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Numbers whose prime factors all have odd exponent
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Let $S$ denote the set of natural numbers $m$ with the property that for all prime powers $p^k || m$ we have $k equiv 1 pmod2$.
What is the asymptotic density of $S$?
Note that $S$ contains all prime numbers and more generally, all square-free numbers, so that $liminf_X rightarrow infty frac# (S cap [1,X])X geq frac6pi^2$.
nt.number-theory analytic-number-theory
asked 4 hours ago
Stanley Yao Xiao
7,69242682
add a comment |Â
up vote
2
down vote
favorite
Let $S$ denote the set of natural numbers $m$ with the property that for all prime powers $p^k || m$ we have $k equiv 1 pmod2$.
What is the asymptotic density of $S$?
Note that $S$ contains all prime numbers and more generally, all square-free numbers, so that $liminf_X rightarrow infty frac# (S cap [1,X])X geq frac6pi^2$.
nt.number-theory analytic-number-theory
asked 4 hours ago
Stanley Yao Xiao
7,69242682
You could try to compute the Dirichlet series $sum_n in S n^-s$.
– François Brunault
3 hours ago
Shouldn't it be something like zeta(3) times the square free density? Gerhard "Being Somewhat Thick About This" Paseman, 2018.10.04.
– Gerhard Paseman
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $S$ denote the set of natural numbers $m$ with the property that for all prime powers $p^k || m$ we have $k equiv 1 pmod2$.
What is the asymptotic density of $S$?
Note that $S$ contains all prime numbers and more generally, all square-free numbers, so that $liminf_X rightarrow infty frac# (S cap [1,X])X geq frac6pi^2$.
nt.number-theory analytic-number-theory
asked 4 hours ago
Stanley Yao Xiao
7,69242682
Let $S$ denote the set of natural numbers $m$ with the property that for all prime powers $p^k || m$ we have $k equiv 1 pmod2$.
What is the asymptotic density of $S$?
Note that $S$ contains all prime numbers and more generally, all square-free numbers, so that $liminf_X rightarrow infty frac# (S cap [1,X])X geq frac6pi^2$.
nt.number-theory analytic-number-theory
nt.number-theory analytic-number-theory
asked 4 hours ago
Stanley Yao Xiao
7,69242682
asked 4 hours ago
Stanley Yao Xiao
7,69242682
asked 4 hours ago
Stanley Yao Xiao
7,69242682
asked 4 hours ago
Stanley Yao Xiao
7,69242682
asked 4 hours ago
Stanley Yao Xiao
7,69242682
7,69242682
You could try to compute the Dirichlet series $sum_n in S n^-s$.
– François Brunault
3 hours ago
Shouldn't it be something like zeta(3) times the square free density? Gerhard "Being Somewhat Thick About This" Paseman, 2018.10.04.
– Gerhard Paseman
2 hours ago
add a comment |Â
You could try to compute the Dirichlet series $sum_n in S n^-s$.
– François Brunault
3 hours ago
Shouldn't it be something like zeta(3) times the square free density? Gerhard "Being Somewhat Thick About This" Paseman, 2018.10.04.
– Gerhard Paseman
2 hours ago
You could try to compute the Dirichlet series $sum_n in S n^-s$.
– François Brunault
3 hours ago
You could try to compute the Dirichlet series $sum_n in S n^-s$.
– François Brunault
3 hours ago
Shouldn't it be something like zeta(3) times the square free density? Gerhard "Being Somewhat Thick About This" Paseman, 2018.10.04.
– Gerhard Paseman
2 hours ago
Shouldn't it be something like zeta(3) times the square free density? Gerhard "Being Somewhat Thick About This" Paseman, 2018.10.04.
– Gerhard Paseman
2 hours ago
add a comment |Â
2 Answers
2
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The Dirichlet series $L (s)=sum_n in S n^-s $ has the Euler product expression
$$L (s) = prod_p frac1+p^-s-p^-2s1-p^-2s.$$
So $S $ has analytic density
$$mu (S) = operatornameRes_s=1 L(s) = prod_p frac1+1/p-1/p^21+1/p = prod_p frac1-2/p^2+1/p^31-1/p^2$$
as can also be seen by an informal argument. I don't expect this Euler product to have a nice expression since the roots of the Euler factor $1+X-X^2$ in the numerator are not roots of unity, so the product cannot be expressed, at least in an obvious way, in terms of values of the Riemann zeta function.
answered 1 hour ago
François Brunault
11.4k23264
2
In fact the last expression you write can be written as $prod_p left(1 - frac2p-1p^3right)left(1 - frac1p^2right)^-1 = zeta(2) C$, where $C = prod_p left(1 - frac2p-1p^3right)$ is known as the care-free constant. Very interesting!
