What proportion of positive integers have two factors that differ by 1?











up vote
4
down vote

favorite












What proportion of positive integers have two factors that differ by 1?



This question occurred to me
while trying to figure out
why there are 7 days in a week.



I looked at 364,
the number of days closest to a year
(there are about 364.2422
days in a year, iirc).
Since
$364 = 2cdot 2 cdot 7 cdot 13$,
the number of possible
number that evenly divide a year
are
2, 4, 7, 13, 14, 26, 28,
and larger.



Given this,
7 looks reasonable -
2 and 4 are too short
and 13 is too long.



Anyway,
I noticed that
13 and 14 are there,
and wondered how often
this happens.



I wasn't able to figure out
a nice way to specify the
probability
(as in a Hardy-Littlewood
product),
and wasn't able to
do it from the inverse direction
(i.e., sort of a sieve
with n(n+1) going into
the array of integers).



Ideally, I would like
an asymptotic function
f(x) such that
$lim_{n to infty} dfrac{text{number of such integers } ge 2 le nx}{n}
=f(x)
$

or find $c$ such that
$lim_{n to infty} dfrac{text{number of such integers } ge 2 le n}{n}
=c
$
.



My guess is that,
in the latter case,
$c = 0$ or 1,
but I have no idea which is true.
Maybe its
$1-frac1{e}$.



Note: I have modified this
to not allow 1 as a divisor.










share|cite|improve this question




















  • 1




    There are $365.2425$ days per year on average when taking leap year into account.
    – JMoravitz
    4 hours ago















up vote
4
down vote

favorite












What proportion of positive integers have two factors that differ by 1?



This question occurred to me
while trying to figure out
why there are 7 days in a week.



I looked at 364,
the number of days closest to a year
(there are about 364.2422
days in a year, iirc).
Since
$364 = 2cdot 2 cdot 7 cdot 13$,
the number of possible
number that evenly divide a year
are
2, 4, 7, 13, 14, 26, 28,
and larger.



Given this,
7 looks reasonable -
2 and 4 are too short
and 13 is too long.



Anyway,
I noticed that
13 and 14 are there,
and wondered how often
this happens.



I wasn't able to figure out
a nice way to specify the
probability
(as in a Hardy-Littlewood
product),
and wasn't able to
do it from the inverse direction
(i.e., sort of a sieve
with n(n+1) going into
the array of integers).



Ideally, I would like
an asymptotic function
f(x) such that
$lim_{n to infty} dfrac{text{number of such integers } ge 2 le nx}{n}
=f(x)
$

or find $c$ such that
$lim_{n to infty} dfrac{text{number of such integers } ge 2 le n}{n}
=c
$
.



My guess is that,
in the latter case,
$c = 0$ or 1,
but I have no idea which is true.
Maybe its
$1-frac1{e}$.



Note: I have modified this
to not allow 1 as a divisor.










share|cite|improve this question




















  • 1




    There are $365.2425$ days per year on average when taking leap year into account.
    – JMoravitz
    4 hours ago













up vote
4
down vote

favorite









up vote
4
down vote

favorite











What proportion of positive integers have two factors that differ by 1?



This question occurred to me
while trying to figure out
why there are 7 days in a week.



I looked at 364,
the number of days closest to a year
(there are about 364.2422
days in a year, iirc).
Since
$364 = 2cdot 2 cdot 7 cdot 13$,
the number of possible
number that evenly divide a year
are
2, 4, 7, 13, 14, 26, 28,
and larger.



Given this,
7 looks reasonable -
2 and 4 are too short
and 13 is too long.



Anyway,
I noticed that
13 and 14 are there,
and wondered how often
this happens.



I wasn't able to figure out
a nice way to specify the
probability
(as in a Hardy-Littlewood
product),
and wasn't able to
do it from the inverse direction
(i.e., sort of a sieve
with n(n+1) going into
the array of integers).



Ideally, I would like
an asymptotic function
f(x) such that
$lim_{n to infty} dfrac{text{number of such integers } ge 2 le nx}{n}
=f(x)
$

or find $c$ such that
$lim_{n to infty} dfrac{text{number of such integers } ge 2 le n}{n}
=c
$
.



My guess is that,
in the latter case,
$c = 0$ or 1,
but I have no idea which is true.
Maybe its
$1-frac1{e}$.



Note: I have modified this
to not allow 1 as a divisor.










share|cite|improve this question















What proportion of positive integers have two factors that differ by 1?



This question occurred to me
while trying to figure out
why there are 7 days in a week.



