Are there any pairs of natural numbers $(m, n)$, with $5 \leq m$, other than $(5, 15128)$ and $(6, 366776)$, that satisfy the condition $(m! + 2)\sigma(n) = 2n \cdot m!$, where $\sigma(n)$ denotes the sum of divisors of $n$ and $m!$ denotes the factorial of $m$?

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 any immaculate group of the type $A_n \times \mathbb{Z}_n$ other than those two?".

I checked this condition for all $n \leq 10000$ and $5 \leq m \leq 7$ but found nothing.

Any help will be appreciated.

A similar question about $M_{11}$: Are there any natural numbers $n$ that satisfy the condition $7921\sigma(n) = 15840n$?

up vote 5 down vote accepted

The equation can be written as $$ \Bigl(\frac{m!}{2}+1\Bigr)\sigma(n)=n\,m!. $$ Since $m!$ and $m!/2+1$ are coprime, $n$ is a multiple of $m!/2+1$. I have looked for solutions of the equation $$ \sigma\Bigl(k\,\Bigl(\frac{m!}{2}+1\Bigr)\Bigr)=k\,m! $$ for $5\le k\le20$, $1\le k\le10^6$. This search provided so far a new solution: $$ m=10,\quad n=691\,816\,586\,092 $$

$$m=7 \; \; , \; \; \; n = 5919262622 = 2 \cdot 7^2 \cdot 13 \cdot 19 \cdot 97 \cdot 2521 $$

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m = 5

n = 15128 =  2^3 31 61  sigma 29760 =  2^6 3 5 31


  multiplier   k   248 =  2^3 31


120 =  2^3 3 5         122 =  2 61

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m = 6

n = 366776 =  2^3 19^2 127  sigma 731520 =  2^7 3^2 5 127


  multiplier   k   1016 =  2^3 127


720 =  2^4 3^2 5         722 =  2 19^2

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m = 7

n = 5919262622 =  2 7^2 13 19 97 2521  sigma 11833829280 =  2^5 3^2 5 7^3 13 19 97


  multiplier   k   2347982 =  2 7^2 13 19 97


5040 =  2^4 3^2 5 7         5042 =  2 2521

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