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
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
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
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
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
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
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 $1/2$.
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
add a comment |
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1 Answer
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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 $1/2$.
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
add a comment |
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 $1/2$.
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
add a comment |
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 $1/2$.
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 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
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
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
add a comment |
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1
There are $365.2425$ days per year on average when taking leap year into account.
– JMoravitz
4 hours ago