Fdtxjr

Would this string work as string?

Has the laser at Magurele, Romania reached a tenth of the Sun's power?

Output visual diagram of picture

Air travel with refrigerated insulin

Turning a hard to access nut?

How can I, as DM, avoid the Conga Line of Death occurring when implementing some form of flanking rule?

Checking @@ROWCOUNT failing

Why does the frost depth increase when the surface temperature warms up?

How to join two vertical cells in latex?

C++ lambda syntax

Friend wants my recommendation but I don't want to give it to him

Is this saw blade faulty?

New Order #2: Turn My Way

What is the purpose of using a decision tree?

What can I do if I am asked to learn different programming languages very frequently?

Why is participating in the European Parliamentary elections used as a threat?

Highest stage count that are used one right after the other?

Why does a 97 / 92 key piano exist by Bosendorfer?

Is there a distance limit for minecart tracks?

What properties make a magic weapon befit a Rogue more than a DEX-based Fighter?

Should a narrator ever describe things based on a character's view instead of facts?

Why didn't Voldemort know what Grindelwald looked like?

Reasons for having MCU pin-states default to pull-up/down out of reset

Does capillary rise violate hydrostatic paradox?

Prove that there exist infinitely many $n$ such that $d(n+1)>d(n)$, where $d(n)$ is number of positive factors

Existence of infinitely many integers $n$ such that $2^n$ ends with $n$Prove that there exist infinitely many squares $a$ such that $sqrtsqrta$ is a squareProve that there are infinitely many natural numbers that can't be written as $a^2+p$Does there exist a positive $k$ s.t. for all $rgeq k$, “$sigma_r(m)<sigma_r(n)$ for every $m<n$” for infinitely many odd positive integers $n$?Does there exist an $a$ such that $a^n+1$ is divisible by $n^3$ for infinitely many $n$?Show that there are infinitely many integers $n$ with a given number of divisorsProve there exists infinitely many real numbers $a$ such that $a(a - 3a)$ is an integerDoes there exist such a positive integer?Infinitely many positive integers of the form $1998k+1$ such that all digits in their decimal representation are equalinfinitely many integers such that …

0

$begingroup$

Prove that there exist infinitely many $n$ such that $d(n+1)>d(n)$, where $d(n)$ is number of positive factors of $n$.

I have think of using proof by contradiction, by setting a largest integer $k$ such that $d(k+1)>d(k)$. Does anyone has any idea or hints?

elementary-number-theory divisor-counting-function

share|cite|improve this question

edited Mar 13 at 19:05

6005

37k751127

asked Mar 13 at 16:59

fiveplustwois7fiveplustwois7

6

New contributor

fiveplustwois7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can you use the fact that there are infinitely many primes?
  $endgroup$
  – Rocket Man
  Mar 13 at 17:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your reply. I don't think so....
  $endgroup$
  – fiveplustwois7
  Mar 13 at 17:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, if $n$ is prime then $n+1$ is not prime. So $d(n) = 2$ and $d(n+1) > 2$. So as there are infinitely "special cases" where $d(n) = 2$ and $d(n+1) > 2$, there are infinitely "general cases" where $d(n+1) > d(n)$. So, yes, you can use the fact that there are infinitely many primes.
  $endgroup$
  – fleablood
  Mar 13 at 19:10
  

  
  
  
  

add a comment | 

0

$begingroup$

Prove that there exist infinitely many $n$ such that $d(n+1)>d(n)$, where $d(n)$ is number of positive factors of $n$.

I have think of using proof by contradiction, by setting a largest integer $k$ such that $d(k+1)>d(k)$. Does anyone has any idea or hints?

elementary-number-theory divisor-counting-function

share|cite|improve this question

edited Mar 13 at 19:05

6005

37k751127

asked Mar 13 at 16:59

fiveplustwois7fiveplustwois7

6

New contributor

fiveplustwois7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can you use the fact that there are infinitely many primes?
  $endgroup$
  – Rocket Man
  Mar 13 at 17:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your reply. I don't think so....
  $endgroup$
  – fiveplustwois7
  Mar 13 at 17:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, if $n$ is prime then $n+1$ is not prime. So $d(n) = 2$ and $d(n+1) > 2$. So as there are infinitely "special cases" where $d(n) = 2$ and $d(n+1) > 2$, there are infinitely "general cases" where $d(n+1) > d(n)$. So, yes, you can use the fact that there are infinitely many primes.
  $endgroup$
  – fleablood
  Mar 13 at 19:10
  

  
  
  
  

add a comment | 

$begingroup$

Prove that there exist infinitely many $n$ such that $d(n+1)>d(n)$, where $d(n)$ is number of positive factors of $n$.

