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Questions tagged [perfect-numbers]

Questions about or involving perfect numbers which are positive integers that are equal to the sum of their proper positive divisors.

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Breaking the barriers at $q=5$ and $q=13$ for $q^k n^2$ an odd perfect number with special prime $q$

Let $\sigma(x)$ be the sum of divisors of the positive integer $x$. If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. The question of existence of odd perfect numbers is the ...
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$n$-th term of the series 1 27 125 1000

What will be the nth term of the series 1 27 125 1000 for $n = 1$, it is 1 for $n = 2$, it is 27 for $n = 3$, it is 125 for $n = 4$, it is 1000
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Improving some known bounds regarding the abundancy index of divisors of odd perfect numbers

Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the sum of aliquot ...
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1answer
43 views

Why aren't these equivalent definitions of a perfect number contradictory?

I was reading the definition of what it means for a number to be perfect and I'm a little confused. from Wikipedia: a perfect number is a positive integer that is equal to the sum of its proper ...
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On GCDs and odd perfect numbers

Let $N=q^k n^2$ be an odd perfect number with special prime $q$. The index $i(q)$ of $N$ at the prime $q$ is then equal to $$i(q):=\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2}=\frac{D(n^2)}{s(q^...
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On the golden ratio and odd perfect numbers

Here is my question: Is $I(n^2) - 1 > 1/I(n^2)$ true when $I(n^2)=\sigma(n^2)/n^2$ is the abundancy index of $n^2$ and $q^k n^2$ is an odd perfect number with special prime $q$ satisfying $k>...
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If $q^k n^2$ is an odd perfect number with special prime $q$, then $n^2 - q^k$ is not a square.

Problem Statement Prove the following proposition. If $q^k n^2$ is an odd perfect number with special prime $q$, then $n^2 - q^k$ is not a square. Motivation Let $q^k n^2$ be an odd perfect ...
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Median divisor of even perfect numbers

I noticed that when divisors of even perfect numbers are listed in ascending order, the middle divisor (I guess the median), is always of the form $2^n$, some power of 2. If true is there a proof for ...
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A consequence of assuming the Descartes-Frenicle-Sorli Conjecture on odd perfect numbers

Let $x$ be a positive integer. Denote the sum of divisors of $x$ by $$\sigma(x) = \sum_{d \mid x}{d},$$ and the deficiency of $x$ by $$D(x) = 2x - \sigma(x).$$ A number $N$ is said to be perfect if $...
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1answer
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Can these bounds in terms of the abundancy index and deficiency functions be improved for deficient-perfect numbers?

Let $$\sigma(x) = \sum_{e \mid x}{e}$$ denote the sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$, and the deficiency of $x$ by $D(x)=2x-\sigma(x)$...
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1answer
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Improving the bound for $\sigma(q^k)/q^k$ where $q^k n^2$ is an odd perfect number given in Eulerian form

Let $x$ be a positive integer. (That is, let $x \in \mathbb{N}$.) We denote the sum of divisors of $x$ as $$\sigma(x) = \sum_{d \mid x}{d}.$$ We also denote the abundancy index of $x$ as $I(x)=\...
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On some constrained partitions of $ k $-multiperfect numbers

Let $ n $ be a $ k $ -multiperfect number. Do we know an upper bound for the number of partitions of $ n $ whose all summands are at the same time multiples of $ k $ and the sum of distinct ...
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1answer
46 views

Can this inequality regarding odd perfect numbers be improved?

Let $\sigma(x)$ denote the sum of the divisors of $x$. Denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$. If $N$ is odd and $\sigma(N)=2N$, then $N$ is called an odd perfect number. Euler showed ...
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On some inequalities relating the special/Euler prime and non-Euler part of odd perfect numbers

Let $N$ be an odd (positive) integer. If $\sigma(N)=2N$ where $\sigma(N)$ is the sum of the divisors of $N$, then $N$ is called an odd perfect number. Let $I(N)=\sigma(N)/N$ denote the abundancy ...
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On Basak's “Bounds On Factors Of Odd Perfect Numbers”

Let $N = q^k n^2$ be an odd perfect number given in Eulerian form, i.e. $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. In what follows, we denote the ...
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1answer
124 views

On Yanqi Xu's 2016 joint undergraduate math research project with Dr. Judy Holdener at Kenyon College

In what follows, we let $\sigma(X)$ denote the sum of the divisors of the positive integer $X$. Denote the abundancy index of $X$ by $I(X)=\sigma(X)/X$, and the deficiency of $X$ by $D(X)=2X-\sigma(X)...
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On an interesting assertion in the OeisWiki page on multiply-perfect numbers

The following (interesting) assertion appears in the OeisWiki page on multiply-perfect numbers: ...
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Divisors of a $ k $-multiperfect number

Let $ n $ be a $ k $-multiperfect number. Denote by $ d_m $ its $ m $ smallest divisor, and $ n_{m} $ the number of divisors of $ n $ divisible by $ d_m $. Is there for all $ 2\leq m\leq\tau(n) $...
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Is there an anomaly in the distribution of Mersenne primes?

