Questions tagged [divisor-sum]
For questions on the divisor sum function and its generalizations.
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Show that $\sum\nolimits_{d|n} \frac{1}{d} = \frac{\sigma (n)}{n}$ for every positive integer $n$.
Show that $\sum\nolimits_{d|n} \frac{1}{d} = \frac{\sigma (n)}{n}$ for every positive integer $n$.
where $\sigma (n)$ is the sum of all the divisors of $n$
and $\sum\nolimits_{d|n} f(d)$ is the ...
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When $p$ is an odd prime, is $(p+2)/p$ an outlaw or an index?
Let $\sigma(x)$ denote the sum of the divisors of $x$, and denote the abundancy index of $x$ as
$$I(x) = \dfrac{\sigma(x)}{x},$$
and the deficiency of $x$ as
$$D(x) = 2x - \sigma(x).$$
If the equation ...
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Asymptotic formula for $\sum_{n\leq x}\sigma(n)$ knowing $\sum_{n\leq x}\frac{\sigma(n)}{n}$
Let $\sigma(n):=\sum_{d|n}d$ be the sum of all divisors of $n$. Find the asymptotic formula for $\sum_{n\leq x}\frac{\sigma(n)}{n}$ and use it to find the one for $\sum_{n\leq x}\sigma(n)$.
Here is ...
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A number-theory question on the deficiency function $2x - \sigma(x)$
Let $\sigma(x)$ be the sum of the divisors of a (positive) integer $x$. (For example, $\sigma(2) = 1 + 2 = 3$.)
Define the deficiency function $D(x)$ to be the number
$$D(x) = 2x - \sigma(x).$$
Let ...
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On the quantity $I(q^k) + I(n^2)$ where $q^k n^2$ is an odd perfect number with special prime $q$
The topic of odd perfect numbers likely needs no introduction.
In what follows, we let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. Let
$$D(x) = 2x - \sigma(x)$$
denote the ...
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The sigma function (sum of divisors) multiplicative proof
I am trying to prove that $\sigma(p_1^a\cdot p_2^b) =\sigma(p_1^a)\cdot\sigma(p_2^b)$ where $p_1$ and $p_2$ are prime numbers.
We know that $\sigma(p_1^a) = \frac{p_1^{a+1}-1}{p_1-1}$ and $\sigma(p_2^...
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Can this bound for the abundancy index of $n$ be improved, given that $q^k n^2$ is an odd perfect number with $k=1$?
In what follows, set $I(x)=\sigma(x)/x$ to be the abundancy index of $x \in \mathbb{N}$, where $\sigma(x)$ is the sum of divisors of $x$. If $I(y)=2$ and $y$ is odd, then $y$ is called an odd perfect ...
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On odd perfect numbers and a GCD - Part III
(Note: This post is an offshoot of this earlier MSE question.)
In what follows, we let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. We also let $D(x)=2x-\sigma(x)$ denote the ...
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Does the following lower bound improve on $I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)}$, where $q^k n^2$ is an odd perfect number? - Part II
Preamble: This question is an offshoot of these earlier posts: (post1), (post2).
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,...
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Find the sum of reciprocals of divisors given the sum of divisors
Let $d_1, d_2, \cdots d_k$ be all the factors of a positive integer '$n$' including $1$ and $n$. Suppose $d_1 + d_2 + d_3+\cdots+d_k = 72$. Then find the value of $\frac{1}{d_1}+\frac{1}{d_2}+\cdots + ...
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How did Descartes come up with the spoof odd perfect number $198585576189$?
We call $n$ a spoof odd perfect number if $n$ is odd and and $n=km$ for two integers $k, m > 1$ such that $\sigma(k)(m + 1) = 2n$, where $\sigma$ is the sum-of-divisors function.
In a letter to ...
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When is the sum of divisors a perfect square?
For $n=3$, $\sigma(n)=4$, a perfect square. Calculating further was not yielding positive results.
I was wondering is there a way to find all such an $n$, like some algorithm?
We know that if $n=p_1^...
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For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$
For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$
My try :
Left hand side :
$\begin{align} \sum_{d|p^k}\sigma (d) &= \sigma(p^0) + \sigma(p^1) + \sigma(p^2) +...
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If $q^k n^2$ is an odd perfect number with Euler factor $q^k$, can $q = 73$ hold?
Call a number $N$ perfect if $\sigma(N)=2N$ where $\sigma$ is the classical sum-of-divisors function.
If $N = q^k n^2$ is an odd perfect number, can $q = 73$ hold?
Here is my attempt:
Since $37=(...
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Evaluating the sum $\sum_{i=1}^n i^2\cdot\lfloor{\frac ni}\rfloor$
I need to evaluate the sum $$\sum_{i=1}^n i^2\cdot\lfloor{\frac ni}\rfloor$$ After a little bit of math I found that the above sum is equal to: $$\sum_{i=1}^n i\cdot n - \sum_{i=1}^ni\cdot (n\space ...
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