An anomaly regarding the sum of all divisors that is square-free
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Let $q(n)$ be the number of integers $m<n$ such that $m$ is square-free.
Let $p(n)$ be the number of integers $m<n$ such that the sum of the prime factors of $m$ is square-free. And let $s(n)$ be the number of integers $m<n$ such that the sum of the divisors of $m$ is square-free. The table shows values of $q$, $p$ and $s$ for some $n$:
n q p s
100 60 68 24
1000 607 660 157
10000 6082 6343 1090
100000 60793 62352 8097
1000000 607925 618969 64306
It's known that the distribution is asymptotically equivalent with a straight line and that $frac{6}{pi^2}cdot 100approx 60.79$ percent of all integers are square-free. So how to explain this anomaly?
Why is the distribution of square-free sums of divisors so low?
number-theory intuition divisor-sum
edited Nov 20 at 1:41
Erick Wong
20.1k22666
asked Nov 19 at 20:47
Lehs
6,87531662
add a comment |
up vote
0
down vote
favorite
Let $q(n)$ be the number of integers $m<n$ such that $m$ is square-free.
Let $p(n)$ be the number of integers $m<n$ such that the sum of the prime factors of $m$ is square-free. And let $s(n)$ be the number of integers $m<n$ such that the sum of the divisors of $m$ is square-free. The table shows values of $q$, $p$ and $s$ for some $n$:
n q p s
100 60 68 24
1000 607 660 157
10000 6082 6343 1090
100000 60793 62352 8097
1000000 607925 618969 64306
It's known that the distribution is asymptotically equivalent with a straight line and that $frac{6}{pi^2}cdot 100approx 60.79$ percent of all integers are square-free. So how to explain this anomaly?
Why is the distribution of square-free sums of divisors so low?
number-theory intuition divisor-sum
edited Nov 20 at 1:41
Erick Wong
20.1k22666
asked Nov 19 at 20:47
Lehs
6,87531662
1
I'm not sure I grasp what you feel is anomalous. Your table shows $q(n)$ approaching the prescribed ratio to $n$. Your final sentence suggests the anomaly is connected to sequence $s(n)$. But clearly the sums of divisors are not distributed like arbitrary integers in relation to the property of being square-free. Have I misunderstood your concern?
– hardmath
Nov 19 at 21:04
1
@hardmath: the word anomaly is maybe not the right word, but there should be an explanation why $sigma(x)$ tends to be a non square-free number.
– Lehs
Nov 19 at 21:14
Do you know how to apply the standard analytic-number-theory tools to your sequences ?
– reuns
Nov 19 at 21:22
@reuns: probably not.
– Lehs
Nov 19 at 21:29
2
The basic fact about the sum of divisors function is that it is multiplicative, i.e. $sigma(mn) = sigma(m) sigma(n)$ when $m,n$ are coprime. It follows that the prime power factors of $n$ determine the factors of $sigma(n)$, and thus a necessary condition for $sigma(n)$ to be square-free is that each $sigma(p^k)$ is square-free (with $p^k$ the generic prime power in a unique prime factorization of $n$).
– hardmath
Nov 20 at 0:45
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $q(n)$ be the number of integers $m<n$ such that $m$ is square-free.
Let $p(n)$ be the number of integers $m<n$ such that the sum of the prime factors of $m$ is square-free. And let $s(n)$ be the number of integers $m<n$ such that the sum of the divisors of $m$ is square-free. The table shows values of $q$, $p$ and $s$ for some $n$:
n q p s
100 60 68 24
1000 607 660 157
10000 6082 6343 1090
100000 60793 62352 8097
1000000 607925 618969 64306
It's known that the distribution is asymptotically equivalent with a straight line and that $frac{6}{pi^2}cdot 100approx 60.79$ percent of all integers are square-free. So how to explain this anomaly?
Why is the distribution of square-free sums of divisors so low?
number-theory intuition divisor-sum
edited Nov 20 at 1:41
Erick Wong
20.1k22666
asked Nov 19 at 20:47
Lehs
6,87531662
Let $q(n)$ be the number of integers $m<n$ such that $m$ is square-free.