In a recent press release off the Great Internet Mersenne Prime Search distributed computing project page, it is announced that $$2^{82589933} - 1$$ is the largest known (Mersenne) prime, ...
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1answer
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In a perfect number $2^{p−1} \times (2^p − 1)$, the ratio of $p$ to the digits in its perfect number approaches $\log(10) / \log(4)$?

I was reading about Mersenne primes and perfect numbers, and how the expression $2^{p−1} \times (2^p − 1)$, where $p$ is any prime number, can be used to generate perfect numbers when $2^{p−1}$ is a ...
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Has it been conjectured that all $k$-multiperfect numbers are multiples of $k$?

A quick glance at the list of the first $ k $ -multiperfect numbers for small $ k $ makes me think that all $ k $ -multiperfect numbers are multiples of $ k $ , which is a generalization of the ...
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1answer
48 views

A question regarding a paper of Ochem and Rao about the radical of an odd perfect number

Let $\operatorname{rad}(n)$ denote the radical or square-free part of the positive integer $n$, that is, $$\operatorname{rad}(n) = \prod_{p \mid n}{p}$$ where $p$ runs over primes. In the paper ...
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1answer
66 views

On the quantity $\sigma(\frac{n^2 \sigma(n^2)}{D(n^2)})$ when $q n^2$ is an odd perfect number with special prime $q$

Denote the sum of the divisors of $x \in \mathbb{N}$ by $\sigma(x)$. Also, denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$. If $m$ is odd and $\sigma(m)=2m$, then $m$ s called an odd perfect ...
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1answer
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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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1answer
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If $p=2^n-1$ is prime, therefore $2^{n-1}·(2^n-1)$ is perfect

If $p=2^n-1$ is prime, therefore $2^{n-1}·(2^n-1)$ is perfect. Well I would want to prove that but I don't really know where to start. Any ideas?
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Could a Fermat prime divide an odd perfect number?

This question is a follow-up to this earlier post. A positive integer $N$ is said to be perfect if $\sigma(N)=2N$, where $\sigma(x)$ is the sum of divisors of $x \in \mathbb{N}$. If $M$ is odd and $\...
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Does mathlove's answer imply that $D(n^2) \mid n^2$?

In what follows, set $\sigma(x)$ to be the sum of divisors of $x \in \mathbb{N}$, and let $$D(x) = 2x - \sigma(x)$$ be the deficiency of $x$, and let $$s(x) = \sigma(x) - x$$ be the sum of the aliquot ...
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1answer
64 views

Why are perfect numbers called perfect numbers?

A perfect number is a number than can be expressed as a sum of its factors. For example, 28 = 1 + 2 + 4 + 7 + 14 Why is this property important? What is so perfect about perfect numbers?
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What is the common (and simplified) value for $D(q^k)D(n^2) = 2s(q^k)s(n^2)$ when $q^k n^2$ is an odd perfect number?

In an answer to an earlier question, it is shown that $$D(2^p - 1)D(2^{p-1}) = 2s(2^p - 1)s(2^{p-1}) = 2^p - 2,$$ if $2^{p-1}(2^p - 1)$ is an even perfect number, $D(x) = 2x - \sigma(x)$ is the ...
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Is $(q^k n^2 \text{ is perfect }) \iff (D(q^k)D(n^2) = 2s(q^k)s(n^2))$ only true for odd perfect numbers $q^k n^2$?

(Preamble: This question is an offshoot of this earlier MSE post.) The title says it all. Is $\bigg(q^k n^2 \text{ is perfect }\bigg) \iff \bigg(D(q^k)D(n^2) = 2s(q^k)s(n^2)\bigg)$ only true for ...
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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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Inquiry about the Wolfram MathWorld entry on Odd Perfect Number

I just have a quick inquiry about the Wolfram MathWorld entry on Odd Perfect Number. Last sentence of the fifth paragraph states that: Hagis (1980) showed that odd perfect numbers must have at ...
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1answer
118 views

Question about a result on odd perfect numbers

In the paper titled Improving the Chen and Chen result for odd perfect numbers (Lemma 8, page 7), Broughan et al. show that if $$\frac{\sigma(n^2)}{q^k}$$ is a square, where $\sigma(x)$ is the sum of ...
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Is there a notion “above” that of perfect numbers?

