Questions tagged [divisor-sum]
For questions on the divisor sum function and its generalizations.
772
questions
0
votes
1
answer
71
views
What conditions on $X$ will guarantee that $\gcd(\text{square part of } X,\text{squarefree part of } X)=1$, if $X$ is neither a square nor squarefree?
The following query is an offshoot of this answer to a closely related post.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
The topic of odd perfect ...
0
votes
0
answers
22
views
Exponential sums with the divisor function
Can anyone explain the first $\ll $ on line 8 on page 188 of "Jutila: On exponential sums involving the divisor function" for me? (https://eudml.org/doc/152693)
Specifically, I think he is ...
1
vote
0
answers
46
views
On Carmichael function and aliquot parts of odd perfect numbers
We denote as $N$ an odd perfect number, and $d\mid N$ one of its divisors. We denote the Carmichael function as $\lambda(x)$, Wikipedia has the article Carmichael function dedicated to this number ...
0
votes
0
answers
44
views
Does $I = \gcd(n,\sigma(n^2)) = (\frac{n}{\sigma(q^k)/2})\cdot\gcd(\sigma(q^k)/2,n)$ imply that $\sigma(q^k)/2 \mid n$ holds?
Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Define the GCDs:
$$G = \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$
$$H = \...
3
votes
1
answer
45
views
Reference request regarding odd 4-perfect numbers
I am reading a paper by Broughan and Zhou (2006) that is dealing with odd 4-perfect numbers. The title of the paper is "Odd multiperfect numbers of abundancy four."
In this paper they ...
2
votes
0
answers
52
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 ...
1
vote
0
answers
31
views
On the equation $\sigma\left(\square\right)=\text{prime}$: propositions that can be potentially interesting or reference request
For integers $A,B\geq 1$ we define the difference $\sigma(A)\sigma(B)-\sigma(AB)$, denoting it as $[A,B]$, where $\sigma(n)=\sum_{1\leq d\mid n}d$ denotes the sum of divisors function. It is possible ...
1
vote
0
answers
47
views
How can we show this relationship between the recursive sum of divisors function and Figurate Number Polynomial on Primes?
Let us define the following recursive function involving the sum of divisors function $\sigma(n)$:
\begin{array}{ l }
r(n,1)=\sigma(n) \\
r(n,2)=\sum_{d|n}r(d,1) \\
r(n,3)=\sum_{d|n}r(d,2) \\
\...
1
vote
0
answers
165
views
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 ...
5
votes
0
answers
238
views
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\ $ ...
0
votes
0
answers
73
views
On odd perfect numbers and a GCD - Part VIII
(This question is an offshoot of earlier posts with a similar title and this recent preprint.)
Let $N = q^k n^2$ be an odd perfect number with special/Eulerian prime $q$ satisfying $q \equiv k \equiv ...
1
vote
1
answer
44
views
Why does this identity with the product $\prod_{p\mid n}(p+k)$ and recursive sum of divisors function is true?
Let us define the following recursive function involving the sum of divisors function $\sigma(n)$:
\begin{array}{ l }
r(n,1)=\sigma(n) \\
r(n,2)=\sum_{d|n}r(d,1) \\
r(n,3)=\sum_{d|n}r(d,2) \\
\...
5
votes
1
answer
51
views
How can we show this relationship between the sum of divisors function and the sum $p^{m}+2p^{m-1}+3p^{m-2}+\ldots+(m+1)$?
The sum of divisors function is commonly denoted by $\sigma(n)$. Now let us introduce a recursive definition of divisor functions:
$r_{n,1}=\sigma(n)$
$r_{n,2}=\sum_{d|n}r_{d,1}$
$r_{n,3}=\sum_{d|n}...
0
votes
1
answer
42
views
If $s = 4m - 3$, then $\sigma(p^s) = (1 + p^{2m-1})(1 + p + \ldots + p^{2m-2})$. Is there a similar factorization for $\sigma(q^t)$ when $t=4n$?
Let $p,q$ be (odd) primes, and let $m,n,s,t$ be positive integers.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
M. A. Nyblom showed that, if $s = 4m - 3$...
1
vote
0
answers
126
views
Why is this alternating summation involving reciprocals of divisors always positive? A conjecture.
Conjecture. For all $n,m \in \Bbb{N}$,
$$
f(n, m) := \sum_{c\mid d\mid n\# \\ \gcd(c, 2m) = 1}\dfrac{(-1)^{\omega(d)}}{d}
$$
is greater than $0$.
Example verification code:
...
0
votes
2
answers
362
views
Does $\sigma(m^2)/p^k$ divide $m^2 - p^k$, if $p^k m^2$ is an odd perfect number with special prime $p$?
The following query is an offshoot of this post 1 and this post 2.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
The topic of odd perfect numbers likely ...
0
votes
2
answers
263
views
What are the remaining cases to consider for this problem, specifically all the possible premises for $i(q)$?
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. (Note that the divisor sum $\sigma$ is a multiplicative function.)