Let $p(n)$ be the number of integers $m<n$ such that the sum of the prime factors of $m$ is square-free. And let $s(n)$ be the number of integers $m<n$ such that the sum of the divisors of $m$ is square-free. The table shows values of $q$, $p$ and $s$ for some $n$:
n q p s
100 60 68 24
1000 607 660 157
10000 6082 6343 1090
100000 60793 62352 8097
1000000 607925 618969 64306
It's known that the distribution is asymptotically equivalent with a straight line and that $frac{6}{pi^2}cdot 100approx 60.79$ percent of all integers are square-free. So how to explain this anomaly?
Why is the distribution of square-free sums of divisors so low?
number-theory intuition divisor-sum
number-theory intuition divisor-sum
edited Nov 20 at 1:41
Erick Wong
20.1k22666
asked Nov 19 at 20:47
Lehs
6,87531662
edited Nov 20 at 1:41
Erick Wong
20.1k22666
asked Nov 19 at 20:47
Lehs
6,87531662
edited Nov 20 at 1:41
Erick Wong
20.1k22666
edited Nov 20 at 1:41
Erick Wong
20.1k22666
edited Nov 20 at 1:41
Erick Wong
20.1k22666
20.1k22666
asked Nov 19 at 20:47
Lehs
6,87531662
asked Nov 19 at 20:47
Lehs
6,87531662
asked Nov 19 at 20:47
Lehs
6,87531662
6,87531662
1
I'm not sure I grasp what you feel is anomalous. Your table shows $q(n)$ approaching the prescribed ratio to $n$. Your final sentence suggests the anomaly is connected to sequence $s(n)$. But clearly the sums of divisors are not distributed like arbitrary integers in relation to the property of being square-free. Have I misunderstood your concern?
– hardmath
Nov 19 at 21:04
1
@hardmath: the word anomaly is maybe not the right word, but there should be an explanation why $sigma(x)$ tends to be a non square-free number.
– Lehs
Nov 19 at 21:14
Do you know how to apply the standard analytic-number-theory tools to your sequences ?
– reuns
Nov 19 at 21:22
@reuns: probably not.
– Lehs
Nov 19 at 21:29
2
The basic fact about the sum of divisors function is that it is multiplicative, i.e. $sigma(mn) = sigma(m) sigma(n)$ when $m,n$ are coprime. It follows that the prime power factors of $n$ determine the factors of $sigma(n)$, and thus a necessary condition for $sigma(n)$ to be square-free is that each $sigma(p^k)$ is square-free (with $p^k$ the generic prime power in a unique prime factorization of $n$).
– hardmath
Nov 20 at 0:45
add a comment |
1
I'm not sure I grasp what you feel is anomalous. Your table shows $q(n)$ approaching the prescribed ratio to $n$. Your final sentence suggests the anomaly is connected to sequence $s(n)$. But clearly the sums of divisors are not distributed like arbitrary integers in relation to the property of being square-free. Have I misunderstood your concern?
– hardmath
Nov 19 at 21:04
1
@hardmath: the word anomaly is maybe not the right word, but there should be an explanation why $sigma(x)$ tends to be a non square-free number.
– Lehs
Nov 19 at 21:14
Do you know how to apply the standard analytic-number-theory tools to your sequences ?
– reuns
Nov 19 at 21:22
@reuns: probably not.
– Lehs
Nov 19 at 21:29
2
The basic fact about the sum of divisors function is that it is multiplicative, i.e. $sigma(mn) = sigma(m) sigma(n)$ when $m,n$ are coprime. It follows that the prime power factors of $n$ determine the factors of $sigma(n)$, and thus a necessary condition for $sigma(n)$ to be square-free is that each $sigma(p^k)$ is square-free (with $p^k$ the generic prime power in a unique prime factorization of $n$).
