error term of primes in arithmetic progression
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Given a positive integer $n>1$, do we have some bound on the sequence
$$a_n=sum_{chineq 1} sum_{p} frac{chi(p)}{p},$$
where $chi$ runs over the nontrivial characters modulo $n$ and $p$ runs over all primes.
This is a real number, since
$$sum_{chineq 1} sum_{p} frac{chi(p)}{p^s}=sum_{pequiv1(n)}frac{phi(n)}{p^s}-sum_{p}frac{1}{p^s}.$$
Here $phi$ is Euler's function. Since by Dirichlet's theorem, ratio of the two terms on the right is $1$ when $srightarrow 1$. So I say this is an error term.
number-theory prime-numbers analytic-number-theory
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up vote
1
down vote
favorite
Given a positive integer $n>1$, do we have some bound on the sequence
$$a_n=sum_{chineq 1} sum_{p} frac{chi(p)}{p},$$
where $chi$ runs over the nontrivial characters modulo $n$ and $p$ runs over all primes.
This is a real number, since
$$sum_{chineq 1} sum_{p} frac{chi(p)}{p^s}=sum_{pequiv1(n)}frac{phi(n)}{p^s}-sum_{p}frac{1}{p^s}.$$
Here $phi$ is Euler's function. Since by Dirichlet's theorem, ratio of the two terms on the right is $1$ when $srightarrow 1$. So I say this is an error term.
number-theory prime-numbers analytic-number-theory
$phi$ is euler's totient function, but was is $chi$?
– unseen_rider
Jan 11 at 19:37
@unseen_rider Yes, $phi$ is Euler's function and $chi$ is a character modulo $n$
– J.Li
Jan 12 at 3:05
en.wikipedia.org/wiki/Siegel_zero
– reuns
Nov 21 at 13:10
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a positive integer $n>1$, do we have some bound on the sequence
$$a_n=sum_{chineq 1} sum_{p} frac{chi(p)}{p},$$
where $chi$ runs over the nontrivial characters modulo $n$ and $p$ runs over all primes.
This is a real number, since
$$sum_{chineq 1} sum_{p} frac{chi(p)}{p^s}=sum_{pequiv1(n)}frac{phi(n)}{p^s}-sum_{p}frac{1}{p^s}.$$
Here $phi$ is Euler's function. Since by Dirichlet's theorem, ratio of the two terms on the right is $1$ when $srightarrow 1$. So I say this is an error term.
number-theory prime-numbers analytic-number-theory
Given a positive integer $n>1$, do we have some bound on the sequence
$$a_n=sum_{chineq 1} sum_{p} frac{chi(p)}{p},$$
where $chi$ runs over the nontrivial characters modulo $n$ and $p$ runs over all primes.
This is a real number, since
$$sum_{chineq 1} sum_{p} frac{chi(p)}{p^s}=sum_{pequiv1(n)}frac{phi(n)}{p^s}-sum_{p}frac{1}{p^s}.$$
Here $phi$ is Euler's function. Since by Dirichlet's theorem, ratio of the two terms on the right is $1$ when $srightarrow 1$. So I say this is an error term.
number-theory prime-numbers analytic-number-theory
number-theory prime-numbers analytic-number-theory
edited Nov 21 at 13:01
Klangen
1,36511131
1,36511131
asked Jan 11 at 12:06
J.Li
406
406
$phi$ is euler's totient function, but was is $chi$?
– unseen_rider
Jan 11 at 19:37
@unseen_rider Yes, $phi$ is Euler's function and $chi$ is a character modulo $n$
– J.Li
Jan 12 at 3:05
en.wikipedia.org/wiki/Siegel_zero
– reuns
Nov 21 at 13:10
add a comment |
$phi$ is euler's totient function, but was is $chi$?
– unseen_rider
Jan 11 at 19:37
@unseen_rider Yes, $phi$ is Euler's function and $chi$ is a character modulo $n$
– J.Li
Jan 12 at 3:05
en.wikipedia.org/wiki/Siegel_zero
– reuns
Nov 21 at 13:10
$phi$ is euler's totient function, but was is $chi$?
– unseen_rider
Jan 11 at 19:37
$phi$ is euler's totient function, but was is $chi$?
– unseen_rider
Jan 11 at 19:37
@unseen_rider Yes, $phi$ is Euler's function and $chi$ is a character modulo $n$
– J.Li
Jan 12 at 3:05
@unseen_rider Yes, $phi$ is Euler's function and $chi$ is a character modulo $n$
– J.Li
Jan 12 at 3:05
en.wikipedia.org/wiki/Siegel_zero
– reuns
Nov 21 at 13:10
en.wikipedia.org/wiki/Siegel_zero
– reuns
Nov 21 at 13:10
add a comment |
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$phi$ is euler's totient function, but was is $chi$?
– unseen_rider
Jan 11 at 19:37
@unseen_rider Yes, $phi$ is Euler's function and $chi$ is a character modulo $n$
– J.Li
Jan 12 at 3:05
en.wikipedia.org/wiki/Siegel_zero
– reuns
Nov 21 at 13:10