Permutations, products, and unit fractions
up vote
1
down vote
favorite
Here's a question motivated by some related MathOverflow questions of Zhi-Wei Sun.
Show that, for any $n ge 1$, there is a permutation of ${1,2,ldots, n}$, i.e., some $pi in S_n$, such that $$sum_{k=1}^n frac{1}{k cdot pi(k)} = 1.$$
combinatorics number-theory permutations egyptian-fractions
asked Nov 25 at 21:57
Brian Hopkins
508615
add a comment |
up vote
1
down vote
favorite
Here's a question motivated by some related MathOverflow questions of Zhi-Wei Sun.
Show that, for any $n ge 1$, there is a permutation of ${1,2,ldots, n}$, i.e., some $pi in S_n$, such that $$sum_{k=1}^n frac{1}{k cdot pi(k)} = 1.$$
combinatorics number-theory permutations egyptian-fractions
asked Nov 25 at 21:57
Brian Hopkins
508615
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here's a question motivated by some related MathOverflow questions of Zhi-Wei Sun.
Show that, for any $n ge 1$, there is a permutation of ${1,2,ldots, n}$, i.e., some $pi in S_n$, such that $$sum_{k=1}^n frac{1}{k cdot pi(k)} = 1.$$
combinatorics number-theory permutations egyptian-fractions
asked Nov 25 at 21:57
Brian Hopkins
508615
Here's a question motivated by some related MathOverflow questions of Zhi-Wei Sun.
Show that, for any $n ge 1$, there is a permutation of ${1,2,ldots, n}$, i.e., some $pi in S_n$, such that $$sum_{k=1}^n frac{1}{k cdot pi(k)} = 1.$$
combinatorics number-theory permutations egyptian-fractions
combinatorics number-theory permutations egyptian-fractions
asked Nov 25 at 21:57
Brian Hopkins
508615
asked Nov 25 at 21:57
Brian Hopkins
508615
asked Nov 25 at 21:57
Brian Hopkins
508615
asked Nov 25 at 21:57
Brian Hopkins
508615
asked Nov 25 at 21:57
Brian Hopkins
508615
508615
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Take $pi(k)=k+1$ except for $pi(n)=1$. Then
$$sum_{k=1}^nfrac1{kpi(k)}=frac1n+sum_{k=1}^{n-1}frac1{k(k+1)}
=frac1n+sum_{k=1}^{n-1}left(frac1k-frac1{k+1}right)=1.
$$
answered Nov 25 at 22:09
Lord Shark the Unknown
99k958131
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
– Brian Hopkins
Nov 26 at 15:13
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',
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
});
}
});
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%2f3013470%2fpermutations-products-and-unit-fractions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Take $pi(k)=k+1$ except for $pi(n)=1$. Then
$$sum_{k=1}^nfrac1{kpi(k)}=frac1n+sum_{k=1}^{n-1}frac1{k(k+1)}
=frac1n+sum_{k=1}^{n-1}left(frac1k-frac1{k+1}right)=1.
$$
answered Nov 25 at 22:09
Lord Shark the Unknown
99k958131
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
– Brian Hopkins
Nov 26 at 15:13
add a comment |
up vote
2
down vote
accepted
Take $pi(k)=k+1$ except for $pi(n)=1$. Then
$$sum_{k=1}^nfrac1{kpi(k)}=frac1n+sum_{k=1}^{n-1}frac1{k(k+1)}
=frac1n+sum_{k=1}^{n-1}left(frac1k-frac1{k+1}right)=1.
$$
answered Nov 25 at 22:09
Lord Shark the Unknown
99k958131
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
– Brian Hopkins
Nov 26 at 15:13
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Take $pi(k)=k+1$ except for $pi(n)=1$. Then
$$sum_{k=1}^nfrac1{kpi(k)}=frac1n+sum_{k=1}^{n-1}frac1{k(k+1)}
=frac1n+sum_{k=1}^{n-1}left(frac1k-frac1{k+1}right)=1.
$$
answered Nov 25 at 22:09
Lord Shark the Unknown
99k958131
Take $pi(k)=k+1$ except for $pi(n)=1$. Then
$$sum_{k=1}^nfrac1{kpi(k)}=frac1n+sum_{k=1}^{n-1}frac1{k(k+1)}
=frac1n+sum_{k=1}^{n-1}left(frac1k-frac1{k+1}right)=1.
$$
answered Nov 25 at 22:09
Lord Shark the Unknown
99k958131
answered Nov 25 at 22:09
Lord Shark the Unknown
99k958131
answered Nov 25 at 22:09
Lord Shark the Unknown
99k958131
answered Nov 25 at 22:09
Lord Shark the Unknown
99k958131
99k958131
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
– Brian Hopkins
Nov 26 at 15:13
add a comment |
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
– Brian Hopkins
Nov 26 at 15:13
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
– Brian Hopkins
Nov 26 at 15:13
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
– Brian Hopkins
Nov 26 at 15:13
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3013470%2fpermutations-products-and-unit-fractions%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
