Tagged Questions

For questions on the divisor sum function and its generalizations.

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If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, and $k=1$, does it follow that $\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$?

Let $\sigma=\sigma_{1}$ be the classical sum-of-divisors function. If $N = q^k n^2$ is an odd perfect number with Euler prime $q$ (i.e., $q$ satisfies $q \equiv k \equiv 1 \pmod 4$), and $k=1$, does ...
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Combining a working hypothesis for odd perfect numbers with an inequality for logarithms

Euler's theorem for odd perfect numbers states that if there exists and odd perfect number, that is an odd positive integer $n$ satisfying $\sigma(n)=2n$, where $\sigma(m)=\sum_{d\mid m}d$ denotes the ...
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What are the mathematical consequences if $10$ is proved to be solitary?

Let $\sigma(x) = \sigma_{1}(x)$ denote the sum of the divisors of $x$, and let $$I(x) = \dfrac{\sigma(x)}{x}$$ be the abundancy index of $x$. For example, $$\sigma(10) = 1 + 2 + 5 + 10 = 18$$ so that ...
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The divisor function is defined as $\sigma_1(n)=\sigma(n)=\sum_{d\mid n}d$. Consider the divisor function over a quadratic $$f(x)=\sigma(a x^2+bx+c)$$ Where $a,b,c \in \mathbb{Z}$ (note we allow $a, b$...
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Project Euler's, Problem #565

Project Euler's, Problem #565 states: Let $\sigma(n)$ be the sum of the divisors of $n$. E.g. the divisors of $4$ are $1, 2$ and $4$, so $\sigma(4)=7$. The numbers $n$ not exceeding $20$ ...
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Justify $\lim_{n\to\infty}n^p\int_0^1\sum_{k=n}^\infty\frac{\sigma(k)e^{x/k}}{k^{p+2}\log\log k} dx=\frac{e^\gamma\int_0^1f(x)dx}{p}$

Inspired in PROBLEM 207, La Gaceta de la Real Sociedad Matemática Española, Vol. 16, N0. 3 (page 507 in spanish, proposed and solved by Furdui), I've tried write examples of this new statement ...
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Explaining an integral involving the divisor function

In a 1973 paper by Martinet, Deshouilliers and Cohen, $A(x)$ is defined as $$A(x)=\lim_{N\to\infty}\frac{\#\{n\leq N\mid \frac{\sigma(n)}{n}≥x \}}{N}$$ where $\sigma(n)$ is the "sum-of-divisors" ...
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Exploring congruences and identities involving Mersenne primes and the terms of Lucas-Lehmer test

When I was exploring congruences$\dagger$ involving the terms $S_k$ defined in Lucas-Lehmer test (this reference is the Wikipedia, but the main reference is Crandall and Pomerance, Prime Numbers: A ...
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Sum of divisors and indices: $\sigma (n)= 2^k$

Are there infinitely many positive integers (say $n$ is one of them), sum of whose divisors are powers of two, i.e, $\sigma (n)= 2^k$ ?
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Explaining a Series Expansion for the Divisor Function

The "sum-of-divisors" function, defined as $\sigma(n)=\sum_{d\,\mid\,n}d$, can be expressed as $$\sigma(n)=\sum_{i=1}^n\sum_{j=1}^i\cos\Big(2\pi n \frac{j}{i}\Big)$$This expansion makes logical sense. ...
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Elementary number theory sum of divisors

Let the sum of the divisors of a number $N$ be equal to $s$(excluding N itself) then show that if $s=N$ then show that N is a perfect number. I tried to use the basic formula for sum of divisors but ...
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Does satisfy $f(n)=\frac{\sigma(n)}{n^2}$ the hypothesis of Halasz’s inequality?

Let $\sigma(n)=\sum_{d\mid n}d$ the sum of divisor function. I would like to know if I can write an example of some of the following Theorem 1 or Theorem 2 from $$f(n)=\frac{\sigma(n)}{n^2}$$ in Tao, ...
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Does “sum of divisors” function attain every value of the form $kn$?

By the "sum of divisors" function I mean the function $\sigma (n)= \sum_{d|n} d$. If we choose $k=1$ then it is not possible that we have $\sigma (n)=n$ because $n$ always has at least two divisors, ...
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Equation involving 'sigma of integer' function

Let $\sigma (n)$ be the sum of all positive divisors of $n\in\mathbb{N}$. Determine for which least $n$: $$\sigma (x) = n$$ has exactly two and exactly three solutions. The problem also asks for the ...
What is wrong with this proof that $p_1 > \omega(x)$, where $p_1$ is the least prime dividing $x$?
Let $x \in \mathbb{N}$, and let $$x = \prod_{i=1}^{\omega(x)}{{p_i}^{\alpha_i}}$$ be the canonical factorization of $x$. (That is, the $p_i$'s are primes with $p_1 < \ldots < p_{\omega(x)}$.) ...
Let $\sigma(m)$ the sum of divisors function and $$S(m)=\sum_{k=1}^m\text{m mod k}$$ the sum of remainders function, then it is know that for integers $m>1$ $$\sigma(m)+S(m)=S(m-1)+2m-1.$$ On the ...