Mixing
Portion of prime congruent to $a$ mod $m$ in the factorial of $n$
$begingroup$
I'm learning analytic number theory and find this interesting question from one textbook:
It is well known in analytic number theory that:
begin{equation}
sum_{substack{pleq x \ pequiv a pmod{a}}} frac{log p}{p} = frac{1}{varphi(m)}log x +O(1) quad (1)
end{equation}
Let $a,minmathbb{N}$ with gcd$(a,m)=1$. Define $n^{(a)}$ as the portion of $n!$ composed of primes congruent to $a$ modulo $m$, that is:
begin{equation*}
n^{(a)}:=prod_{substack{p^{k}||n! \ p equiv a pmod{m}}} p^{k}.
end{equation*}
Using (1), prove that $log n^{(a)}sim frac{1}{varphi(m)}log n!$
Taking the logarithm on both of the sides gives us:
begin{equation*}
log n^{(a)}=sum_{substack{p^{k}||n!\ pequiv a pmod{m}}} klog p,
end{equation*}
and
begin{equation*}
sum_{substack{pleq n! \ pequiv a pmod{a}}} frac{log p}{p} = frac{1}{varphi(m)}log n! +O(1)
end{equation*}
It implies us to establish an equation between the two parts above. However, intuitively the two expressions includes variously different items since the primes $pleq n!$ and $p^{k}||n!$ are very different, hence how to build up this equation ?
prime-numbers analytic-number-theory
asked Jan 27 at 14:44
Xin HuXin Hu
547
$endgroup$
add a comment |
$begingroup$
I'm learning analytic number theory and find this interesting question from one textbook:
It is well known in analytic number theory that:
begin{equation}
sum_{substack{pleq x \ pequiv a pmod{a}}} frac{log p}{p} = frac{1}{varphi(m)}log x +O(1) quad (1)
end{equation}
Let $a,minmathbb{N}$ with gcd$(a,m)=1$. Define $n^{(a)}$ as the portion of $n!$ composed of primes congruent to $a$ modulo $m$, that is:
begin{equation*}
n^{(a)}:=prod_{substack{p^{k}||n! \ p equiv a pmod{m}}} p^{k}.
end{equation*}
Using (1), prove that $log n^{(a)}sim frac{1}{varphi(m)}log n!$
Taking the logarithm on both of the sides gives us:
begin{equation*}
log n^{(a)}=sum_{substack{p^{k}||n!\ pequiv a pmod{m}}} klog p,
end{equation*}
and
begin{equation*}
sum_{substack{pleq n! \ pequiv a pmod{a}}} frac{log p}{p} = frac{1}{varphi(m)}log n! +O(1)
end{equation*}
It implies us to establish an equation between the two parts above. However, intuitively the two expressions includes variously different items since the primes $pleq n!$ and $p^{k}||n!$ are very different, hence how to build up this equation ?
prime-numbers analytic-number-theory
asked Jan 27 at 14:44
Xin HuXin Hu
547
$endgroup$
add a comment |
$begingroup$
I'm learning analytic number theory and find this interesting question from one textbook:
It is well known in analytic number theory that:
begin{equation}
sum_{substack{pleq x \ pequiv a pmod{a}}} frac{log p}{p} = frac{1}{varphi(m)}log x +O(1) quad (1)
end{equation}
Let $a,minmathbb{N}$ with gcd$(a,m)=1$. Define $n^{(a)}$ as the portion of $n!$ composed of primes congruent to $a$ modulo $m$, that is:
begin{equation*}
n^{(a)}:=prod_{substack{p^{k}||n! \ p equiv a pmod{m}}} p^{k}.
end{equation*}
Using (1), prove that $log n^{(a)}sim frac{1}{varphi(m)}log n!$
Taking the logarithm on both of the sides gives us:
begin{equation*}
log n^{(a)}=sum_{substack{p^{k}||n!\ pequiv a pmod{m}}} klog p,
end{equation*}
and
begin{equation*}
sum_{substack{pleq n! \ pequiv a pmod{a}}} frac{log p}{p} = frac{1}{varphi(m)}log n! +O(1)
end{equation*}
It implies us to establish an equation between the two parts above. However, intuitively the two expressions includes variously different items since the primes $pleq n!$ and $p^{k}||n!$ are very different, hence how to build up this equation ?
prime-numbers analytic-number-theory
asked Jan 27 at 14:44
Xin HuXin Hu
547
$endgroup$
I'm learning analytic number theory and find this interesting question from one textbook:
It is well known in analytic number theory that:
begin{equation}
sum_{substack{pleq x \ pequiv a pmod{a}}} frac{log p}{p} = frac{1}{varphi(m)}log x +O(1) quad (1)
end{equation}
Let $a,minmathbb{N}$ with gcd$(a,m)=1$. Define $n^{(a)}$ as the portion of $n!$ composed of primes congruent to $a$ modulo $m$, that is:
begin{equation*}
n^{(a)}:=prod_{substack{p^{k}||n! \ p equiv a pmod{m}}} p^{k}.
end{equation*}
Using (1), prove that $log n^{(a)}sim frac{1}{varphi(m)}log n!$
Taking the logarithm on both of the sides gives us:
begin{equation*}
log n^{(a)}=sum_{substack{p^{k}||n!\ pequiv a pmod{m}}} klog p,
end{equation*}
and
begin{equation*}
sum_{substack{pleq n! \ pequiv a pmod{a}}} frac{log p}{p} = frac{1}{varphi(m)}log n! +O(1)
end{equation*}
It implies us to establish an equation between the two parts above. However, intuitively the two expressions includes variously different items since the primes $pleq n!$ and $p^{k}||n!$ are very different, hence how to build up this equation ?
prime-numbers analytic-number-theory
prime-numbers analytic-number-theory
asked Jan 27 at 14:44
Xin HuXin Hu
547
asked Jan 27 at 14:44
Xin HuXin Hu
547
asked Jan 27 at 14:44
Xin HuXin Hu
547
asked Jan 27 at 14:44
Xin HuXin Hu
547
asked Jan 27 at 14:44
Xin HuXin Hu
547
547
add a comment |
add a comment |
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