All Questions
Tagged with arithmetic-functions gcd-and-lcm
24 questions
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Does the following GCD divisibility constraint imply that $\sigma(m^2)/p^k \mid m$, if $p^k m^2$ is an odd perfect number with special prime $p$?
The topic of odd perfect numbers likely needs no introduction.
In what follows, denote the classical sum of divisors of the positive integer $x$ by $$\sigma(x)=\sigma_1(x).$$
Let $p^k m^2$ be an odd ...
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On the prime factorization of $n$ and the quantity $J = \frac{n}{\gcd(n,\sigma(q^k)/2)}$, where $q^k n^2$ is an odd perfect number
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 $\...
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On a consequence of $G \mid I \iff \gcd(G, I) = G$ (Re: Odd Perfect Numbers and GCDs)
Let $N = q^k n^2$ be an odd perfect number given in the so-called Eulerian form, where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the classical sum of ...
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Are these valid proofs for the equation $\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))$, 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 $q \equiv k \equiv 1 \pmod 4$.
It is known that
$$i(q)=\gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/...
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1
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1
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Are the (Bezout) coefficients for $\gcd(n^2, \sigma(n^2)) = q\sigma(n^2) - 2(q - 1)n^2$ (where $q^k n^2$ is an odd perfect number) unique?
In what follows, denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the aliquot sum of $x$ by $s(x)=\sigma(x)...
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Does $\sigma(m^2)/p^k$ divide $m^2 - p^k$, if $p^k m^2$ is an odd perfect number with special prime $p$?
The following query is an offshoot of this post 1 and this post 2.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
The topic of odd perfect numbers likely ...
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How to compute sum of sum of gcd of factor pairs of a number upto a large number efficiently?
Define
$$f(n)=\sum_{d|n}gcd(d,\frac{n}{d})$$
$$F(x)=\sum_{n=1}^xf(n)$$ for natural numbers d,n and x.
I would like to know if there is some simplified form of $F(x)$ in terms of arithmetic functions, ...
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If an arithmetic function $f$ satisfies $f(mn) \leq f(m)f(n)$ (whenever $\gcd(m,n)=1$), is $f$ weakly multiplicative or submultiplicative?
From the preprint On sums of the small divisors of a natural number (Lemma 1, page 2) by Douglas E. Iannucci:
We observe here that the function $a(n)$ is not multiplicative. It is, however, ...
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1
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138
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When does $\gcd(m,\sigma(m^2))$ equal $\gcd(m^2,\sigma(m^2))$? What are the exceptions?
(This question is related to this earlier one.)
Let $\sigma(x)$ be the sum of divisors of the positive integer $x$. The greatest common divisor of the integers $a$ and $b$ is denoted by $\gcd(a,b)$.
...
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On odd perfect numbers and a GCD - Part II
(Note: This has been cross-posted to MO.)
Note that $\gcd(\sigma(q^k),\sigma(n^2))=i(q)=\gcd(n^2,\sigma(n^2))$ if and only if $\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))$. I have therefore undeleted ...
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Does $\gcd(r, \sigma(r)) = 2r - \sigma(r)$ always hold when $r$ is deficient-perfect? [duplicate]
Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. A number $r$ is called deficient-perfect if $(2r - \sigma(r)) \mid r$.
Here is my question:
Does $\gcd(r, \sigma(r)) = 2r - ...
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3
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2
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175
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If $\sigma(n)/n = 5/3$, then $5 \nmid n$. Does it also follow that $3 \nmid \sigma(n)$?
Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. If $\sigma(N)=2N$ (equivalently, when $I(N)=2$) then $N$ is called a ...
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2
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If $\sigma(n) = 2n - d$ and $d \mid n$, is it true that $d = \gcd(n,\sigma(n))$?
In what follows, assume that $d > 0$.
Let
$$\sigma(x)=\sum_{e \mid x}{e}$$
denote the classical sum-of-divisors function, and denote the deficiency of $x \in \mathbb{N}$ by
$$D(x)=2x-\sigma(x).$$
...
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1
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1
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If $q \equiv k \equiv 1 \pmod 4$, is it necessarily true that $\gcd\bigg(\sigma(q^k),\sigma(q^{(k-1)/2})\bigg)=1$?
