Questions tagged [perfect-numbers]

Question 1

The problem For each $n\in \mathbb N^*$ determine the first decimal place of the number $a_n = \sqrt{4n^2+13n+10}$. My idea So I tried different values for $n$ and calculated the first decimal, and I ...

Question 2

Is $37$ the only prime number $p$ such that the repeating part of $\dfrac{1}{p}$ is one less than an even perfect number? This question was asked in our school contest that ended a few months ago. I ...

Question 3

Consider the sequence defined as follows: Start with a number N. Compute the prime factors of N with multiplicity, and add 1. Then, sum this list together to get N'. Iterate this procedure until you ...

Question 4

TLDR; My question is if this my assumptions regarding convolution steps, etc. are correct, and if there are further asymptotic expansions I could make for $\psi(x)$. I'm almost certain of a mistake in ...

Question 5

Let $I(a)=\sigma(a)/a$ denote the abundancy index of the positive integer $a$, where $\sigma(a)=\sigma_1(a)$ is the classical sum of divisors of $a$. If $I(a)=b$ where $b$ is an integer, then $a$ is ...

Question 6

The topic of odd perfect numbers likely needs no introduction. In what follows, we shall denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Let $q^k n^2$ be ...

Question 7

In what follows, let $\sigma(x)$ denote the classical sum of divisors of the positive integer $x$, and let $D(x) = 2x-\sigma(x)$ denote the deficiency of $x$. If $p^k m^2$ is an odd perfect number ...

Question 8

Let $p$ be any prime integer greater than $2$. How might we go about finding the sum of a finite geometric sequence $p^0+p^1+p^2...+p^n$ which is $p$-smooth, i.e. all of its factors are less than $p$ ?...

Question 9

Let $K:=\sum_{k\ge 0}\left(\frac{\sqrt{5}-1}{2}\right)^{2^{k}}$. The observation that the sequence of divisors of a large enough (even) perfect number is "close" to a sequence of consecutive ...

Question 10

I got bored in maths class and came up with this theorem, can someone tell me how to prove it/if it's even possible to prove it at all? Assuming $n$ is a perfect number If you order the divisors of $n$...

Question 11

I’ve been delving into an intriguing property related to perfect numbers and their proper divisors. Specifically, I’m investigating numbers $N$ for which the product of their proper divisors equals ...

Question 12

Let $p^k m^2$ be an odd perfect number with special prime $p$. Since $p \equiv k \equiv 1 \pmod 4$ must hold, then $m^2 - p^k \equiv 0 \pmod 4$. It follows that $$m^2 - p^k = a^2 - b^2 = (a - b)(a + b)...

Question 13

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 ...

Question 14

Using the divisor_sigma[n, k] function from the python sympy library where n is the positive integer which is having its divisors added and k is the power each factor is raised to, I was looking for ...

Question 15

where even perfect numbers are of the form $2^{p-1}(2^{p} - 1)$ ( $p$ and $2^{p} - 1$ are prime numbers ) My attempt $\phi(n)$ = ($2^{p - 1} - 2$)($2^{p} - 2$) So, we need to prove that. $2^{p - 1}$($...

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Questions tagged [perfect-numbers]

For each $n\in \mathbb N^*$, determine the first decimal place of the number $a_n = \sqrt{4n^2+13n+10}$.

Is $37$ the only prime number $p$ such that the repeating part of $\dfrac{1}{p}$ is one less than an even perfect number?

How quickly does the sum of prime factors chain grow?

Result Involving Perfect Numbers

Is this an example of a vacuous implication?

If $\frac{\sigma(q^k)}{2}$ is squarefree, does it follow that $\frac{n^2}{\sigma(q^k)/2}$ is also squarefree where $q^k n^2$ is an odd perfect number?

Does $\frac{\sigma\left(m^2\right)}p = D\left(p^{k-1} m^2\right)$ hold if $p^k m^2$ is an odd perfect number with special prime $p$? [closed]

Finding sums of finite geometric sequences with common ratio $p$ that are $p-$smooth numbers.

Is $n>6$ perfect if $\vert\sum_{d\mid n}(\frac{\sqrt{5}-1}{2})^{d}-K\vert\lt\frac{1}{2n}$?

Can someone tell me how to proof this/tell me if this is even possible to proof

Exploring Perfect Cubic Numbers: When Does the Product of Proper Divisors Equal $N^3$? [closed]

Proof Verification: If $p^k m^2$ is an odd perfect number with special prime $p$, then $p \equiv 1 \pmod 8$ holds.

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 $p$?

Generalized "perfect numbers" using different n,k values of divisorSum[n, k]

prove that $n - \phi(n)$ is a square where $n$ is an even perfect number

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