Number of primes between $n$ and $2n$
$begingroup$
What is a good lower bound on $pi(2n)-pi(n)$? Bertrand's postulate gives $1$. It is expected to be as I understand of form $frac{ccdot n}{log n}$ from Prime Number Theorem.
Does the ratio always hold for all large enough $n$ with some $c$ always between $0$ and $c_0$ for some absolute constant $c_0$?
How often does it fail as far as we know?
number-theory prime-numbers analytic-number-theory prime-gaps
asked Dec 30 '18 at 16:14
BroutBrout
2,6161431
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add a comment |
$begingroup$
What is a good lower bound on $pi(2n)-pi(n)$? Bertrand's postulate gives $1$. It is expected to be as I understand of form $frac{ccdot n}{log n}$ from Prime Number Theorem.
Does the ratio always hold for all large enough $n$ with some $c$ always between $0$ and $c_0$ for some absolute constant $c_0$?
How often does it fail as far as we know?
number-theory prime-numbers analytic-number-theory prime-gaps
asked Dec 30 '18 at 16:14
BroutBrout
2,6161431
$endgroup$
add a comment |
$begingroup$
What is a good lower bound on $pi(2n)-pi(n)$? Bertrand's postulate gives $1$. It is expected to be as I understand of form $frac{ccdot n}{log n}$ from Prime Number Theorem.
Does the ratio always hold for all large enough $n$ with some $c$ always between $0$ and $c_0$ for some absolute constant $c_0$?
How often does it fail as far as we know?
number-theory prime-numbers analytic-number-theory prime-gaps
asked Dec 30 '18 at 16:14
BroutBrout
2,6161431
$endgroup$
What is a good lower bound on $pi(2n)-pi(n)$? Bertrand's postulate gives $1$. It is expected to be as I understand of form $frac{ccdot n}{log n}$ from Prime Number Theorem.
Does the ratio always hold for all large enough $n$ with some $c$ always between $0$ and $c_0$ for some absolute constant $c_0$?
How often does it fail as far as we know?
number-theory prime-numbers analytic-number-theory prime-gaps
number-theory prime-numbers analytic-number-theory prime-gaps
asked Dec 30 '18 at 16:14
BroutBrout
2,6161431
asked Dec 30 '18 at 16:14
BroutBrout
2,6161431
asked Dec 30 '18 at 16:14
BroutBrout
2,6161431
asked Dec 30 '18 at 16:14
BroutBrout
2,6161431
asked Dec 30 '18 at 16:14
BroutBrout
2,6161431
2,6161431
add a comment |
add a comment |
1 Answer
1
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$begingroup$
This is partly answered here:
Primes between $n$ and $2n$
It follows from the prime-number theorem that
$$ lim_{n to infty} frac{pi(2n) - pi(n)}{n/log n} = 2 - 1 = 1,$$ so the number of primes between $n$ and $2n$, which is $pi(2n) - pi(n)$, is actually asymptotic to $frac{n}{log n}$ which gets arbitrarily large.
answered Dec 30 '18 at 16:16
Dietrich BurdeDietrich Burde
82.2k649107
$endgroup$
$begingroup$
However the best gap seems to be of form $n^{0.525}$ and with that I get only $O(n^{0.475})$. What do I miss?
$endgroup$
– Brout
Dec 30 '18 at 16:18
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
This is partly answered here:
Primes between $n$ and $2n$
It follows from the prime-number theorem that
$$ lim_{n to infty} frac{pi(2n) - pi(n)}{n/log n} = 2 - 1 = 1,$$ so the number of primes between $n$ and $2n$, which is $pi(2n) - pi(n)$, is actually asymptotic to $frac{n}{log n}$ which gets arbitrarily large.
answered Dec 30 '18 at 16:16
Dietrich BurdeDietrich Burde
82.2k649107
$endgroup$
$begingroup$
However the best gap seems to be of form $n^{0.525}$ and with that I get only $O(n^{0.475})$. What do I miss?
$endgroup$
– Brout
Dec 30 '18 at 16:18
add a comment |
$begingroup$
This is partly answered here:
Primes between $n$ and $2n$
It follows from the prime-number theorem that
$$ lim_{n to infty} frac{pi(2n) - pi(n)}{n/log n} = 2 - 1 = 1,$$ so the number of primes between $n$ and $2n$, which is $pi(2n) - pi(n)$, is actually asymptotic to $frac{n}{log n}$ which gets arbitrarily large.
answered Dec 30 '18 at 16:16
Dietrich BurdeDietrich Burde
82.2k649107
$endgroup$
$begingroup$
However the best gap seems to be of form $n^{0.525}$ and with that I get only $O(n^{0.475})$. What do I miss?
$endgroup$
– Brout
Dec 30 '18 at 16:18
add a comment |
$begingroup$
This is partly answered here:
Primes between $n$ and $2n$
It follows from the prime-number theorem that
$$ lim_{n to infty} frac{pi(2n) - pi(n)}{n/log n} = 2 - 1 = 1,$$ so the number of primes between $n$ and $2n$, which is $pi(2n) - pi(n)$, is actually asymptotic to $frac{n}{log n}$ which gets arbitrarily large.
answered Dec 30 '18 at 16:16
Dietrich BurdeDietrich Burde
82.2k649107
$endgroup$
This is partly answered here:
Primes between $n$ and $2n$
It follows from the prime-number theorem that
$$ lim_{n to infty} frac{pi(2n) - pi(n)}{n/log n} = 2 - 1 = 1,$$ so the number of primes between $n$ and $2n$, which is $pi(2n) - pi(n)$, is actually asymptotic to $frac{n}{log n}$ which gets arbitrarily large.
answered Dec 30 '18 at 16:16
Dietrich BurdeDietrich Burde
82.2k649107
answered Dec 30 '18 at 16:16
Dietrich BurdeDietrich Burde
82.2k649107
answered Dec 30 '18 at 16:16
Dietrich BurdeDietrich Burde
82.2k649107
answered Dec 30 '18 at 16:16
Dietrich BurdeDietrich Burde
82.2k649107
82.2k649107
$begingroup$
However the best gap seems to be of form $n^{0.525}$ and with that I get only $O(n^{0.475})$. What do I miss?
$endgroup$
– Brout
Dec 30 '18 at 16:18
add a comment |
$begingroup$
However the best gap seems to be of form $n^{0.525}$ and with that I get only $O(n^{0.475})$. What do I miss?
$endgroup$
– Brout
Dec 30 '18 at 16:18
$begingroup$
However the best gap seems to be of form $n^{0.525}$ and with that I get only $O(n^{0.475})$. What do I miss?
$endgroup$
– Brout
Dec 30 '18 at 16:18
$begingroup$
However the best gap seems to be of form $n^{0.525}$ and with that I get only $O(n^{0.475})$. What do I miss?
$endgroup$
– Brout
Dec 30 '18 at 16:18
add a comment |
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