Mixing
The sum of all divisors of $N$ is a 2 potency iff $N$ is a product of different Mersenne primes
As far as I have controlled:
$sigma(a)=2^n$, for some $ninmathbb N iff $ $a$ is a product of different Mersenne primes.
The $Leftarrow$-part is an immediate consequence of that $sigma$ is multiplicative, but the other part seems more complicated, if it is true.
Maybe this is interesting in the view of that it is unknown if there are only a finite number of Mersenne primes or not.
elementary-number-theory prime-numbers conjectures divisor-sum mersenne-numbers
asked Nov 20 '18 at 14:37
Lehs
6,94431662
add a comment |
As far as I have controlled:
$sigma(a)=2^n$, for some $ninmathbb N iff $ $a$ is a product of different Mersenne primes.
The $Leftarrow$-part is an immediate consequence of that $sigma$ is multiplicative, but the other part seems more complicated, if it is true.
Maybe this is interesting in the view of that it is unknown if there are only a finite number of Mersenne primes or not.
elementary-number-theory prime-numbers conjectures divisor-sum mersenne-numbers
asked Nov 20 '18 at 14:37
Lehs
6,94431662
3
math.stackexchange.com/a/1836974/78967
– mathlove
Nov 20 '18 at 15:22
add a comment |
As far as I have controlled:
$sigma(a)=2^n$, for some $ninmathbb N iff $ $a$ is a product of different Mersenne primes.
The $Leftarrow$-part is an immediate consequence of that $sigma$ is multiplicative, but the other part seems more complicated, if it is true.
Maybe this is interesting in the view of that it is unknown if there are only a finite number of Mersenne primes or not.
elementary-number-theory prime-numbers conjectures divisor-sum mersenne-numbers
asked Nov 20 '18 at 14:37
Lehs
6,94431662
As far as I have controlled:
$sigma(a)=2^n$, for some $ninmathbb N iff $ $a$ is a product of different Mersenne primes.
The $Leftarrow$-part is an immediate consequence of that $sigma$ is multiplicative, but the other part seems more complicated, if it is true.
Maybe this is interesting in the view of that it is unknown if there are only a finite number of Mersenne primes or not.
elementary-number-theory prime-numbers conjectures divisor-sum mersenne-numbers
elementary-number-theory prime-numbers conjectures divisor-sum mersenne-numbers
asked Nov 20 '18 at 14:37
Lehs
6,94431662
asked Nov 20 '18 at 14:37
Lehs
6,94431662
edited Nov 20 '18 at 14:43
asked Nov 20 '18 at 14:37
Lehs
6,94431662
asked Nov 20 '18 at 14:37
Lehs
6,94431662
asked Nov 20 '18 at 14:37
Lehs
6,94431662
6,94431662
3
math.stackexchange.com/a/1836974/78967
– mathlove
Nov 20 '18 at 15:22
add a comment |
3
math.stackexchange.com/a/1836974/78967
– mathlove
Nov 20 '18 at 15:22
3
3
math.stackexchange.com/a/1836974/78967
– mathlove
Nov 20 '18 at 15:22
math.stackexchange.com/a/1836974/78967
– mathlove
Nov 20 '18 at 15:22
add a comment |
1 Answer
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Here is a solution for $n$ squarefree:
Assume the lhs is true. Then you can split $a$ up into its primes. By definition you have $sigma(p)=p+1$ for each, thus $$sigma(prod_{i=1}^r p_i)=prod_{i=1}^r (p_i+1)=2^n$$
Since $2$ is the only prime factor of the right product, every factor is at least of the "Mersenne form" $2^m-1$.
If $m$ is a prime, we're done. If $m=pq$ ($p$ prime, $q$ integer) is computed then consider
$$2^m-1=(2^p-1)(2^{(q-1)p}+2^{(q-2)p} +...+2^{p+1})$$
Repeat the procedure with $2^{n/p}$.
The generalization could come from the fact that $sigma(p_n#)=2^n$ where $p_n#$ is the primorial, which only consists of square free primes.
answered Nov 20 '18 at 14:57
Nodt Greenish
29813
1
I don't think we know that each prime occurs only to the first power. You need to account for the possibility that $sigma(p^k)=2^m$ for some prime $p$ with $k>1$.
– paw88789
Nov 20 '18 at 15:21
1
Thanks for your comment, I restricted the scope of my answer to square free numbers.
– Nodt Greenish
Nov 20 '18 at 15:23
add a comment |
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1 Answer
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Here is a solution for $n$ squarefree:
Assume the lhs is true. Then you can split $a$ up into its primes. By definition you have $sigma(p)=p+1$ for each, thus $$sigma(prod_{i=1}^r p_i)=prod_{i=1}^r (p_i+1)=2^n$$
Since $2$ is the only prime factor of the right product, every factor is at least of the "Mersenne form" $2^m-1$.
If $m$ is a prime, we're done. If $m=pq$ ($p$ prime, $q$ integer) is computed then consider
$$2^m-1=(2^p-1)(2^{(q-1)p}+2^{(q-2)p} +...+2^{p+1})$$
Repeat the procedure with $2^{n/p}$.
