Median divisor of even perfect numbers
$begingroup$
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 this, or does it happen all the time? I only checked up to the 8th perfect number. Thank you and apologies for the possible silliness of the question.
perfect-numbers
edited 3 hours ago
Parcly Taxel
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asked 4 hours ago
SoulisSoulis
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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 this, or does it happen all the time? I only checked up to the 8th perfect number. Thank you and apologies for the possible silliness of the question.
perfect-numbers
edited 3 hours ago
Parcly Taxel
41.9k1372101
asked 4 hours ago
SoulisSoulis
161
New contributor
Soulis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
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 this, or does it happen all the time? I only checked up to the 8th perfect number. Thank you and apologies for the possible silliness of the question.
perfect-numbers
edited 3 hours ago
Parcly Taxel
41.9k1372101
asked 4 hours ago
SoulisSoulis
161
New contributor
Soulis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 this, or does it happen all the time? I only checked up to the 8th perfect number. Thank you and apologies for the possible silliness of the question.
perfect-numbers
perfect-numbers
edited 3 hours ago
Parcly Taxel
41.9k1372101
asked 4 hours ago
SoulisSoulis
161
New contributor
Soulis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Parcly Taxel
41.9k1372101
asked 4 hours ago
SoulisSoulis
161
New contributor
Soulis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 3 hours ago
Parcly Taxel
41.9k1372101
edited 3 hours ago
Parcly Taxel
41.9k1372101
edited 3 hours ago
Parcly Taxel
41.9k1372101
41.9k1372101
asked 4 hours ago
SoulisSoulis
161
New contributor
Soulis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
SoulisSoulis
161
asked 4 hours ago
SoulisSoulis
161
161
New contributor
Soulis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Soulis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
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$begingroup$
We know that all even perfect numbers are of the form $(2^n-1)2^{n-1}$ where $2^n-1$ is prime, so there are $2n$ divisors. Ignoring the divisor that is the perfect number itself (so that the remaining $2n-1$ divisors sum to the perfect number), it is easy to see that those divisors have a simple ordering too:
$$underbrace{1,2,dots,2^{n-1}}_{n},2^n-1,2(2^n-1),dots,2^{n-2}(2^n-1)$$
The median is the $n$th divisor in this ordering, which we see is $2^{n-1}$. Therefore your claim is correct.
answered 3 hours ago
Parcly TaxelParcly Taxel
41.9k1372101
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$begingroup$
Every even perfect number is of the form $n=2^{p-1}(2^p-1)$ with $2^p-1$ prime.
Therefore, the factors of $n$ are
$1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), 2^3(2^p-1), ...2^{p-2}(2^p-1),$
and the middle one is $2^{p-1}.$
answered 3 hours ago
J. W. TannerJ. W. Tanner
2,0691117
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2 Answers
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2 Answers
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$begingroup$
We know that all even perfect numbers are of the form $(2^n-1)2^{n-1}$ where $2^n-1$ is prime, so there are $2n$ divisors. Ignoring the divisor that is the perfect number itself (so that the remaining $2n-1$ divisors sum to the perfect number), it is easy to see that those divisors have a simple ordering too:
$$underbrace{1,2,dots,2^{n-1}}_{n},2^n-1,2(2^n-1),dots,2^{n-2}(2^n-1)$$
The median is the $n$th divisor in this ordering, which we see is $2^{n-1}$. Therefore your claim is correct.
answered 3 hours ago
Parcly TaxelParcly Taxel
41.9k1372101
$endgroup$
add a comment |
$begingroup$
We know that all even perfect numbers are of the form $(2^n-1)2^{n-1}$ where $2^n-1$ is prime, so there are $2n$ divisors. Ignoring the divisor that is the perfect number itself (so that the remaining $2n-1$ divisors sum to the perfect number), it is easy to see that those divisors have a simple ordering too:
$$underbrace{1,2,dots,2^{n-1}}_{n},2^n-1,2(2^n-1),dots,2^{n-2}(2^n-1)$$
The median is the $n$th divisor in this ordering, which we see is $2^{n-1}$. Therefore your claim is correct.
answered 3 hours ago
Parcly TaxelParcly Taxel
41.9k1372101
$endgroup$
add a comment |
$begingroup$
We know that all even perfect numbers are of the form $(2^n-1)2^{n-1}$ where $2^n-1$ is prime, so there are $2n$ divisors. Ignoring the divisor that is the perfect number itself (so that the remaining $2n-1$ divisors sum to the perfect number), it is easy to see that those divisors have a simple ordering too:
$$underbrace{1,2,dots,2^{n-1}}_{n},2^n-1,2(2^n-1),dots,2^{n-2}(2^n-1)$$
The median is the $n$th divisor in this ordering, which we see is $2^{n-1}$. Therefore your claim is correct.
answered 3 hours ago
Parcly TaxelParcly Taxel
41.9k1372101
$endgroup$
We know that all even perfect numbers are of the form $(2^n-1)2^{n-1}$ where $2^n-1$ is prime, so there are $2n$ divisors. Ignoring the divisor that is the perfect number itself (so that the remaining $2n-1$ divisors sum to the perfect number), it is easy to see that those divisors have a simple ordering too:
$$underbrace{1,2,dots,2^{n-1}}_{n},2^n-1,2(2^n-1),dots,2^{n-2}(2^n-1)$$
The median is the $n$th divisor in this ordering, which we see is $2^{n-1}$. Therefore your claim is correct.
answered 3 hours ago
Parcly TaxelParcly Taxel
41.9k1372101
answered 3 hours ago
Parcly TaxelParcly Taxel
41.9k1372101
answered 3 hours ago
Parcly TaxelParcly Taxel
41.9k1372101
answered 3 hours ago
Parcly TaxelParcly Taxel
41.9k1372101
41.9k1372101
add a comment |
add a comment |
$begingroup$
Every even perfect number is of the form $n=2^{p-1}(2^p-1)$ with $2^p-1$ prime.
Therefore, the factors of $n$ are
$1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), 2^3(2^p-1), ...2^{p-2}(2^p-1),$
and the middle one is $2^{p-1}.$
answered 3 hours ago
J. W. TannerJ. W. Tanner
2,0691117
$endgroup$
add a comment |
$begingroup$
Every even perfect number is of the form $n=2^{p-1}(2^p-1)$ with $2^p-1$ prime.
Therefore, the factors of $n$ are
$1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), 2^3(2^p-1), ...2^{p-2}(2^p-1),$
and the middle one is $2^{p-1}.$
answered 3 hours ago
J. W. TannerJ. W. Tanner
2,0691117
$endgroup$
add a comment |
$begingroup$
Every even perfect number is of the form $n=2^{p-1}(2^p-1)$ with $2^p-1$ prime.
Therefore, the factors of $n$ are
$1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), 2^3(2^p-1), ...2^{p-2}(2^p-1),$
and the middle one is $2^{p-1}.$
answered 3 hours ago
J. W. TannerJ. W. Tanner
2,0691117
$endgroup$
Every even perfect number is of the form $n=2^{p-1}(2^p-1)$ with $2^p-1$ prime.
Therefore, the factors of $n$ are
$1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), 2^3(2^p-1), ...2^{p-2}(2^p-1),$
and the middle one is $2^{p-1}.$
answered 3 hours ago
J. W. TannerJ. W. Tanner
2,0691117
answered 3 hours ago
J. W. TannerJ. W. Tanner
2,0691117
answered 3 hours ago
J. W. TannerJ. W. Tanner
2,0691117
answered 3 hours ago
J. W. TannerJ. W. Tanner
2,0691117
2,0691117
add a comment |
add a comment |
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