– hardmath
Nov 20 at 0:45
1
1
I'm not sure I grasp what you feel is anomalous. Your table shows $q(n)$ approaching the prescribed ratio to $n$. Your final sentence suggests the anomaly is connected to sequence $s(n)$. But clearly the sums of divisors are not distributed like arbitrary integers in relation to the property of being square-free. Have I misunderstood your concern?
– hardmath
Nov 19 at 21:04
I'm not sure I grasp what you feel is anomalous. Your table shows $q(n)$ approaching the prescribed ratio to $n$. Your final sentence suggests the anomaly is connected to sequence $s(n)$. But clearly the sums of divisors are not distributed like arbitrary integers in relation to the property of being square-free. Have I misunderstood your concern?
– hardmath
Nov 19 at 21:04
1
1
@hardmath: the word anomaly is maybe not the right word, but there should be an explanation why $sigma(x)$ tends to be a non square-free number.
– Lehs
Nov 19 at 21:14
@hardmath: the word anomaly is maybe not the right word, but there should be an explanation why $sigma(x)$ tends to be a non square-free number.
– Lehs
Nov 19 at 21:14
Do you know how to apply the standard analytic-number-theory tools to your sequences ?
– reuns
Nov 19 at 21:22
Do you know how to apply the standard analytic-number-theory tools to your sequences ?
– reuns
Nov 19 at 21:22
@reuns: probably not.
– Lehs
Nov 19 at 21:29
@reuns: probably not.
– Lehs
Nov 19 at 21:29
2
2
The basic fact about the sum of divisors function is that it is multiplicative, i.e. $sigma(mn) = sigma(m) sigma(n)$ when $m,n$ are coprime. It follows that the prime power factors of $n$ determine the factors of $sigma(n)$, and thus a necessary condition for $sigma(n)$ to be square-free is that each $sigma(p^k)$ is square-free (with $p^k$ the generic prime power in a unique prime factorization of $n$).
– hardmath
Nov 20 at 0:45
The basic fact about the sum of divisors function is that it is multiplicative, i.e. $sigma(mn) = sigma(m) sigma(n)$ when $m,n$ are coprime. It follows that the prime power factors of $n$ determine the factors of $sigma(n)$, and thus a necessary condition for $sigma(n)$ to be square-free is that each $sigma(p^k)$ is square-free (with $p^k$ the generic prime power in a unique prime factorization of $n$).
– hardmath
Nov 20 at 0:45
add a comment |
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Expanding on hardmath's insightful comment, another necessary condition for $sigma(n)$ to be square-free is that the values of $sigma(p^k)$ are pairwise relatively prime. For instance, almost all of the prime power factors of $n$ would need to have even exponents in order for $sigma(n)$ to avoid being divisible by $4$. Consider also that the majority of $n$ have many prime divisors (about $log log n$ typically). So $n$ needs to be "almost" a perfect square (except for one allowable exceptional odd prime and the prime $2$).
This is really quite rare and it would be unreasonable to expect any positive proportion of $n$ to satisfy such constraints. It’s fairly easy to prove that the density of such $n$ is only $O(x/log x)$ up to $x$, similar to that of the primes (but with a larger constant in front).
answered Nov 20 at 1:39
Erick Wong
20.1k22666
add a comment |
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Expanding on hardmath's insightful comment, another necessary condition for $sigma(n)$ to be square-free is that the values of $sigma(p^k)$ are pairwise relatively prime. For instance, almost all of the prime power factors of $n$ would need to have even exponents in order for $sigma(n)$ to avoid being divisible by $4$. Consider also that the majority of $n$ have many prime divisors (about $log log n$ typically). So $n$ needs to be "almost" a perfect square (except for one allowable exceptional odd prime and the prime $2$).
