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











up vote
14
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







share|cite|improve this question




















  • 8




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






  • 1




    A list of such numbers can be found at oeis.org/A088723
    – Dan
    10 hours ago






  • 1




    I presume this doesn't constitute an answer for what you wanted, but I wrote a program to do a brute-force count and going up to $10^6$ I find 221944 numbers with this property; or up to $10^7$ I find 2219451. Perhaps that number will give someone an idea.
    – David Z
    10 hours ago










  • History of the seven-day week (probably because it's close to 1/4 a lunar month, rather than 1/52 an astronomical year): en.wikipedia.org/wiki/Week#History
    – aschepler
    3 hours ago










  • Slightly better approximation: 0.22194811. I approximately measured the density in the range [154e12, 154e12 + 4e9).
    – Veedrac
    26 secs ago

















up vote
14
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







share|cite|improve this question




















  • 8




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






  • 1




    A list of such numbers can be found at oeis.org/A088723
    – Dan
    10 hours ago






  • 1




    I presume this doesn't constitute an answer for what you wanted, but I wrote a program to do a brute-force count and going up to $10^6$ I find 221944 numbers with this property; or up to $10^7$ I find 2219451. Perhaps that number will give someone an idea.
    – David Z
    10 hours ago










  • History of the seven-day week (probably because it's close to 1/4 a lunar month, rather than 1/52 an astronomical year): en.wikipedia.org/wiki/Week#History
    – aschepler
    3 hours ago










  • Slightly better approximation: 0.22194811. I approximately measured the density in the range [154e12, 154e12 + 4e9).
    – Veedrac
    26 secs ago















up vote
14
down vote

favorite









up vote
14
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







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




number-theory asymptotics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 13 hours ago

























asked 16 hours ago









marty cohen

71.8k546125




71.8k546125








  • 8




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






  • 1




    A list of such numbers can be found at oeis.org/A088723
    – Dan
    10 hours ago






  • 1




    I presume this doesn't constitute an answer for what you wanted, but I wrote a program to do a brute-force count and going up to $10^6$ I find 221944 numbers with this property; or up to $10^7$ I find 2219451. Perhaps that number will give someone an idea.
    – David Z
    10 hours ago










  • History of the seven-day week (probably because it's close to 1/4 a lunar month, rather than 1/52 an astronomical year): en.wikipedia.org/wiki/Week#History
    – aschepler
    3 hours ago










  • Slightly better approximation: 0.22194811. I approximately measured the density in the range [154e12, 154e12 + 4e9).
    – Veedrac
    26 secs ago
















  • 8




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






  • 1




    A list of such numbers can be found at oeis.org/A088723
    – Dan
    10 hours ago






  • 1




    I presume this doesn't constitute an answer for what you wanted, but I wrote a program to do a brute-force count and going up to $10^6$ I find 221944 numbers with this property; or up to $10^7$ I find 2219451. Perhaps that number will give someone an idea.
    – David Z
    10 hours ago










  • History of the seven-day week (probably because it's close to 1/4 a lunar month, rather than 1/52 an astronomical year): en.wikipedia.org/wiki/Week#History
    – aschepler
    3 hours ago










  • Slightly better approximation: 0.22194811. I approximately measured the density in the range [154e12, 154e12 + 4e9).
    – Veedrac
    26 secs ago










8




8




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




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




1




1




A list of such numbers can be found at oeis.org/A088723
– Dan
10 hours ago




A list of such numbers can be found at oeis.org/A088723
– Dan
10 hours ago




1




1




I presume this doesn't constitute an answer for what you wanted, but I wrote a program to do a brute-force count and going up to $10^6$ I find 221944 numbers with this property; or up to $10^7$ I find 2219451. Perhaps that number will give someone an idea.
– David Z
10 hours ago




I presume this doesn't constitute an answer for what you wanted, but I wrote a program to do a brute-force count and going up to $10^6$ I find 221944 numbers with this property; or up to $10^7$ I find 2219451. Perhaps that number will give someone an idea.
– David Z
10 hours ago












History of the seven-day week (probably because it's close to 1/4 a lunar month, rather than 1/52 an astronomical year): en.wikipedia.org/wiki/Week#History
– aschepler
3 hours ago




History of the seven-day week (probably because it's close to 1/4 a lunar month, rather than 1/52 an astronomical year): en.wikipedia.org/wiki/Week#History
– aschepler
3 hours ago












Slightly better approximation: 0.22194811. I approximately measured the density in the range [154e12, 154e12 + 4e9).
– Veedrac
26 secs ago






Slightly better approximation: 0.22194811. I approximately measured the density in the range [154e12, 154e12 + 4e9).
– Veedrac
26 secs ago












2 Answers
2






active

oldest

votes

















up vote
34
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

















  • 2




    So. Freaking. Clever.
    – Lucas Henrique
    14 hours ago






  • 10




    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
    13 hours ago






  • 2




    This is a much more interesting problem if one specifies '1' may not be a factor.
    – user121330
    1 hour ago






  • 1




    OK, so now how many even numbers have N pairs? (evil grin)
    – Carl Witthoft
    35 mins ago


















up vote
6
down vote













What kind of numbers have this property?




