Mixing
For what numbers $x$ is $D(x) = 2x - sigma_1(x)$ equal to $varphi(x)$?
1
$begingroup$
Let $sigma_1$ be the classical sum-of-divisors function.
Let
$$D(x) = 2x - sigma_1(x)$$
be the deficiency of $x$.
Let $varphi(x)$ denote the Euler totient.
Here is my question:
For what numbers $x$ is $D(x) = 2x - sigma_1(x)$ equal to $varphi(x)$?
Trivially, $x=1$ answers my question. But are there any others?
divisor-sum arithmetic-functions
asked Aug 28 '16 at 15:27
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
$endgroup$
add a comment |
1
$begingroup$
Let $sigma_1$ be the classical sum-of-divisors function.
Let
$$D(x) = 2x - sigma_1(x)$$
be the deficiency of $x$.
Let $varphi(x)$ denote the Euler totient.
Here is my question:
For what numbers $x$ is $D(x) = 2x - sigma_1(x)$ equal to $varphi(x)$?
Trivially, $x=1$ answers my question. But are there any others?
divisor-sum arithmetic-functions
asked Aug 28 '16 at 15:27
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
$endgroup$
2
$begingroup$
Only $x=1$ and $x=p$ for some prime natural number $p$ are the equality case. Your other thread on a "sharp" bound of $D(x)$ should provide a clue on why that is. You can first show that $x$ must be squarefree, and then $x$ can have at most one prime factor.
$endgroup$
– Batominovski
Aug 28 '16 at 15:49
2
$begingroup$
$sigma(n) = sum_{d | n} d$ and $phi(n) = sum_{d | n} d mu(n/d)$ so $frac{sigma(n)+phi(n)}{2} = sum_{d | n} d frac{1+mu(n/d)}{2} ge n$ and it is $= n$ iff $frac{1+mu(n/d)}{2} = 0$ whenever $d > 1$ i.e. iff $n$ is prime
$endgroup$
– reuns
Aug 28 '16 at 17:25
add a comment |
1
1
1
1
$begingroup$
Let $sigma_1$ be the classical sum-of-divisors function.
Let
$$D(x) = 2x - sigma_1(x)$$
be the deficiency of $x$.
Let $varphi(x)$ denote the Euler totient.
Here is my question:
For what numbers $x$ is $D(x) = 2x - sigma_1(x)$ equal to $varphi(x)$?
Trivially, $x=1$ answers my question. But are there any others?
divisor-sum arithmetic-functions
asked Aug 28 '16 at 15:27
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
$endgroup$
Let $sigma_1$ be the classical sum-of-divisors function.
Let
$$D(x) = 2x - sigma_1(x)$$
be the deficiency of $x$.
Let $varphi(x)$ denote the Euler totient.
Here is my question:
For what numbers $x$ is $D(x) = 2x - sigma_1(x)$ equal to $varphi(x)$?
Trivially, $x=1$ answers my question. But are there any others?
divisor-sum arithmetic-functions
divisor-sum arithmetic-functions
asked Aug 28 '16 at 15:27
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
asked Aug 28 '16 at 15:27
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
edited Jan 15 at 15:28
Jose Arnaldo Bebita Dris
asked Aug 28 '16 at 15:27
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
asked Aug 28 '16 at 15:27
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
asked Aug 28 '16 at 15:27
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
5,43641944
2
$begingroup$
Only $x=1$ and $x=p$ for some prime natural number $p$ are the equality case. Your other thread on a "sharp" bound of $D(x)$ should provide a clue on why that is. You can first show that $x$ must be squarefree, and then $x$ can have at most one prime factor.
$endgroup$
– Batominovski
Aug 28 '16 at 15:49
2
$begingroup$
$sigma(n) = sum_{d | n} d$ and $phi(n) = sum_{d | n} d mu(n/d)$ so $frac{sigma(n)+phi(n)}{2} = sum_{d | n} d frac{1+mu(n/d)}{2} ge n$ and it is $= n$ iff $frac{1+mu(n/d)}{2} = 0$ whenever $d > 1$ i.e. iff $n$ is prime
$endgroup$
– reuns
Aug 28 '16 at 17:25
add a comment |
2
$begingroup$
Only $x=1$ and $x=p$ for some prime natural number $p$ are the equality case. Your other thread on a "sharp" bound of $D(x)$ should provide a clue on why that is. You can first show that $x$ must be squarefree, and then $x$ can have at most one prime factor.
$endgroup$
– Batominovski
Aug 28 '16 at 15:49
2
$begingroup$
$sigma(n) = sum_{d | n} d$ and $phi(n) = sum_{d | n} d mu(n/d)$ so $frac{sigma(n)+phi(n)}{2} = sum_{d | n} d frac{1+mu(n/d)}{2} ge n$ and it is $= n$ iff $frac{1+mu(n/d)}{2} = 0$ whenever $d > 1$ i.e. iff $n$ is prime
$endgroup$
– reuns
Aug 28 '16 at 17:25
2
2
$begingroup$
Only $x=1$ and $x=p$ for some prime natural number $p$ are the equality case. Your other thread on a "sharp" bound of $D(x)$ should provide a clue on why that is. You can first show that $x$ must be squarefree, and then $x$ can have at most one prime factor.
$endgroup$
– Batominovski
Aug 28 '16 at 15:49
$begingroup$
Only $x=1$ and $x=p$ for some prime natural number $p$ are the equality case. Your other thread on a "sharp" bound of $D(x)$ should provide a clue on why that is. You can first show that $x$ must be squarefree, and then $x$ can have at most one prime factor.
$endgroup$
– Batominovski
Aug 28 '16 at 15:49
2
2
$begingroup$
$sigma(n) = sum_{d | n} d$ and $phi(n) = sum_{d | n} d mu(n/d)$ so $frac{sigma(n)+phi(n)}{2} = sum_{d | n} d frac{1+mu(n/d)}{2} ge n$ and it is $= n$ iff $frac{1+mu(n/d)}{2} = 0$ whenever $d > 1$ i.e. iff $n$ is prime
$endgroup$
– reuns
Aug 28 '16 at 17:25
$begingroup$
$sigma(n) = sum_{d | n} d$ and $phi(n) = sum_{d | n} d mu(n/d)$ so $frac{sigma(n)+phi(n)}{2} = sum_{d | n} d frac{1+mu(n/d)}{2} ge n$ and it is $= n$ iff $frac{1+mu(n/d)}{2} = 0$ whenever $d > 1$ i.e. iff $n$ is prime
$endgroup$
– reuns
Aug 28 '16 at 17:25
add a comment |
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2
$begingroup$
Only $x=1$ and $x=p$ for some prime natural number $p$ are the equality case. Your other thread on a "sharp" bound of $D(x)$ should provide a clue on why that is. You can first show that $x$ must be squarefree, and then $x$ can have at most one prime factor.
$endgroup$
– Batominovski
Aug 28 '16 at 15:49
2
$begingroup$
$sigma(n) = sum_{d | n} d$ and $phi(n) = sum_{d | n} d mu(n/d)$ so $frac{sigma(n)+phi(n)}{2} = sum_{d | n} d frac{1+mu(n/d)}{2} ge n$ and it is $= n$ iff $frac{1+mu(n/d)}{2} = 0$ whenever $d > 1$ i.e. iff $n$ is prime
$endgroup$
– reuns
Aug 28 '16 at 17:25