The generalization could come from the fact that $sigma(p_n#)=2^n$ where $p_n#$ is the primorial, which only consists of square free primes.
answered Nov 20 '18 at 14:57
Nodt Greenish
29813
1
I don't think we know that each prime occurs only to the first power. You need to account for the possibility that $sigma(p^k)=2^m$ for some prime $p$ with $k>1$.
– paw88789
Nov 20 '18 at 15:21
1
Thanks for your comment, I restricted the scope of my answer to square free numbers.
– Nodt Greenish
Nov 20 '18 at 15:23
add a comment |
Here is a solution for $n$ squarefree:
Assume the lhs is true. Then you can split $a$ up into its primes. By definition you have $sigma(p)=p+1$ for each, thus $$sigma(prod_{i=1}^r p_i)=prod_{i=1}^r (p_i+1)=2^n$$
Since $2$ is the only prime factor of the right product, every factor is at least of the "Mersenne form" $2^m-1$.
If $m$ is a prime, we're done. If $m=pq$ ($p$ prime, $q$ integer) is computed then consider
$$2^m-1=(2^p-1)(2^{(q-1)p}+2^{(q-2)p} +...+2^{p+1})$$
Repeat the procedure with $2^{n/p}$.
The generalization could come from the fact that $sigma(p_n#)=2^n$ where $p_n#$ is the primorial, which only consists of square free primes.
answered Nov 20 '18 at 14:57
Nodt Greenish
29813
1
I don't think we know that each prime occurs only to the first power. You need to account for the possibility that $sigma(p^k)=2^m$ for some prime $p$ with $k>1$.
– paw88789
Nov 20 '18 at 15:21
1
Thanks for your comment, I restricted the scope of my answer to square free numbers.
– Nodt Greenish
Nov 20 '18 at 15:23
add a comment |
Here is a solution for $n$ squarefree:
Assume the lhs is true. Then you can split $a$ up into its primes. By definition you have $sigma(p)=p+1$ for each, thus $$sigma(prod_{i=1}^r p_i)=prod_{i=1}^r (p_i+1)=2^n$$
Since $2$ is the only prime factor of the right product, every factor is at least of the "Mersenne form" $2^m-1$.
If $m$ is a prime, we're done. If $m=pq$ ($p$ prime, $q$ integer) is computed then consider
$$2^m-1=(2^p-1)(2^{(q-1)p}+2^{(q-2)p} +...+2^{p+1})$$
Repeat the procedure with $2^{n/p}$.
The generalization could come from the fact that $sigma(p_n#)=2^n$ where $p_n#$ is the primorial, which only consists of square free primes.
answered Nov 20 '18 at 14:57
Nodt Greenish
29813
Here is a solution for $n$ squarefree:
Assume the lhs is true. Then you can split $a$ up into its primes. By definition you have $sigma(p)=p+1$ for each, thus $$sigma(prod_{i=1}^r p_i)=prod_{i=1}^r (p_i+1)=2^n$$
Since $2$ is the only prime factor of the right product, every factor is at least of the "Mersenne form" $2^m-1$.
If $m$ is a prime, we're done. If $m=pq$ ($p$ prime, $q$ integer) is computed then consider
$$2^m-1=(2^p-1)(2^{(q-1)p}+2^{(q-2)p} +...+2^{p+1})$$
Repeat the procedure with $2^{n/p}$.
The generalization could come from the fact that $sigma(p_n#)=2^n$ where $p_n#$ is the primorial, which only consists of square free primes.
answered Nov 20 '18 at 14:57
Nodt Greenish
29813
edited Nov 20 '18 at 15:28
answered Nov 20 '18 at 14:57
Nodt Greenish
29813
answered Nov 20 '18 at 14:57
Nodt Greenish
29813
answered Nov 20 '18 at 14:57
Nodt Greenish
29813
29813
1
I don't think we know that each prime occurs only to the first power. You need to account for the possibility that $sigma(p^k)=2^m$ for some prime $p$ with $k>1$.
– paw88789
Nov 20 '18 at 15:21
1
Thanks for your comment, I restricted the scope of my answer to square free numbers.
– Nodt Greenish
Nov 20 '18 at 15:23
add a comment |
1
I don't think we know that each prime occurs only to the first power. You need to account for the possibility that $sigma(p^k)=2^m$ for some prime $p$ with $k>1$.
– paw88789
Nov 20 '18 at 15:21
1
Thanks for your comment, I restricted the scope of my answer to square free numbers.
– Nodt Greenish
Nov 20 '18 at 15:23
1
1
I don't think we know that each prime occurs only to the first power. You need to account for the possibility that $sigma(p^k)=2^m$ for some prime $p$ with $k>1$.
– paw88789
Nov 20 '18 at 15:21
I don't think we know that each prime occurs only to the first power. You need to account for the possibility that $sigma(p^k)=2^m$ for some prime $p$ with $k>1$.
– paw88789
Nov 20 '18 at 15:21
1
1
Thanks for your comment, I restricted the scope of my answer to square free numbers.
– Nodt Greenish
Nov 20 '18 at 15:23
Thanks for your comment, I restricted the scope of my answer to square free numbers.
– Nodt Greenish
Nov 20 '18 at 15:23
add a comment |
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math.stackexchange.com/a/1836974/78967
– mathlove
Nov 20 '18 at 15:22