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
41.9k1372101
asked 3 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 3 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 3 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 3 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 3 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
edited 3 hours ago
Parcly Taxel
41.9k1372101
edited 3 hours ago
Parcly Taxel
41.9k1372101
41.9k1372101
asked 3 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 3 hours ago
SoulisSoulis
161
asked 3 hours ago
SoulisSoulis
161
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.
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.
Soulis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 |
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Soulis is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3117077%2fmedian-divisor-of-even-perfect-numbers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 |
Soulis is a new contributor. Be nice, and check out our Code of Conduct.
Soulis is a new contributor. Be nice, and check out our Code of Conduct.
Soulis is a new contributor. Be nice, and check out our Code of Conduct.
Soulis is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3117077%2fmedian-divisor-of-even-perfect-numbers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
