All Questions
Tagged with divisor-sum totient-function
63
questions
3
votes
2
answers
491
views
Square of prime numbers
This conjecture is inspired by the comment of @Eric Snyder: Prime numbers which end with 03, 23, 43, 63 or 83
$n$ is a natural number $>1$, $\varphi(n)$ denotes the Euler's totient function, $P_n$ ...
- 87
3
votes
0
answers
158
views
Prime numbers which end with 03, 23, 43, 63 or 83
This is inspired from this post: Prime numbers which end with $19, 39, 59, 79$ or $99$
Here I found a new formula:
$n$ is a natural number $>1$, $\varphi(n)$ denotes the Euler's totient function, $...
- 87
-1
votes
1
answer
121
views
Prime numbers which end with $59$ or $79$ [closed]
This is the weak conjecture of https://math.stackexchange.com/questions/4834936/prime-numbers-which-end-with-19-39-59-79-or-99.\
$\varphi(n)$ denotes the Euler’s totient function, $n$ denotes a ...
- 87
6
votes
1
answer
258
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A question about prime numbers, totient function $ \phi(n) $ and sum of divisors function $ \sigma(n) $
I noticed something interesting with the totient function $ \phi(n) $ and sum of divisors function $ \sigma(n) $ when $n > 1$.
It seems than :
$ \sigma(4n^2-1) \equiv 0 \pmod{\phi(2n^2)}$ only if ...
- 141
1
vote
1
answer
144
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On a conjecture involving multiplicative functions and the integers $1836$ and $137$
We denote the Euler's totient as $\varphi(x)$, the Dedekind psi function as $\psi(x)$ and the sum of divisors function as $\sigma(x)$. Are well-known arithmetic functions, see the corresponding ...
4
votes
1
answer
107
views
On a conjecture involving multiplicative functions and the integers $1836$ and $136$
We denote the Euler's totient as $\varphi(x)$, the Dedekind psi function as $\psi(x)$ and the sum of divisors function as $\sigma(x)$. Are well-known arithmetic functions, see Wikipedia. I would like ...
2
votes
0
answers
68
views
Largest possible prime factor for given $k$?
Let $k$ be a positive integer.
What is the largest possible prime factor of a squarefree positive integer $\ n\ $ with $\ \omega(n)=k\ $ (That is, it has exactly $\ k\ $ prime factors) satisfying the ...
- 85.1k
2
votes
0
answers
182
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Are there infinite many squarefree numbers with $\ \varphi(n)\mid \sigma(n)\ $?
This question is inspired by this post.
I wonder whether the number of squarefree integers $\ n\ge 2\ $ with $\ \varphi(n)\mid \sigma(n)\ $ is still infinite. As in the link , $\ \varphi(n)\ $ is the ...
- 85.1k
5
votes
0
answers
254
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Are there infinite many positive integers $\ n\ $ satisfying $\ \varphi(n)|\sigma(n)\ $?
Today I suddenly discovered that many positive integers $\ n\ $ satisfy $\ \varphi(n)\mid \sigma(n)\ $. This leads to the following question :
Are there infinitely many postive integers $\ n\ $ ...
- 93.6k
1
vote
2
answers
68
views
Is it possible to improve on these bounds for $\frac{\varphi(n)}{n}$, if $q^k n^2$ is an odd perfect number with special prime $q$?
Let $N = q^k n^2$ be an odd perfect number with special/Eulerian prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the Euler-totient function of the positive integer $x$ by ...
- 8,010
1
vote
0
answers
97
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Improving $\dfrac{120}{217\zeta(3)} < \dfrac{\varphi(m)}{m}$ to $\dfrac{1}{2} < \dfrac{\varphi(m)}{m}$, where $p^k m^2$ is an odd perfect number
Preamble: This question is an offshoot of this earlier MSE post.
Consider a hypothetical odd perfect number $N=p^k m^2$ with special/Euler prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(...
- 8,010
3
votes
1
answer
233
views
Abundant products of iterations of Euler's totient function
Let $a_0(n) = n$ and $a_{i+1}(n) = \varphi(a_i(n))$ for $i\geq 0$, where $\varphi(n)$ is Euler's totient function (the number of positive integers less than or equal to $n$ and coprime with $n$). ...
- 2,803
0
votes
0
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47
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Divisor sum of totient function squared
I was wondering if it's possible to express the following sum as a function of $n$:
$$
\sum_{d|n} [\phi(d)]^2
$$
similarly to the known relation:
$$
\sum_{d|n} \phi(d) = n
$$
where $\phi$ is the ...
- 1,082
3
votes
0
answers
48
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Question about $\varphi(n)+n\mid \sigma(n)$
I want to prove that $$\varphi(n)+n\mid \sigma(n)$$ is impossible , if $\ \omega(n)=2\ $ , in other words , $\ n\ $ has exactly two distinct prime factors.
$\ \varphi(n)\ $ is the totient function ...
- 85.1k
10
votes
1
answer
308
views
Is there an odd solution of $\varphi(n)+n=\sigma(n)$?
I want to show that the only solution of $$\varphi(n)+n=\sigma(n)$$ for a positive integer $n$ is $n=2$.
What I worked out is that we must have $$\varphi(n)>\frac{n}{2}$$
To show this assume $n$ ...
- 85.1k