I have think of using proof by contradiction, by setting a largest integer $k$ such that $d(k+1)>d(k)$. Does anyone has any idea or hints?

elementary-number-theory divisor-counting-function

share|cite|improve this question

edited Mar 13 at 19:05

6005

37k751127

asked Mar 13 at 16:59

fiveplustwois7fiveplustwois7

6

New contributor

fiveplustwois7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 13 at 19:05

6005

37k751127

asked Mar 13 at 16:59

fiveplustwois7fiveplustwois7

6

New contributor

fiveplustwois7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 13 at 19:05

6005

37k751127

asked Mar 13 at 16:59

fiveplustwois7fiveplustwois7

6

New contributor

fiveplustwois7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 13 at 19:05

6005

37k751127

edited Mar 13 at 19:05

6005

37k751127

asked Mar 13 at 16:59

fiveplustwois7fiveplustwois7

6

New contributor

fiveplustwois7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 13 at 16:59

fiveplustwois7fiveplustwois7

6

2 Answers
2

active

oldest

votes

1

$begingroup$

The number of divisors of $2^n$ is $n+1$, so the number of divisors can become arbitary large. If from some point on , $d(n+1)le d(n)$ would always hold, the number of divisors would be bounded for all positive integers. This is a contradiction.

share|cite|improve this answer

answered Mar 13 at 18:38

PeterPeter

48.8k1239136

$endgroup$

add a comment | 

1

$begingroup$

I have think of using proof by contradiction, by setting a largest integer $k$ such that $d(k+1)>d(k)$.

This approach is good, and will work. Let's continue it. So $k$ is the largest integer such that $d(k+1) > d(k)$. What does that mean about $d(k+1)$, $d(k+2)$, $d(k+3)$, and so on? Well, it means that

• $d(k+2)$ is NOT larger than $d(k+1)$, that is, $d(k+2) le d(k+1)$;




• $d(k+3)$ is NOT larger than $d(k+2)$, that is, $d(k+3) le d(k+2)$;

and so on. So we conclude that
$$
d(k+1) ge d(k+2) ge d(k+3) ge d(k+4) ge d(k+5) ge cdots
$$
and so on.

What this means is that $d(n)$ is "weakly decreasing" for $n > k$, or in simpler terms, $d(n)$ never gets larger than $d(k)$.

So to get a contradiction, we should come up with some $n$ such that $d(n)$ keeps getting larger and larger. Peter's answer gives one idea: $n$ is a power of two $(n = 2^m$). Another idea would be to consider
$$
n = p_1 p_2 p_3 ldots p_m,
$$
the product of the first $m$ prime numbers.

share|cite|improve this answer

answered Mar 13 at 19:03

60056005

37k751127

$endgroup$

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
);



);

fiveplustwois7 is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3146856%2fprove-that-there-exist-infinitely-many-n-such-that-dn1dn-where-dn%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

The number of divisors of $2^n$ is $n+1$, so the number of divisors can become arbitary large. If from some point on , $d(n+1)le d(n)$ would always hold, the number of divisors would be bounded for all positive integers. This is a contradiction.

share|cite|improve this answer

answered Mar 13 at 18:38

PeterPeter

48.8k1239136

$endgroup$

add a comment | 

1

$begingroup$

The number of divisors of $2^n$ is $n+1$, so the number of divisors can become arbitary large. If from some point on , $d(n+1)le d(n)$ would always hold, the number of divisors would be bounded for all positive integers. This is a contradiction.

share|cite|improve this answer

answered Mar 13 at 18:38

PeterPeter

48.8k1239136

$endgroup$

add a comment | 

$begingroup$

The number of divisors of $2^n$ is $n+1$, so the number of divisors can become arbitary large. If from some point on , $d(n+1)le d(n)$ would always hold, the number of divisors would be bounded for all positive integers. This is a contradiction.

share|cite|improve this answer

answered Mar 13 at 18:38

PeterPeter

48.8k1239136

$endgroup$

share|cite|improve this answer

answered Mar 13 at 18:38

PeterPeter

48.8k1239136

answered Mar 13 at 18:38

PeterPeter

48.8k1239136

answered Mar 13 at 18:38

PeterPeter

48.8k1239136

1

$begingroup$

I have think of using proof by contradiction, by setting a largest integer $k$ such that $d(k+1)>d(k)$.