– Stanley Yao Xiao
1 hour ago
@StanleyYaoXiao I didn't know this constant had a name. In fact, I don't know anything about these curious Euler products (are they periods? and so on). I also like the alternative expression $mu(S) = prod 1-1/(p (p+1)) $.
– François Brunault
20 mins ago
add a comment |Â
up vote
0
down vote
As I see it, the desired set is the disjoint union of (the rad(n) multiples of) squarefree numbers times n^2 as n ranges over all positive integers, which is like (zeta(3) + 1/64 + 3/512 + 2/729 + 1/1728 + 7/4096 + ...)/zeta(2) for the density.
Gerhard "Doing This On Minimal Caffeine" Paseman, 2018.10.04.
answered 1 hour ago
Gerhard Paseman
7,95111845
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
The Dirichlet series $L (s)=sum_n in S n^-s $ has the Euler product expression
$$L (s) = prod_p frac1+p^-s-p^-2s1-p^-2s.$$
So $S $ has analytic density
$$mu (S) = operatornameRes_s=1 L(s) = prod_p frac1+1/p-1/p^21+1/p = prod_p frac1-2/p^2+1/p^31-1/p^2$$
as can also be seen by an informal argument. I don't expect this Euler product to have a nice expression since the roots of the Euler factor $1+X-X^2$ in the numerator are not roots of unity, so the product cannot be expressed, at least in an obvious way, in terms of values of the Riemann zeta function.
answered 1 hour ago
François Brunault
11.4k23264
2
In fact the last expression you write can be written as $prod_p left(1 - frac2p-1p^3right)left(1 - frac1p^2right)^-1 = zeta(2) C$, where $C = prod_p left(1 - frac2p-1p^3right)$ is known as the care-free constant. Very interesting!
– Stanley Yao Xiao
1 hour ago
@StanleyYaoXiao I didn't know this constant had a name. In fact, I don't know anything about these curious Euler products (are they periods? and so on). I also like the alternative expression $mu(S) = prod 1-1/(p (p+1)) $.
– François Brunault
20 mins ago
add a comment |Â
up vote
5
down vote
accepted
The Dirichlet series $L (s)=sum_n in S n^-s $ has the Euler product expression
$$L (s) = prod_p frac1+p^-s-p^-2s1-p^-2s.$$
So $S $ has analytic density
$$mu (S) = operatornameRes_s=1 L(s) = prod_p frac1+1/p-1/p^21+1/p = prod_p frac1-2/p^2+1/p^31-1/p^2$$
as can also be seen by an informal argument. I don't expect this Euler product to have a nice expression since the roots of the Euler factor $1+X-X^2$ in the numerator are not roots of unity, so the product cannot be expressed, at least in an obvious way, in terms of values of the Riemann zeta function.
answered 1 hour ago
François Brunault
11.4k23264
2
In fact the last expression you write can be written as $prod_p left(1 - frac2p-1p^3right)left(1 - frac1p^2right)^-1 = zeta(2) C$, where $C = prod_p left(1 - frac2p-1p^3right)$ is known as the care-free constant. Very interesting!
– Stanley Yao Xiao
1 hour ago
@StanleyYaoXiao I didn't know this constant had a name. In fact, I don't know anything about these curious Euler products (are they periods? and so on). I also like the alternative expression $mu(S) = prod 1-1/(p (p+1)) $.
– François Brunault
20 mins ago
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The Dirichlet series $L (s)=sum_n in S n^-s $ has the Euler product expression
$$L (s) = prod_p frac1+p^-s-p^-2s1-p^-2s.$$
So $S $ has analytic density
$$mu (S) = operatornameRes_s=1 L(s) = prod_p frac1+1/p-1/p^21+1/p = prod_p frac1-2/p^2+1/p^31-1/p^2$$
as can also be seen by an informal argument. I don't expect this Euler product to have a nice expression since the roots of the Euler factor $1+X-X^2$ in the numerator are not roots of unity, so the product cannot be expressed, at least in an obvious way, in terms of values of the Riemann zeta function.
answered 1 hour ago
François Brunault
11.4k23264
The Dirichlet series $L (s)=sum_n in S n^-s $ has the Euler product expression
$$L (s) = prod_p frac1+p^-s-p^-2s1-p^-2s.$$
So $S $ has analytic density
$$mu (S) = operatornameRes_s=1 L(s) = prod_p frac1+1/p-1/p^21+1/p = prod_p frac1-2/p^2+1/p^31-1/p^2$$
as can also be seen by an informal argument. I don't expect this Euler product to have a nice expression since the roots of the Euler factor $1+X-X^2$ in the numerator are not roots of unity, so the product cannot be expressed, at least in an obvious way, in terms of values of the Riemann zeta function.
answered 1 hour ago
François Brunault
11.4k23264
answered 1 hour ago
François Brunault
11.4k23264
answered 1 hour ago
François Brunault
11.4k23264
answered 1 hour ago
François Brunault
11.4k23264
11.4k23264
2
In fact the last expression you write can be written as $prod_p left(1 - frac2p-1p^3right)left(1 - frac1p^2right)^-1 = zeta(2) C$, where $C = prod_p left(1 - frac2p-1p^3right)$ is known as the care-free constant. Very interesting!
