Question

If $q^k n^2$ is an odd perfect number with special prime $q$, is $\sigma(q^k)$ coprime to $\sigma(n^2)$?

score 270 · Answer 1 · 2020-12-27 08:15:38

270

Views

If $q^k n^2$ is an odd perfect number with special prime $q$, is $\sigma(q^k)$ coprime to $\sigma(n^2)$?

Published on 27 Dec 2020 - 8:15

score 420 · Answer 2 · 2026-02-23 11:08:22

420

Views

Odd perfect numbers and egyptian fraction conjecture

Published on 23 Feb 2026 - 11:08

score 253 · Answer 3 · 2021-01-07 04:53:32

253

Views

Revisiting questions 3888565 and 3894925

Published on 07 Jan 2021 - 4:53

score 266 · Answer 4 · 2021-01-13 06:21:39

266

Views

What numerical lower bound on the index of an odd perfect number is implied by the results in F.-J. Chen and Y.-G. Chen's 2014 paper?

Published on 13 Jan 2021 - 6:21

score 438 · Answer 5 · 2021-01-25 10:05:04

438

Views

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?

Published on 25 Jan 2021 - 10:05

score 86 · Answer 6 · 2021-02-16 15:19:47

86

Views

On the inequality $\frac{\sigma(n^2)}{q^k} < \frac{n^2 - q^k}{C}$ where $C>1$ and $q^k n^2$ is an odd perfect number - Part II

Published on 16 Feb 2021 - 15:19

score 553 · Answer 7 · 2021-02-18 05:41:21

553

Views

On $\frac{D(n^2)}{s(q)} \geq \frac{2n^2}{\sigma(q)} \geq \frac{\sigma(n^2)}{q} \geq \frac{2s(n^2)}{D(q)}$ where $q^k n^2$ is an odd perfect number

Published on 18 Feb 2021 - 5:41

score 34 · Answer 8 · 2021-03-01 21:45:44

34

Views

Lower bound of $(m,k)$-perfect numbers

Published on 01 Mar 2021 - 21:45

score 124 · Answer 9 · 2021-03-09 06:00:04

124

Views

On a conjectured upper bound for $k=\nu_q(N)$, if $N=q^k n^2$ is an odd perfect number with special prime $q$

Published on 09 Mar 2021 - 6:00

score 94 · Answer 10 · 2021-03-24 12:00:34

94

Views

How to check if number is harmonic divisor or not

Published on 24 Mar 2021 - 12:00

score 57 · Answer 11 · 2021-03-26 03:18:12

57

Views

If $k + 1$ is prime and $(k + 1) \mid (q - 1)$, then $\sigma(q^k)$ is divisible by $k + 1$, but not by $(k + 1)^2$ (unless $k+1=2$).

Published on 26 Mar 2021 - 3:18

score 71 · Answer 12 · 2021-04-08 08:22:42

71

Views

Is there an odd $x$ such that $2x^2 \equiv 0 \pmod {\sigma(x)}$ and $\sigma(x^2) \equiv 0 \pmod {\sigma(x) - 1}$?

Published on 08 Apr 2021 - 8:22

score 141 · Answer 13 · 2021-04-11 07:05:30

141

Views

On bounds for the quantity $n^2 / D(n^2)$ when $q^k n^2$ is an odd perfect number with special prime $q$ and $k > 1$

Published on 11 Apr 2021 - 7:05

score 67 · Answer 14 · 2021-04-11 13:29:35

67

Views

A proof (?) for $k = 1 \implies q \neq 5$, if $q^k n^2$ is an odd perfect number with special prime $q$

Published on 11 Apr 2021 - 13:29

score 124 · Answer 15 · 2021-04-16 05:15:36

124

Views

Applying a criterion on deficient numbers to the proper factors of an odd perfect number

Published on 16 Apr 2021 - 5:15

List Question

If $q^k n^2$ is an odd perfect number with special prime $q$, is $\sigma(q^k)$ coprime to $\sigma(n^2)$?

Odd perfect numbers and egyptian fraction conjecture

Revisiting questions 3888565 and 3894925

What numerical lower bound on the index of an odd perfect number is implied by the results in F.-J. Chen and Y.-G. Chen's 2014 paper?

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?

On the inequality $\frac{\sigma(n^2)}{q^k} < \frac{n^2 - q^k}{C}$ where $C>1$ and $q^k n^2$ is an odd perfect number - Part II

On $\frac{D(n^2)}{s(q)} \geq \frac{2n^2}{\sigma(q)} \geq \frac{\sigma(n^2)}{q} \geq \frac{2s(n^2)}{D(q)}$ where $q^k n^2$ is an odd perfect number

Lower bound of $(m,k)$-perfect numbers

On a conjectured upper bound for $k=\nu_q(N)$, if $N=q^k n^2$ is an odd perfect number with special prime $q$

How to check if number is harmonic divisor or not

If $k + 1$ is prime and $(k + 1) \mid (q - 1)$, then $\sigma(q^k)$ is divisible by $k + 1$, but not by $(k + 1)^2$ (unless $k+1=2$).

Is there an odd $x$ such that $2x^2 \equiv 0 \pmod {\sigma(x)}$ and $\sigma(x^2) \equiv 0 \pmod {\sigma(x) - 1}$?

On bounds for the quantity $n^2 / D(n^2)$ when $q^k n^2$ is an odd perfect number with special prime $q$ and $k > 1$

A proof (?) for $k = 1 \implies q \neq 5$, if $q^k n^2$ is an odd perfect number with special prime $q$

Applying a criterion on deficient numbers to the proper factors of an odd perfect number

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