This is really quite rare and it would be unreasonable to expect any positive proportion of $n$ to satisfy such constraints. It’s fairly easy to prove that the density of such $n$ is only $O(x/log x)$ up to $x$, similar to that of the primes (but with a larger constant in front).
answered Nov 20 at 1:39
Erick Wong
20.1k22666
add a comment |
up vote
1
down vote
Expanding on hardmath's insightful comment, another necessary condition for $sigma(n)$ to be square-free is that the values of $sigma(p^k)$ are pairwise relatively prime. For instance, almost all of the prime power factors of $n$ would need to have even exponents in order for $sigma(n)$ to avoid being divisible by $4$. Consider also that the majority of $n$ have many prime divisors (about $log log n$ typically). So $n$ needs to be "almost" a perfect square (except for one allowable exceptional odd prime and the prime $2$).
This is really quite rare and it would be unreasonable to expect any positive proportion of $n$ to satisfy such constraints. It’s fairly easy to prove that the density of such $n$ is only $O(x/log x)$ up to $x$, similar to that of the primes (but with a larger constant in front).
answered Nov 20 at 1:39
Erick Wong
20.1k22666
add a comment |
up vote
1
down vote
up vote
1
down vote
Expanding on hardmath's insightful comment, another necessary condition for $sigma(n)$ to be square-free is that the values of $sigma(p^k)$ are pairwise relatively prime. For instance, almost all of the prime power factors of $n$ would need to have even exponents in order for $sigma(n)$ to avoid being divisible by $4$. Consider also that the majority of $n$ have many prime divisors (about $log log n$ typically). So $n$ needs to be "almost" a perfect square (except for one allowable exceptional odd prime and the prime $2$).
This is really quite rare and it would be unreasonable to expect any positive proportion of $n$ to satisfy such constraints. It’s fairly easy to prove that the density of such $n$ is only $O(x/log x)$ up to $x$, similar to that of the primes (but with a larger constant in front).
answered Nov 20 at 1:39
Erick Wong
20.1k22666
Expanding on hardmath's insightful comment, another necessary condition for $sigma(n)$ to be square-free is that the values of $sigma(p^k)$ are pairwise relatively prime. For instance, almost all of the prime power factors of $n$ would need to have even exponents in order for $sigma(n)$ to avoid being divisible by $4$. Consider also that the majority of $n$ have many prime divisors (about $log log n$ typically). So $n$ needs to be "almost" a perfect square (except for one allowable exceptional odd prime and the prime $2$).
This is really quite rare and it would be unreasonable to expect any positive proportion of $n$ to satisfy such constraints. It’s fairly easy to prove that the density of such $n$ is only $O(x/log x)$ up to $x$, similar to that of the primes (but with a larger constant in front).
answered Nov 20 at 1:39
Erick Wong
20.1k22666
edited Nov 20 at 17:30
answered Nov 20 at 1:39
Erick Wong
20.1k22666
answered Nov 20 at 1:39
Erick Wong
20.1k22666
answered Nov 20 at 1:39
Erick Wong
20.1k22666
20.1k22666
add a comment |
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I'm not sure I grasp what you feel is anomalous. Your table shows $q(n)$ approaching the prescribed ratio to $n$. Your final sentence suggests the anomaly is connected to sequence $s(n)$. But clearly the sums of divisors are not distributed like arbitrary integers in relation to the property of being square-free. Have I misunderstood your concern?
– hardmath
Nov 19 at 21:04
1
@hardmath: the word anomaly is maybe not the right word, but there should be an explanation why $sigma(x)$ tends to be a non square-free number.
– Lehs
Nov 19 at 21:14
Do you know how to apply the standard analytic-number-theory tools to your sequences ?
– reuns
Nov 19 at 21:22
@reuns: probably not.
– Lehs
Nov 19 at 21:29
2
The basic fact about the sum of divisors function is that it is multiplicative, i.e. $sigma(mn) = sigma(m) sigma(n)$ when $m,n$ are coprime. It follows that the prime power factors of $n$ determine the factors of $sigma(n)$, and thus a necessary condition for $sigma(n)$ to be square-free is that each $sigma(p^k)$ is square-free (with $p^k$ the generic prime power in a unique prime factorization of $n$).
– hardmath
Nov 20 at 0:45