I looked at 364,
the number of days closest to a year
(there are about 364.2422
days in a year, iirc).
Since
$364 = 2cdot 2 cdot 7 cdot 13$,
the number of possible
number that evenly divide a year
are
2, 4, 7, 13, 14, 26, 28,
and larger.



Given this,
7 looks reasonable -
2 and 4 are too short
and 13 is too long.



Anyway,
I noticed that
13 and 14 are there,
and wondered how often
this happens.



I wasn't able to figure out
a nice way to specify the
probability
(as in a Hardy-Littlewood
product),
and wasn't able to
do it from the inverse direction
(i.e., sort of a sieve
with n(n+1) going into
the array of integers).



Ideally, I would like
an asymptotic function
f(x) such that
$lim_{n to infty} dfrac{text{number of such integers } ge 2 le nx}{n}
=f(x)
$

or find $c$ such that
$lim_{n to infty} dfrac{text{number of such integers } ge 2 le n}{n}
=c
$
.



My guess is that,
in the latter case,
$c = 0$ or 1,
but I have no idea which is true.
Maybe its
$1-frac1{e}$.



Note: I have modified this
to not allow 1 as a divisor.







number-theory asymptotics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 44 mins ago

























asked 4 hours ago









marty cohen

71.8k546124




71.8k546124








  • 1




    There are $365.2425$ days per year on average when taking leap year into account.
    – JMoravitz
    4 hours ago














  • 1




    There are $365.2425$ days per year on average when taking leap year into account.
    – JMoravitz
    4 hours ago








1




1




There are $365.2425$ days per year on average when taking leap year into account.
– JMoravitz
4 hours ago




There are $365.2425$ days per year on average when taking leap year into account.
– JMoravitz
4 hours ago










1 Answer
1






active

oldest

votes

















up vote
7
down vote













Every even number has consecutive factors: $1$ and $2$.



No odd number has, because all its factors are odd.



The probability is $1/2$.






share|cite|improve this answer





















  • So. Freaking. Clever.
    – Lucas Henrique
    1 hour ago










  • Good point. I have become my evil twin by not allowing 1 as a divisor. To make up for this, I have upvoted you.
    – marty cohen
    42 mins ago











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038712%2fwhat-proportion-of-positive-integers-have-two-factors-that-differ-by-1%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
7
down vote













Every even number has consecutive factors: $1$ and $2$.



No odd number has, because all its factors are odd.



The probability is $1/2$.






share|cite|improve this answer





















  • So. Freaking. Clever.
    – Lucas Henrique
    1 hour ago










  • Good point. I have become my evil twin by not allowing 1 as a divisor. To make up for this, I have upvoted you.
    – marty cohen
    42 mins ago















up vote
7
down vote













Every even number has consecutive factors: $1$ and $2$.



No odd number has, because all its factors are odd.



The probability is $1/2$.






share|cite|improve this answer





















  • So. Freaking. Clever.
    – Lucas Henrique
    1 hour ago










  • Good point. I have become my evil twin by not allowing 1 as a divisor. To make up for this, I have upvoted you.
    – marty cohen
    42 mins ago













up vote
7
down vote










up vote
7
down vote









Every even number has consecutive factors: $1$ and $2$.



No odd number has, because all its factors are odd.



The probability is $1/2$.






share|cite|improve this answer












Every even number has consecutive factors: $1$ and $2$.



No odd number has, because all its factors are odd.



The probability is $1/2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









ajotatxe

52.5k23789




52.5k23789












  • So. Freaking. Clever.
    – Lucas Henrique
    1 hour ago










  • Good point. I have become my evil twin by not allowing 1 as a divisor. To make up for this, I have upvoted you.
    – marty cohen
    42 mins ago


















  • So. Freaking. Clever.
    – Lucas Henrique
    1 hour ago










  • Good point. I have become my evil twin by not allowing 1 as a divisor. To make up for this, I have upvoted you.
    – marty cohen
    42 mins ago
















So. Freaking. Clever.
– Lucas Henrique
1 hour ago




So. Freaking. Clever.
– Lucas Henrique
1 hour ago












Good point. I have become my evil twin by not allowing 1 as a divisor. To make up for this, I have upvoted you.
– marty cohen
42 mins ago




Good point. I have become my evil twin by not allowing 1 as a divisor. To make up for this, I have upvoted you.
– marty cohen
42 mins ago


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038712%2fwhat-proportion-of-positive-integers-have-two-factors-that-differ-by-1%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror

Mangá

Eduardo VII do Reino Unido