A number $P$ is said to be perfect if $\...
0
votes
0
answers
30
views
Is this divisors related sequence asymptotically linear?
In order to study this Riemann hypothesis equivalent here, I have been looking at the first differences of this sequence:
$$a(n) = \sum _{m=1}^{n^2} \left(\sum _{k=1}^{n^2} [k|m][k\leq n][m\leq k \...
1
vote
1
answer
108
views
Does there exist a nontrivial prime power $q^k$ such that $\sigma(n^2)/n = q^k$ for some $n$?
Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of the positive integer $x$.
My question in the present post is closely related to this one in MO:
QUESTION
Does there exist a nontrivial ...
0
votes
1
answer
19
views
R.T.P. σ(p^n)=(p^(n+1)-1)/(p-1). (where σ denotes the divisor function) For a prime p. [closed]
I need to prove that for a prime p, σ(p^n) is (p^(n+1)-1)/(p-1).
E.g.
σ(3^3) is (3^(3+1)-1)/(3-1).
=3^(4)-1/2
=80/2
=40
Therefore σ(3^3)=40.
Let me know for any suggestions,proofs or references.
0
votes
1
answer
116
views
On odd perfect numbers and a GCD - Part VII
(Pardon me for being somewhat stubborn, but this question will be the last for this week. This post is an offshoot of this one.)
Let $N = q^k n^2$ be an odd perfect number be an odd perfect number ...
0
votes
1
answer
38
views
Does $G \mid I$ and $I \mid H$ still hold if $\sigma(q^k)/2$ is not squarefree, where $q^k n^2$ is an odd perfect number with special prime $q$?
This question is an offshoot of this post #1 and this post #2.
Let $N = q^k n^2$ be an odd perfect number be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\...
0
votes
0
answers
26
views
Please check my proof: $\gcd(\sigma(q^k),\sigma(n^2)) \mid \gcd(n^2,\sigma(n^2))$ holds if $q^k n^2$ is an odd perfect number with special prime $q$.
Let $q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
It is known that
$$\gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)\cdot\gcd\bigg(n^2,\...
0
votes
0
answers
40
views
Please check my proof: If $q^k n^2$ is an odd perfect number with special prime $q$, then $\sigma(q^k)/2 \mid n$ if and only if $n \mid \sigma(n^2)$.
Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Denote the classical sum of divisors of the positive integer $x$ by $\...
0
votes
0
answers
33
views
(Applied to OPN) Let $c$ and $d$ be integers, not both equal to zero. If $q$ and $r$ are integers such that $c = dq + r$, then $\gcd(c,d)=\gcd(d,r)$.
I was reading the following web resource, and found the following Lemma 8.1, which I think would be immensely useful for a computational task on odd perfect numbers (hereinafter abbreviated as OPN):
...
1
vote
0
answers
32
views
If $km$ is a Descartes number with quasi-Euler prime $m$, must $\sqrt{k}$ be a squarefree palindrome?
Let $$\sigma(x) = \sum_{d \mid x}{d}$$ denote the sum of divisors of $x \in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers or positive integers.
Recall that a Descartes number is an odd ...
3
votes
2
answers
106
views
The map $m \to m\sigma(m)$ is not injective.
Let
$$\tau(n) = \sum_{d \mid n}{1}$$
be the divisor function,
$$\omega(n) = \sum_{p \mid n}{1}$$
be the prime divisor function,
$$\varphi(n) = \#\{1 \leqslant k \leqslant n : \gcd(k,n) = 1\}$$
be ...
0
votes
4
answers
86
views
Does $\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} < 10$ hold in general for an odd perfect number $q^k n^2$ with special prime $q$?
This question is an offshoot of this MSE answer.
Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$. (Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.)
If $\sigma(M) = 2M$...
0
votes
0
answers
15
views
sum of unbiased dice
Two unbiased dice are rolled and the numbers obtained are added. In this way, you have a better chance of getting numbers whose sum is:
a} 11
b} 10
c} 9
d} 8
e} 7
I did my sums and it is giving ...
1
vote
0
answers
67
views
On the reciprocal of $I(n^2) - \frac{2(q - 1)}{q}$, if $q^k n^2$ is a(n) (odd) perfect number with special prime $q$
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
A number $N$ is said to be perfect if $\sigma(N)=...
1
vote
3
answers
107
views
Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ improve $I(q^k) + I(n^2) < 3$, 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 prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/...
1
vote
1
answer
91
views
Bounds for the abundancy index of divisors of odd perfect numbers in terms of the deficiency function - Part II
This post is an offshoot of this MSE question.
Motivation
Let $x, y$ and $z$ be positive integers.
Denote the sum of divisors of $x$ by $\sigma(x)$. Also, denote the deficiency of $y$ by $D(y)=2y-\...
-1
votes
1
answer
73
views
An equation about divisor function [closed]
We define the 2th divisor function as follows: $\sigma_2(n)=\sum_{d\mid n}d^2$?
Is there infinite positive integer $n$ such that $\sigma_2(n)=(n+3)^2$? If there is, how can we find the least $n$?