This approach is good, and will work. Let's continue it. So $k$ is the largest integer such that $d(k+1) > d(k)$. What does that mean about $d(k+1)$, $d(k+2)$, $d(k+3)$, and so on? Well, it means that

• $d(k+2)$ is NOT larger than $d(k+1)$, that is, $d(k+2) le d(k+1)$;




• $d(k+3)$ is NOT larger than $d(k+2)$, that is, $d(k+3) le d(k+2)$;

and so on. So we conclude that
$$
d(k+1) ge d(k+2) ge d(k+3) ge d(k+4) ge d(k+5) ge cdots
$$
and so on.

What this means is that $d(n)$ is "weakly decreasing" for $n > k$, or in simpler terms, $d(n)$ never gets larger than $d(k)$.

So to get a contradiction, we should come up with some $n$ such that $d(n)$ keeps getting larger and larger. Peter's answer gives one idea: $n$ is a power of two $(n = 2^m$). Another idea would be to consider
$$
n = p_1 p_2 p_3 ldots p_m,
$$
the product of the first $m$ prime numbers.

share|cite|improve this answer

answered Mar 13 at 19:03

60056005

37k751127

$endgroup$

add a comment | 

1

$begingroup$

I have think of using proof by contradiction, by setting a largest integer $k$ such that $d(k+1)>d(k)$.

This approach is good, and will work. Let's continue it. So $k$ is the largest integer such that $d(k+1) > d(k)$. What does that mean about $d(k+1)$, $d(k+2)$, $d(k+3)$, and so on? Well, it means that

• $d(k+2)$ is NOT larger than $d(k+1)$, that is, $d(k+2) le d(k+1)$;




• $d(k+3)$ is NOT larger than $d(k+2)$, that is, $d(k+3) le d(k+2)$;

and so on. So we conclude that
$$
d(k+1) ge d(k+2) ge d(k+3) ge d(k+4) ge d(k+5) ge cdots
$$
and so on.

What this means is that $d(n)$ is "weakly decreasing" for $n > k$, or in simpler terms, $d(n)$ never gets larger than $d(k)$.

So to get a contradiction, we should come up with some $n$ such that $d(n)$ keeps getting larger and larger. Peter's answer gives one idea: $n$ is a power of two $(n = 2^m$). Another idea would be to consider
$$
n = p_1 p_2 p_3 ldots p_m,
$$
the product of the first $m$ prime numbers.

share|cite|improve this answer

answered Mar 13 at 19:03

60056005

37k751127

$endgroup$

add a comment | 

$begingroup$

I have think of using proof by contradiction, by setting a largest integer $k$ such that $d(k+1)>d(k)$.

This approach is good, and will work. Let's continue it. So $k$ is the largest integer such that $d(k+1) > d(k)$. What does that mean about $d(k+1)$, $d(k+2)$, $d(k+3)$, and so on? Well, it means that

• $d(k+2)$ is NOT larger than $d(k+1)$, that is, $d(k+2) le d(k+1)$;




• $d(k+3)$ is NOT larger than $d(k+2)$, that is, $d(k+3) le d(k+2)$;

and so on. So we conclude that
$$
d(k+1) ge d(k+2) ge d(k+3) ge d(k+4) ge d(k+5) ge cdots
$$
and so on.

What this means is that $d(n)$ is "weakly decreasing" for $n > k$, or in simpler terms, $d(n)$ never gets larger than $d(k)$.

So to get a contradiction, we should come up with some $n$ such that $d(n)$ keeps getting larger and larger. Peter's answer gives one idea: $n$ is a power of two $(n = 2^m$). Another idea would be to consider
$$
n = p_1 p_2 p_3 ldots p_m,
$$
the product of the first $m$ prime numbers.

share|cite|improve this answer

answered Mar 13 at 19:03

60056005

37k751127

$endgroup$

share|cite|improve this answer

answered Mar 13 at 19:03

60056005

37k751127

answered Mar 13 at 19:03

60056005

37k751127

answered Mar 13 at 19:03

60056005

37k751127