– Stanley Yao Xiao
1 hour ago
@StanleyYaoXiao I didn't know this constant had a name. In fact, I don't know anything about these curious Euler products (are they periods? and so on). I also like the alternative expression $mu(S) = prod 1-1/(p (p+1)) $.
– François Brunault
20 mins ago
add a comment |Â
2
In fact the last expression you write can be written as $prod_p left(1 - frac2p-1p^3right)left(1 - frac1p^2right)^-1 = zeta(2) C$, where $C = prod_p left(1 - frac2p-1p^3right)$ is known as the care-free constant. Very interesting!
– Stanley Yao Xiao
1 hour ago
@StanleyYaoXiao I didn't know this constant had a name. In fact, I don't know anything about these curious Euler products (are they periods? and so on). I also like the alternative expression $mu(S) = prod 1-1/(p (p+1)) $.
– François Brunault
20 mins ago
2
2
In fact the last expression you write can be written as $prod_p left(1 - frac2p-1p^3right)left(1 - frac1p^2right)^-1 = zeta(2) C$, where $C = prod_p left(1 - frac2p-1p^3right)$ is known as the care-free constant. Very interesting!
– Stanley Yao Xiao
1 hour ago
In fact the last expression you write can be written as $prod_p left(1 - frac2p-1p^3right)left(1 - frac1p^2right)^-1 = zeta(2) C$, where $C = prod_p left(1 - frac2p-1p^3right)$ is known as the care-free constant. Very interesting!
– Stanley Yao Xiao
1 hour ago
@StanleyYaoXiao I didn't know this constant had a name. In fact, I don't know anything about these curious Euler products (are they periods? and so on). I also like the alternative expression $mu(S) = prod 1-1/(p (p+1)) $.
– François Brunault
20 mins ago
@StanleyYaoXiao I didn't know this constant had a name. In fact, I don't know anything about these curious Euler products (are they periods? and so on). I also like the alternative expression $mu(S) = prod 1-1/(p (p+1)) $.
– François Brunault
20 mins ago
add a comment |Â
up vote
0
down vote
As I see it, the desired set is the disjoint union of (the rad(n) multiples of) squarefree numbers times n^2 as n ranges over all positive integers, which is like (zeta(3) + 1/64 + 3/512 + 2/729 + 1/1728 + 7/4096 + ...)/zeta(2) for the density.
Gerhard "Doing This On Minimal Caffeine" Paseman, 2018.10.04.
answered 1 hour ago
Gerhard Paseman
7,95111845
add a comment |Â
up vote
0
down vote
As I see it, the desired set is the disjoint union of (the rad(n) multiples of) squarefree numbers times n^2 as n ranges over all positive integers, which is like (zeta(3) + 1/64 + 3/512 + 2/729 + 1/1728 + 7/4096 + ...)/zeta(2) for the density.
Gerhard "Doing This On Minimal Caffeine" Paseman, 2018.10.04.
answered 1 hour ago
Gerhard Paseman
7,95111845
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As I see it, the desired set is the disjoint union of (the rad(n) multiples of) squarefree numbers times n^2 as n ranges over all positive integers, which is like (zeta(3) + 1/64 + 3/512 + 2/729 + 1/1728 + 7/4096 + ...)/zeta(2) for the density.
Gerhard "Doing This On Minimal Caffeine" Paseman, 2018.10.04.
answered 1 hour ago
Gerhard Paseman
7,95111845
As I see it, the desired set is the disjoint union of (the rad(n) multiples of) squarefree numbers times n^2 as n ranges over all positive integers, which is like (zeta(3) + 1/64 + 3/512 + 2/729 + 1/1728 + 7/4096 + ...)/zeta(2) for the density.
Gerhard "Doing This On Minimal Caffeine" Paseman, 2018.10.04.
answered 1 hour ago
Gerhard Paseman
7,95111845
edited 1 hour ago
answered 1 hour ago
Gerhard Paseman
7,95111845
answered 1 hour ago
Gerhard Paseman
7,95111845
answered 1 hour ago
Gerhard Paseman
7,95111845
7,95111845
add a comment |Â
add a comment |Â
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You could try to compute the Dirichlet series $sum_n in S n^-s$.
– François Brunault
3 hours ago
Shouldn't it be something like zeta(3) times the square free density? Gerhard "Being Somewhat Thick About This" Paseman, 2018.10.04.
– Gerhard Paseman
2 hours ago