  • All multiples of 6 (because 6 = 2 × 3). So that's 1/6 of the integers.

  • All multiples of 12 (12 = 3 × 4), but these have already been counted as multiples of 6.

  • All multiples of 20 (20 = 4 × 5), so add 1/20 of the integers. But we've double-counted multiples of 60 (LCD of 6 and 20), so subtract 1/60. This gives us 1/6 + 1/20 - 1/60 = 1/5.

  • All multiples of 30 (5 × 6) or 42 (6 × 7), but again, these have already been counted as multiples of 6.

  • All multiples of 56 (7 × 8), but don't double-count the ones that are also multiples of 6 or 20. If I did the arithmetic correctly, this brings us up to 22/105.

  • All multiples of 72 (8 × 9) or 90 (9 × 10), but these are already multiples of 6.

  • All multiples of 110 (10 × 11), being careful not to double-count multiples of 6, 20, or 56. We're now at 491/2310.


Continue the pattern to get a lower bound on the probability. I bet it converges to something, but I haven't bothered to compute what.






share|cite|improve this answer

















  • 1




    Don't forget that 1/2 gives you the other bound, for the same reason as the original answer
    – Eric
    7 hours ago






  • 1




    True: The product of two consecutive integers is always even, so no odd number can have the property we're looking for.
    – Dan
    2 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
34
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

















  • 2




    So. Freaking. Clever.
    – Lucas Henrique
    14 hours ago






  • 10




    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
    13 hours ago






  • 2




    This is a much more interesting problem if one specifies '1' may not be a factor.
    – user121330
    1 hour ago






  • 1




    OK, so now how many even numbers have N pairs? (evil grin)
    – Carl Witthoft
    35 mins ago















up vote
34
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

















  • 2




    So. Freaking. Clever.
    – Lucas Henrique
    14 hours ago






  • 10




    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
    13 hours ago






  • 2




    This is a much more interesting problem if one specifies '1' may not be a factor.
    – user121330
    1 hour ago






  • 1




    OK, so now how many even numbers have N pairs? (evil grin)
    – Carl Witthoft
    35 mins ago













up vote
34
down vote










up vote
34
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 16 hours ago









ajotatxe

52.7k23890




52.7k23890








  • 2




    So. Freaking. Clever.
    – Lucas Henrique
    14 hours ago






  • 10




    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
    13 hours ago






  • 2




    This is a much more interesting problem if one specifies '1' may not be a factor.
    – user121330
    1 hour ago






  • 1




    OK, so now how many even numbers have N pairs? (evil grin)
    – Carl Witthoft
    35 mins ago














  • 2




    So. Freaking. Clever.
    – Lucas Henrique
    14 hours ago






  • 10




    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
    13 hours ago






  • 2




    This is a much more interesting problem if one specifies '1' may not be a factor.
    – user121330
    1 hour ago






  • 1




    OK, so now how many even numbers have N pairs? (evil grin)
    – Carl Witthoft
    35 mins ago








2




2




So. Freaking. Clever.
– Lucas Henrique
14 hours ago




So. Freaking. Clever.
– Lucas Henrique
14 hours ago




10




10




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
13 hours 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
13 hours ago




2




2




This is a much more interesting problem if one specifies '1' may not be a factor.
– user121330
1 hour ago




This is a much more interesting problem if one specifies '1' may not be a factor.
– user121330
1 hour ago




1




1




OK, so now how many even numbers have N pairs? (evil grin)
– Carl Witthoft
35 mins ago




OK, so now how many even numbers have N pairs? (evil grin)
– Carl Witthoft
35 mins ago










up vote
6
down vote













What kind of numbers have this property?




  • All multiples of 6 (because 6 = 2 × 3). So that's 1/6 of the integers.

  • All multiples of 12 (12 = 3 × 4), but these have already been counted as multiples of 6.

  • All multiples of 20 (20 = 4 × 5), so add 1/20 of the integers. But we've double-counted multiples of 60 (LCD of 6 and 20), so subtract 1/60. This gives us 1/6 + 1/20 - 1/60 = 1/5.

  • All multiples of 30 (5 × 6) or 42 (6 × 7), but again, these have already been counted as multiples of 6.

  • All multiples of 56 (7 × 8), but don't double-count the ones that are also multiples of 6 or 20. If I did the arithmetic correctly, this brings us up to 22/105.

  • All multiples of 72 (8 × 9) or 90 (9 × 10), but these are already multiples of 6.

  • All multiples of 110 (10 × 11), being careful not to double-count multiples of 6, 20, or 56. We're now at 491/2310.