When trying to understand a notion, it often gives great insight to see it as the "shadow" of something bigger, carrying more information. The notion of categorification relies on this idea. A basic ...
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1answer
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On the quantity ${n^2}/D(n^2)$ where $n^2$ is the non-Euler part of members of the OEIS sequence A228059

Let $\sigma(x)$ denote the sum of the divisors of the number $x \in \mathbb{N}$, the set of positive integers. Denote the deficiency of $x$ as $D(x):=2x-\sigma(x)$. This afternoon I noticed some ...
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Is $198585576189$ a member of OEIS sequence A228059?

I currently do not have enough computing power, so please pardon me for my question, which occurred just recently to me. So here it goes: Is the Descartes spoof $$\mathscr{D} = {3^2}\cdot{7^2}\...
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Specific evidence for or against the Descartes-Frenicle-Sorli conjecture on odd perfect numbers

This answer to an earlier (and related) MSE question summarizes what appears to be the first documented "evidence" against the Descartes-Frenicle-Sorli conjecture that $k=1$, if $q^k n^2$ is an odd ...
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Is $\left( {{2}^{x}}-1 \right)\left( {{5}^{x}}-1 \right)$ a square number for integer $x>1$

Motivated by this question. How to prove that $\left( {{2}^{x}}-1 \right)\left( {{5}^{x}}-1 \right)$ is not a square number for integer $x>1$? Thanks for any suggestions. Edition by the ...
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1answer
60 views

Is ${n^2}/D(n^2) \in \mathbb{N}$, if $q^k n^2$ is an odd perfect number?

Let $x \in \mathbb{N}$, the set of positive integers. The sum of the divisors of $x$ is denoted by $\sigma(x)$. Denote the deficiency of $x$ by $D(x):=2x-\sigma(x)$, and the sum of the aliquot parts ...
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2answers
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Does every perfect number have p*(1) and (p-1)*0 when p is prime at base 2?

For example 6, convert to the base 2 = 110 number of 1's is 2(prime), number of zero's is 1(prime - 1) or 496 = (111110000)2 5(prime) times 1 and 4 times 0 Is this always correct?
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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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3answers
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On the equation $\varphi(n)=\left(\frac{1+\sqrt{1+8n}}{8}\right)\cdot\left(\operatorname{rad}(n)-\frac{1+\sqrt{1+8n}}{2}\right)$

An integer is said to be an even perfect number if satisifies $\sigma(n)=2n$, where $\sigma(n)$ is the sum of the positive divisors of $n$. The first few even perfect numbers are $6,28,496$ and $8128$....
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A conjecture regarding odd perfect numbers

(Note: This question has now been cross-posted to MO.) Let $\sigma(z)$ denote the sum of the divisors of $z \in \mathbb{N}$, the set of positive integers. Denote the deficiency of $z$ by $D(z):=2z-\...
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On $\text{Lower bound}\leq \operatorname{rad}(n)$, where $n$ is an odd perfect number: reference request or what work can be done about it

For integers $n\geq 1$ we denote the square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is the product of distinct primes dividing an integer $n>1$ ...
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Solve for $p$ in this equation $\;2^{p−1}(2^p − 1) = X$.

Solve for $p$ in this equation $2^{p−1}(2^p − 1) = X$. This is a general formula for finding $X$ (an even perfect number), where $p$ is any prime number. I want to find $p$ when $X$ is given, so that ...
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1answer
119 views

Could a Mersenne prime divide an odd perfect number?

The relationship between Mersenne primes $2^r-1$ and even perfect numbers $2^{r-1}(2^r-1)$ is well-known (Euclid, Euler). In a video on the web I heard the statement that it is known that a Mersenne ...
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1answer
53 views

On a conjecture that $P_n^{\,2}+5^2+2^k=(P_n-1)^2+l^2$.

I was looking at perfect numbers and came across something that might serve a little interesting. Denote by $P_n$ the $n^\text{th}$ perfect number, then there appears to always exist $k\in\mathbb{W}...
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2answers
49 views

Would like to get numerical (lower [and upper?]) bounds for $p$

This question is an offshoot of this earlier MSE question. Let $\sigma(z)$ denote the sum of divisors of $z \in \mathbb{N}$, the set of positive integers. Denote the abundancy index of $z$ by $I(z) :...
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1answer
63 views

Confirmation of Proof: $\pi_n + p_a + p_b \geqslant \sum_{i=1}^4 x_i$.

I was messing around with numbers and I made the following conjecture: Conjecture: Let $\pi_n$ be the $n^{\text{th}}$ perfect number; $p_a$ be the prime after $\pi_n$ and $p_b$ be the ...
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0answers
91 views

Equations involving the Euler's totient function and Mersenne primes

In this post we denote the Euler's totient function as $\varphi(n)$, first we show a claim related to Mersenne primes, see for example this Wikipedia and secondly we are going to ask a related ...