3
votes
1
answer
114
views
If $\gcd(n^2, \sigma(n^2)) = q\sigma(n^2) - 2(q - 1) n^2$, does it follow that $q n^2$ is perfect?
Let $q$ be an (odd) prime, and let $\gcd(q,n)=1$.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
A number $N$ is said to be perfect if $\sigma(N)=2N$.
...
1
vote
1
answer
121
views
On the quantity $I(q^k) + I(n^2)$ considered as functions of $q$ and $k$, when $q^k n^2$ is an odd perfect number with special prime $q$ - Part II
Hereinafter, let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Denote the abundancy index of the positive integer $x$ ...
0
votes
3
answers
184
views
On the conjectured inequality $q > k$, where $q^k n^2$ 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 $\gcd(q,n)=1$.
Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ ...
2
votes
1
answer
127
views
On odd perfect numbers and a GCD - Part VI
(Note: This post is closely related to this earlier MSE question.)
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 $\gcd(q,n)=1$....
2
votes
1
answer
82
views
On odd perfect numbers and a GCD - Part V
(Note: This post is tangentially related to this earlier MSE question.)
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,...
1
vote
0
answers
38
views
Additive divisor problems
There is plenty of literature on the binary additive divisor problem, that's evaluating
\[ \sum _{n\leq x}d(n)d(n+h)\]
for various $h$. Why is there nothing (or I don't find) on
\[ \sum _{n\leq x}d(n)...
3
votes
2
answers
135
views
Help with "A Simpler Dense Proof regarding the Abundancy Index."
I'm reading Richard Ryan's article "A Simpler Dense Proof regarding the Abundancy Index" and got stuck in his proof for Theorem 2. The Theorem is stated as follows:
Suppose we have a ...
2
votes
1
answer
81
views
If $k=1$, then the index of an odd perfect number $p^k n^2$ (as defined by Chen and Chen (2014)) is not squarefree.
In what follows, we let $\sigma=\sigma_1$ denote the classical sum-of-divisors function. Denote the deficiency function of the positive integer $x$ by $D(x)=2(x)-\sigma(x)$, and the aliquot sum of $x$...
0
votes
0
answers
57
views
Is it true that $l_1(q,n) \geq g(k)$, if $q^k n^2$ is an odd perfect number with special prime $q$?
(Note: This post is an offshoot of these earlier questions: (post 1) and (post 2).)
Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(...
1
vote
1
answer
108
views
On the inequality $I(q^k)+I(n^2) \leq \frac{3q^{2k} + 2q^k + 1}{q^k (q^k + 1)}$ where $q^k n^2$ is an odd perfect number
(Note: This post is an offshoot of this earlier MSE question.)
Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Define the ...
1
vote
2
answers
51
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 ...
0
votes
0
answers
58
views
If $q^k n^2$ is an odd perfect number with special prime $q$, is $\gcd(\sigma(q^k),\sigma(n^2))=1$ equivalent to $k=1$?
Let $q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Here is my:
QUESTION
If $q^k n^2$ is an odd perfect number with special prime ...
4
votes
2
answers
143
views
Is $\limsup_n \frac{\sigma(n)}{n \log p(n)} <\infty$, where $p(n)$ is the greatest prime factor of $n$ and $\sigma(n)=\sum_{d | n} d$?
Let $\sigma(n)=\sum_{d | n} d$ and $p(n)$ be the greatest prime factor of $n$.
Can we prove that
$$\limsup_n \frac{\sigma(n)}{n \log p(n)} <\infty ?$$
I know that
$$\limsup_n \frac{\sigma(n)}{n \...
0
votes
0
answers
161
views
A proposed unconditional factor-chain proof for the inequality $I(n) > \frac{3}{2}$, where $q^k n^2$ is an odd perfect number with special prime $q$
Let $N = q^k n^2$ be a hypothetical odd perfect number given in Eulerian form. (That is, $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.)
Denote the ...
1
vote
0
answers
31
views
Correctness of $\sigma_s(n) (1-\frac{1}{n^s}) = 0 \pmod{k}$ (divisor sum congruence)
I've played around with some identities and came up with:
$\sigma_s(n)(1-\frac{1}{n^s}) = 0 \pmod{k}$
(for $n$ and $k$ positive integers, and $s$ an integer)
With the conditions that 1) $n$ is ...
2
votes
1
answer
167
views
On odd perfect numbers $q^k n^2$ and the deficient-perfect divisor $q^{\frac{k-1}{2}} n^2$ - Part III
(Preamble: This question is an offshoot of this answer by mathlove to an earlier post.)
Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of the divisors of the positive integer $x$.
If $\sigma(m)=...
2
votes
2
answers
77
views
On improving the upper bound $I(m^2) \leq \frac{2p}{p+1}$, if $p^k m^2$ is an odd perfect number with special prime $p$
(Preamble: This question is an offshoot of this earlier MSE post.)
Let $p^k m^2$ be an odd perfect number (OPN) with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
...