Continue the pattern to get a lower bound on the probability. I bet it converges to something, but I haven't bothered to compute what.






share|cite|improve this answer

















  • 1




    Don't forget that 1/2 gives you the other bound, for the same reason as the original answer
    – Eric
    7 hours ago






  • 1




    True: The product of two consecutive integers is always even, so no odd number can have the property we're looking for.
    – Dan
    2 hours ago















up vote
6
down vote













What kind of numbers have this property?




  • All multiples of 6 (because 6 = 2 × 3). So that's 1/6 of the integers.

  • All multiples of 12 (12 = 3 × 4), but these have already been counted as multiples of 6.

  • All multiples of 20 (20 = 4 × 5), so add 1/20 of the integers. But we've double-counted multiples of 60 (LCD of 6 and 20), so subtract 1/60. This gives us 1/6 + 1/20 - 1/60 = 1/5.

  • All multiples of 30 (5 × 6) or 42 (6 × 7), but again, these have already been counted as multiples of 6.

  • All multiples of 56 (7 × 8), but don't double-count the ones that are also multiples of 6 or 20. If I did the arithmetic correctly, this brings us up to 22/105.

  • All multiples of 72 (8 × 9) or 90 (9 × 10), but these are already multiples of 6.

  • All multiples of 110 (10 × 11), being careful not to double-count multiples of 6, 20, or 56. We're now at 491/2310.


Continue the pattern to get a lower bound on the probability. I bet it converges to something, but I haven't bothered to compute what.






share|cite|improve this answer

















  • 1




    Don't forget that 1/2 gives you the other bound, for the same reason as the original answer
    – Eric
    7 hours ago






  • 1




    True: The product of two consecutive integers is always even, so no odd number can have the property we're looking for.
    – Dan
    2 hours ago













up vote
6
down vote










up vote
6
down vote









What kind of numbers have this property?




  • All multiples of 6 (because 6 = 2 × 3). So that's 1/6 of the integers.

  • All multiples of 12 (12 = 3 × 4), but these have already been counted as multiples of 6.

  • All multiples of 20 (20 = 4 × 5), so add 1/20 of the integers. But we've double-counted multiples of 60 (LCD of 6 and 20), so subtract 1/60. This gives us 1/6 + 1/20 - 1/60 = 1/5.

  • All multiples of 30 (5 × 6) or 42 (6 × 7), but again, these have already been counted as multiples of 6.

  • All multiples of 56 (7 × 8), but don't double-count the ones that are also multiples of 6 or 20. If I did the arithmetic correctly, this brings us up to 22/105.

  • All multiples of 72 (8 × 9) or 90 (9 × 10), but these are already multiples of 6.

  • All multiples of 110 (10 × 11), being careful not to double-count multiples of 6, 20, or 56. We're now at 491/2310.


Continue the pattern to get a lower bound on the probability. I bet it converges to something, but I haven't bothered to compute what.






share|cite|improve this answer












What kind of numbers have this property?




  • All multiples of 6 (because 6 = 2 × 3). So that's 1/6 of the integers.

  • All multiples of 12 (12 = 3 × 4), but these have already been counted as multiples of 6.

  • All multiples of 20 (20 = 4 × 5), so add 1/20 of the integers. But we've double-counted multiples of 60 (LCD of 6 and 20), so subtract 1/60. This gives us 1/6 + 1/20 - 1/60 = 1/5.

  • All multiples of 30 (5 × 6) or 42 (6 × 7), but again, these have already been counted as multiples of 6.

  • All multiples of 56 (7 × 8), but don't double-count the ones that are also multiples of 6 or 20. If I did the arithmetic correctly, this brings us up to 22/105.

  • All multiples of 72 (8 × 9) or 90 (9 × 10), but these are already multiples of 6.

  • All multiples of 110 (10 × 11), being careful not to double-count multiples of 6, 20, or 56. We're now at 491/2310.


Continue the pattern to get a lower bound on the probability. I bet it converges to something, but I haven't bothered to compute what.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 10 hours ago









Dan

4,20511416




4,20511416








  • 1




    Don't forget that 1/2 gives you the other bound, for the same reason as the original answer
    – Eric
    7 hours ago






  • 1




    True: The product of two consecutive integers is always even, so no odd number can have the property we're looking for.
    – Dan
    2 hours ago














  • 1




    Don't forget that 1/2 gives you the other bound, for the same reason as the original answer
    – Eric
    7 hours ago






  • 1




    True: The product of two consecutive integers is always even, so no odd number can have the property we're looking for.
    – Dan
    2 hours ago








1




1




Don't forget that 1/2 gives you the other bound, for the same reason as the original answer
– Eric
7 hours ago




Don't forget that 1/2 gives you the other bound, for the same reason as the original answer
– Eric
7 hours ago




1




1




True: The product of two consecutive integers is always even, so no odd number can have the property we're looking for.
– Dan
2 hours ago




True: The product of two consecutive integers is always even, so no odd number can have the property we're looking for.
– Dan
2 hours ago


















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