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.
number-theory asymptotics
asked 4 hours ago
marty cohen
71.8k546124
add a comment |
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.
number-theory asymptotics
asked 4 hours ago
marty cohen
71.8k546124
1
There are $365.2425$ days per year on average when taking leap year into account.
– JMoravitz
4 hours ago
add a comment |
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.
number-theory asymptotics
asked 4 hours ago
marty cohen
71.8k546124
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
number-theory asymptotics
asked 4 hours ago
marty cohen
71.8k546124
asked 4 hours ago
marty cohen
71.8k546124
edited 27 mins ago
asked 4 hours ago
marty cohen
71.8k546124
asked 4 hours ago
marty cohen
71.8k546124
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
up vote
5
down vote
Every even number has consecutive factors: $1$ and $2$.
No odd number has, because all its factors are odd.
The probability is $1/2$.
answered 3 hours ago
ajotatxe
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
26 mins ago
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%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
5
down vote
Every even number has consecutive factors: $1$ and $2$.
No odd number has, because all its factors are odd.
The probability is $1/2$.
answered 3 hours ago
ajotatxe
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
26 mins ago
add a comment |
up vote
5
down vote
Every even number has consecutive factors: $1$ and $2$.
No odd number has, because all its factors are odd.
The probability is $1/2$.
answered 3 hours ago
ajotatxe
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
26 mins ago
add a comment |
up vote
5
down vote
up vote
5
down vote
Every even number has consecutive factors: $1$ and $2$.
No odd number has, because all its factors are odd.
The probability is $1/2$.
answered 3 hours ago
ajotatxe
52.5k23789
Every even number has consecutive factors: $1$ and $2$.
No odd number has, because all its factors are odd.
The probability is $1/2$.
answered 3 hours ago
ajotatxe
52.5k23789
answered 3 hours ago
ajotatxe
52.5k23789
answered 3 hours ago
ajotatxe
52.5k23789
answered 3 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
26 mins ago
add a comment |
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
26 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
26 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
26 mins ago
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%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
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
1
There are $365.2425$ days per year on average when taking leap year into account.
– JMoravitz
4 hours ago