Question

Show that if an odd perfect number exists, it must be divisible by at least 3 different primes

score 136 · Answer 1 · 2013-10-25 04:45:17

136

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Show that if an odd perfect number exists, it must be divisible by at least 3 different primes

Published on 25 Oct 2013 - 4:45

score 435 · Answer 2 · 2013-11-02 00:41:25

435

Views

On odd perfect numbers $N$ given in the Eulerian form $N = {q^k}{n^2}$

Published on 02 Nov 2013 - 0:41

score 307 · Answer 3 · 2013-11-25 08:59:48

307

Views

Proof that an odd perfect number must be in the form $ 36k+9$ or $12k+1$

Published on 25 Nov 2013 - 8:59

score 3.8k · Answer 4 · 2013-12-05 16:54:13

3.8k

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Sum of the reciprocals of divisors of a perfect number is $2$?

Published on 05 Dec 2013 - 16:54

score 76 · Answer 5 · 2013-12-06 17:52:43

76

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Let $a_n=2^{2n}\left(2^{2n+1}-1 \right)$. Show induction of $n$

Published on 06 Dec 2013 - 17:52

score 139 · Answer 6 · 2013-12-18 21:52:26

139

Views

Are there any Perfectly Abundant Numbers?

Published on 18 Dec 2013 - 21:52

score 856 · Answer 7 · 2013-12-19 20:00:49

856

Views

Euler's phi function and perfect numbers

Published on 19 Dec 2013 - 20:00

score 125 · Answer 8 · 2013-12-20 16:55:13

125

Views

What is a cofactor in this case?

Published on 20 Dec 2013 - 16:55

score 233 · Answer 9 · 2013-12-30 22:29:16

233

Views

Generalized perfect numbers

Published on 30 Dec 2013 - 22:29

score 155 · Answer 10 · 2014-02-17 14:49:16

155

Views

Combinatorial interpretation of Euclid's form for even perfect numbers

Published on 17 Feb 2014 - 14:49

score 87 · Answer 11 · 2014-02-26 18:09:16

87

Views

Divisibility of even perfect numbers

Published on 26 Feb 2014 - 18:09

score 10.3k · Answer 12 · 2014-03-05 23:53:01

10.3k

Views

Applications of Perfect Numbers

Published on 05 Mar 2014 - 23:53

score 105 · Answer 13 · 2017-04-26 13:08:41

105

Views

If $N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}}$ is an odd perfect number, then ${p_i}^{\alpha_i} < \sqrt{N}$.

Published on 26 Apr 2017 - 13:08

score 73 · Answer 14 · 2017-05-26 20:41:35

73

Views

Closed form of sum involving binomial coefficients having to do with odd perfect numbers

Published on 26 May 2017 - 20:41

score 85 · Answer 15 · 2017-06-02 08:18:18

85

Views

If $N=q^k n^2$ is an odd perfect number, does $q \leq 97$ imply that $I(q^k) + I(n^2) \leq 2.99$?

Published on 02 Jun 2017 - 8:18

List Question

Show that if an odd perfect number exists, it must be divisible by at least 3 different primes

On odd perfect numbers $N$ given in the Eulerian form $N = {q^k}{n^2}$

Proof that an odd perfect number must be in the form $ 36k+9$ or $12k+1$

Sum of the reciprocals of divisors of a perfect number is $2$?

Let $a_n=2^{2n}\left(2^{2n+1}-1 \right)$. Show induction of $n$

Are there any Perfectly Abundant Numbers?

Euler's phi function and perfect numbers

What is a cofactor in this case?

Generalized perfect numbers

Combinatorial interpretation of Euclid's form for even perfect numbers

Divisibility of even perfect numbers

Applications of Perfect Numbers

If $N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}}$ is an odd perfect number, then ${p_i}^{\alpha_i} < \sqrt{N}$.

Closed form of sum involving binomial coefficients having to do with odd perfect numbers

If $N=q^k n^2$ is an odd perfect number, does $q \leq 97$ imply that $I(q^k) + I(n^2) \leq 2.99$?

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