Let $\sigma$ denote the classical sum-of-divisors function. In what follows, we let $q$ be a prime number.
Here is my question:
If $q \equiv k \equiv 1 \pmod 4$, is it necessarily true that $\gcd\...
- 8,228
2
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1
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82
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What proportion of the positive integers satisfies $\gcd(n^2,\sigma(n^2))>\sigma(n)$, where $\sigma$ is the sum-of-divisors function?
The title says it all.
What proportion of the positive integers satisfies $\gcd(n^2,\sigma(n^2))>\sigma(n)$, where $\sigma$ is the sum-of-divisors function?
I tried using Sage Cell Server to ...
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4
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If $x$ and $y$ are solitary numbers satisfying $\gcd(x,y)=1$, under what conditions does it follow that $xy$ is also solitary?
Let $\sigma(z)$ denote the sum of divisors of $z \in \mathbb{N}$. Denote the abundancy index of $z$ by $I(z) = \sigma(z)/z$.
If the equation $I(z)=I(a)$ has the lone solution $z=a$, then $a$ is said ...
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2
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71
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Are there any other squares $n^2$ for which $\gcd(n^2, \sigma(n^2)) = 2n^2 - \sigma(n^2)$?
Let $\sigma(x)$ denote the sum of the divisors of the positive integer $x$.
Denote the deficiency of $x$ by
$$D(x)=2x-\sigma(x).$$
I am interested in solutions to the equation
$$\gcd(n^2, \sigma(n^2)...
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4
votes
1
answer
349
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Is it possible to simplify this expression even further?
(Preamble: This question is tangentially related to this earlier one.)
Let $\sigma(z)$ denote the sum of the divisors of $z \in \mathbb{N}$, the set of positive integers. Denote the deficiency of $...
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4
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213
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Do odd numbers $n$ satisfying $\gcd(n, \sigma(n)) > \sqrt{n}$ have a special form?
Let $\sigma(n)$ denote the sum of divisors of the positive integer $n$.
Using Sage Cell Server, I was able to get the following odd numbers $n < 5000$ satisfying $\gcd(n, \sigma(n)) > \sqrt{n}$:...
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Does $D(n)$ always depend on $\gcd(n,\sigma(n))$ when $\sigma(N)=aN+b$, $N=xn$, and $n$ is a square?
Does $D(n)$ always depend on $\gcd(n,\sigma(n))$ when $\sigma(N)=aN+b$, $N=xn$, and $n$ is a square (with $\gcd(x,n)=1$)?
Here, $\sigma$ is the sum of divisors and $D(n) := 2n - \sigma(n)$ is the ...
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7
votes
1
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711
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A GCD summation $\sum_{\gcd(i,n)=1} \gcd(i-1,n)$
Let $$\ n=p_1^{a_1}p_2^{a_2}p_3^{a_3}\ldots$$ where $p_1,p_2\ldots$ are prime factors of $n$.
Show that
$$\sum_{i=1,\,\gcd(i,n)=1}^n \gcd(i-1,n) = \prod_{} (a_i+1)(p_i-1)p_i^{a_i-1}.$$
I was ...
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1
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78
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Is the following statement true if $N = q^k n^2$ is an odd perfect number given in Eulerian form?
A number $N$ is said to be perfect when $\sigma(N)=2N$, where $\sigma$ is the classical sum-of-divisors function.
An odd perfect number $N = {q^k}{n^2}$ is said to be given in Eulerian form if $q$ is ...
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4
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LCM of binomial coefficients and related functions
I know about the following identity: $$\displaystyle \text{lcm} \left( {n \choose 0}, {n \choose 1}, ... {n \choose n} \right) = \frac{\text{lcm}(1, 2, ... n+1)}{n+1}$$
1) Is there any method to find $...
user125368
3
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1
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325
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Looking an asymptotic for $\sum_{k\leq x}\Lambda(k)e^k$, where $\Lambda (n)$ is the von Mangoldt function
Using Abel's identity (see Theorem 4.2 in page 77 of [1]) and Prime Number Theorem (Theorem 4.4 in page 75) I compute
$$\frac{1}{x}\sum_{k\leq x}\Lambda(k)e^k\sim 1\cdot e^x-\frac{1}{x}\int_1^x \psi(...
user243301