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Solutions of the equation $(m! + 2)sigma(n) = 2n cdot m!$ where $5 leq m$
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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 $AtimesmathbbZ_n$ is immaculate iff $(|A|+1)sigma(n) = 2|A|n$. Two well known examples of immaculate groups of that form are $A_5timesmathbbZ_15128$ and $A_6timesmathbbZ_366776$. In terms of immaculate groups this question thus can be reworded as:
"Does there exist any immaculate group of the type $A_n times mathbbZ_n$ 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 $7921sigma(n) = 15840n$?
group-theory elementary-number-theory finite-groups normal-subgroups divisor-sum
asked 17 hours ago
Yanior Weg
1,0421426
add a comment |Â
up vote
3
down vote
favorite
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 $AtimesmathbbZ_n$ is immaculate iff $(|A|+1)sigma(n) = 2|A|n$. Two well known examples of immaculate groups of that form are $A_5timesmathbbZ_15128$ and $A_6timesmathbbZ_366776$. In terms of immaculate groups this question thus can be reworded as:
"Does there exist any immaculate group of the type $A_n times mathbbZ_n$ 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 $7921sigma(n) = 15840n$?
group-theory elementary-number-theory finite-groups normal-subgroups divisor-sum
asked 17 hours ago
Yanior Weg
1,0421426
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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 $AtimesmathbbZ_n$ is immaculate iff $(|A|+1)sigma(n) = 2|A|n$. Two well known examples of immaculate groups of that form are $A_5timesmathbbZ_15128$ and $A_6timesmathbbZ_366776$. In terms of immaculate groups this question thus can be reworded as:
"Does there exist any immaculate group of the type $A_n times mathbbZ_n$ 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 $7921sigma(n) = 15840n$?
group-theory elementary-number-theory finite-groups normal-subgroups divisor-sum
asked 17 hours ago
Yanior Weg
1,0421426
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 $AtimesmathbbZ_n$ is immaculate iff $(|A|+1)sigma(n) = 2|A|n$. Two well known examples of immaculate groups of that form are $A_5timesmathbbZ_15128$ and $A_6timesmathbbZ_366776$. In terms of immaculate groups this question thus can be reworded as:
"Does there exist any immaculate group of the type $A_n times mathbbZ_n$ 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 $7921sigma(n) = 15840n$?
group-theory elementary-number-theory finite-groups normal-subgroups divisor-sum
asked 17 hours ago
Yanior Weg
1,0421426
asked 17 hours ago
Yanior Weg
1,0421426
asked 17 hours ago
Yanior Weg
1,0421426
asked 17 hours ago
Yanior Weg
1,0421426
1,0421426
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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4
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accepted
The equation can be written as
$$
Bigl(fracm!2+1Bigr)sigma(n)=n,m!.
$$
Since $m!$ and $m!/2+1$ are coprime, $n$ is a multiple of $m!/2+1$. I have looked for solutions of the equation
$$
sigmaBigl(k,Bigl(fracm!2+1Bigr)Bigr)=k,m!
$$
for $5le kle20$, $1le kle10^6$. This search provided so far a new solution:
$$
m=10,quad n=691,816,586,092
$$
answered 16 hours ago
Julián Aguirre
64.3k23894
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0
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$$m=7 ; ; , ; ; ; n = 5919262622 = 2 cdot 7^2 cdot 13 cdot 19 cdot 97 cdot 2521 $$
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
Sun Aug 5 12:42:11 PDT 2018
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
answered 12 hours ago
Will Jagy
96.7k594195
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The equation can be written as
$$
Bigl(fracm!2+1Bigr)sigma(n)=n,m!.
$$
Since $m!$ and $m!/2+1$ are coprime, $n$ is a multiple of $m!/2+1$. I have looked for solutions of the equation
$$
sigmaBigl(k,Bigl(fracm!2+1Bigr)Bigr)=k,m!
$$
for $5le kle20$, $1le kle10^6$. This search provided so far a new solution:
$$
m=10,quad n=691,816,586,092
$$
answered 16 hours ago
Julián Aguirre
64.3k23894
add a comment |Â
up vote
4
down vote
accepted
The equation can be written as
$$
Bigl(fracm!2+1Bigr)sigma(n)=n,m!.
$$
Since $m!$ and $m!/2+1$ are coprime, $n$ is a multiple of $m!/2+1$. I have looked for solutions of the equation
$$
sigmaBigl(k,Bigl(fracm!2+1Bigr)Bigr)=k,m!
$$
for $5le kle20$, $1le kle10^6$. This search provided so far a new solution:
$$
m=10,quad n=691,816,586,092
$$
answered 16 hours ago
Julián Aguirre
64.3k23894
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The equation can be written as
$$
Bigl(fracm!2+1Bigr)sigma(n)=n,m!.
$$
Since $m!$ and $m!/2+1$ are coprime, $n$ is a multiple of $m!/2+1$. I have looked for solutions of the equation
$$
sigmaBigl(k,Bigl(fracm!2+1Bigr)Bigr)=k,m!
$$
for $5le kle20$, $1le kle10^6$. This search provided so far a new solution:
$$
m=10,quad n=691,816,586,092
$$
answered 16 hours ago
Julián Aguirre
64.3k23894
The equation can be written as
$$
Bigl(fracm!2+1Bigr)sigma(n)=n,m!.
$$
Since $m!$ and $m!/2+1$ are coprime, $n$ is a multiple of $m!/2+1$. I have looked for solutions of the equation
$$
sigmaBigl(k,Bigl(fracm!2+1Bigr)Bigr)=k,m!
$$
for $5le kle20$, $1le kle10^6$. This search provided so far a new solution:
$$
m=10,quad n=691,816,586,092
$$
answered 16 hours ago
Julián Aguirre
64.3k23894
answered 16 hours ago
Julián Aguirre
64.3k23894
answered 16 hours ago
Julián Aguirre
64.3k23894
answered 16 hours ago
Julián Aguirre
64.3k23894
64.3k23894
add a comment |Â
add a comment |Â
up vote
0
down vote
$$m=7 ; ; , ; ; ; n = 5919262622 = 2 cdot 7^2 cdot 13 cdot 19 cdot 97 cdot 2521 $$
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
Sun Aug 5 12:42:11 PDT 2018
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
answered 12 hours ago
Will Jagy
96.7k594195
add a comment |Â
up vote
0
down vote
$$m=7 ; ; , ; ; ; n = 5919262622 = 2 cdot 7^2 cdot 13 cdot 19 cdot 97 cdot 2521 $$
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
Sun Aug 5 12:42:11 PDT 2018
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
answered 12 hours ago
Will Jagy
96.7k594195
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$m=7 ; ; , ; ; ; n = 5919262622 = 2 cdot 7^2 cdot 13 cdot 19 cdot 97 cdot 2521 $$
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
Sun Aug 5 12:42:11 PDT 2018
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
answered 12 hours ago
Will Jagy
96.7k594195
$$m=7 ; ; , ; ; ; n = 5919262622 = 2 cdot 7^2 cdot 13 cdot 19 cdot 97 cdot 2521 $$
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
Sun Aug 5 12:42:11 PDT 2018
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
answered 12 hours ago
Will Jagy
96.7k594195
edited 11 hours ago
answered 12 hours ago
Will Jagy
96.7k594195
answered 12 hours ago
Will Jagy
96.7k594195
answered 12 hours ago
Will Jagy
96.7k594195
96.7k594195
add a comment |Â
add a comment |Â
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