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






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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











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active

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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


















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