6 = 3$, $k_6 = 1$ or $k_6 = 2$ to $n$ in the second-to-last row (to $55/64$ for $k_6 = 1$, and to $165/208$ for $k_6 = 2$) leads nowhere.

It looks to (very non-mathematician) me that an exhaustive search over primes $p$ is possible, due to terms $f_i$ having a power of a prime in the numerator, as specified in $\eqref{NA4}$. Whether an exhaustive search is possible or not is an open question (and is important for those looking for proof), but the efficient numerical brute force search strategies are straightforward; especially if one is looking for some other ratios than $\frac{n}{\sigma(n)} = \frac{7921}{15840}$.

To continue the search above, I'd need a prime $p$ and a positive integer $k$ such that $\sum_{j=0}^{k} p^j = 55$ (to yield a factor with denominator $55$). No such pair exists, so the search strategies I've come up with thus far are exhausted.


Hopefully, one of the math sages here can take this further from here.

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    $\begingroup$ This is the way to the jungle. For example, the equality $2^{11}-1=23\cdot89$ requires consideration of the variant $k_0=3$ etc. $\endgroup$ – Yuri Negometyanov Jul 29 '18 at 3:52
  • $\begingroup$ @YuriNegometyanov: At first glance, using $2^{10}$ does indeed look promising. Unfortunately, however, $\frac{89^3}{\sigma(89^3)}\cdot\frac{2^{10}}{\sigma(2^{10})}$ is already below the target ratio of $\frac{7921}{15840}$. In fact, if you use $k_0=3$, then the largest power of $2$ that may appear in $n$ is $2^5$. $\endgroup$ – jpvee Jul 30 '18 at 8:36
  • $\begingroup$ I do believe my efforts stalled, because rewriting $\eqref{NA2}$ as $\eqref{NA3}$ places an additional constraint on the coefficients. I really should rewrite $\eqref{NA2}$ to use another index variable, say $j$, on the right side. Similarly, the indexing in the numerator and denominator in $\eqref{NA3}$ do not need to be the same. @YuriNegometyanov: Is this what you meant by the way to the jungle (off the track)? $\endgroup$ – Nominal Animal Jul 30 '18 at 9:45
  • $\begingroup$ @NominalAnimal I mean the jungles of variants in the situation where solution doesn't exist. $\endgroup$ – Yuri Negometyanov Jul 30 '18 at 10:55
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    $\begingroup$ @YuriNegometyanov: I don't think your proof works - see my comment regarding the right-hand side in equation (3). $\endgroup$ – jpvee Jul 30 '18 at 11:36
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This is not an answer either: I have written a program doing a non-exhaustive search, and so far, no solutions have come up for the Mathieu group $M_{11}$; however, the program did find a solution e.g. for the larger Mathieu group $M_{22}$ of order $443520=2^7\cdot3^2\cdot5\cdot7\cdot11$:

Let $n=55009909630=2\cdot5\cdot13\cdot79\cdot109\cdot157\cdot313$, then since $$ \begin{align} (|M_{22}|+1)\cdot\sigma(n)&=443521\cdot\sigma(2\cdot5\cdot13\cdot79\cdot109\cdot157\cdot313) \\ &=13\cdot109\cdot313\cdot3\cdot6\cdot14\cdot80\cdot110\cdot158\cdot314 \\ &=2^9\cdot3^2\cdot5^2\cdot7\cdot11\cdot13\cdot79\cdot109\cdot157\cdot313 \\ &=2\cdot|M_{22}|\cdot n\text{,}\end{align}$$ $M_{22}\times\mathbb{Z}_{55009909630}$ must be immaculate.

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Perhaps the solution to this problem might be realized via an application of the latest results of Holdener, et. al. on abundancy outlaws?

Holdener, Weiner (2014) - Searching for and Characterizing Abundancy Outlaws

Holdener, Moore (2011) - A Geometric Representation of the Abundancy Index

Holdener, Czarnecki (2008) - The Abundancy Index: Tracking Down Outlaws

Holdener, Stanton (2007) - Abundancy "Outlaws" of the Form $\frac{\sigma(N) + t}{N}$

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