Estimate related to the Möbius functionAsymptotic density of k-almost primesWalsh Fourier Transform of the Möbius functionAsymptotics for the number of ways to sum primes such that the sum is <= nHow does Yitang Zhang use Cauchy's inequality and Theorem 2 to obtain the error term coming from the $S_2$ sumReferences to proofs of upper and lower bounds on the number of coprimes in an interval?Small quotients of smooth numbersAsymptotic estimate for a random model of primesChecking Mertens and the like in less than linear time or less than $sqrtx$ spaceAsymptotic formula for the average number of zeros of a polynomial modulo pIdelic/Hom representation of locally free class groupsQuestion on sums of multiplicative functions twisted by the Mobius function
Estimate related to the Möbius function
Asymptotic density of k-almost primesWalsh Fourier Transform of the Möbius functionAsymptotics for the number of ways to sum primes such that the sum is <= nHow does Yitang Zhang use Cauchy's inequality and Theorem 2 to obtain the error term coming from the $S_2$ sumReferences to proofs of upper and lower bounds on the number of coprimes in an interval?Small quotients of smooth numbersAsymptotic estimate for a random model of primesChecking Mertens and the like in less than linear time or less than $sqrtx$ spaceAsymptotic formula for the average number of zeros of a polynomial modulo pIdelic/Hom representation of locally free class groupsQuestion on sums of multiplicative functions twisted by the Mobius function
$begingroup$
I need to know, or at least have a good bound for, the asymptotic behaviour on $x$ of amount of integers less o equal than $x$ that are square free and with exactly $k$ primes on its decomposition. That is the cardinal of the following set
$$
mathcalJ_T(x,k) = n in mathbbN : n le x, Omega(n)=k , n mbox is square free .$$
Other way to describe this cardinal, using the Möbius function, is
$$ |mathcalJ_T(x,k) | = sum_Omega(n) = k, n le x |mu(n)|.$$
I am looking for the asymptotic behaviour on $x$, but this will depend also on $k$ in some way. The bound given using
$$
sum_Omega(n) = k, n le x |mu(n)| le sum_n le x |mu(n)| le frac6xpi^2 + O(sqrtx) ll x,
$$
is not good enought for my purposes.
Thanks in advanced, any reference or idea is helpful
nt.number-theory analytic-number-theory prime-numbers asymptotics
New contributor
$endgroup$
add a comment |
$begingroup$
I need to know, or at least have a good bound for, the asymptotic behaviour on $x$ of amount of integers less o equal than $x$ that are square free and with exactly $k$ primes on its decomposition. That is the cardinal of the following set
$$
mathcalJ_T(x,k) = n in mathbbN : n le x, Omega(n)=k , n mbox is square free .$$
Other way to describe this cardinal, using the Möbius function, is
$$ |mathcalJ_T(x,k) | = sum_Omega(n) = k, n le x |mu(n)|.$$
I am looking for the asymptotic behaviour on $x$, but this will depend also on $k$ in some way. The bound given using
$$
sum_Omega(n) = k, n le x |mu(n)| le sum_n le x |mu(n)| le frac6xpi^2 + O(sqrtx) ll x,
$$
is not good enought for my purposes.
Thanks in advanced, any reference or idea is helpful
nt.number-theory analytic-number-theory prime-numbers asymptotics
New contributor
$endgroup$
$begingroup$
Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using sum_d^2 mu(d)) the number of such squarefree integers leq x with omega(n) = k as an alternating sum of, for each dleq x^1/2, the number of integers leq x/d^2 with exactly k - omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below:
$endgroup$
– alpoge
3 hours ago
$begingroup$
mathoverflow.net/questions/35927/…
$endgroup$
– alpoge
3 hours ago
$begingroup$
Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers leq x with two prime factors seems to grow like x loglogx / logx. Who among us, even with our computers, has seen loglogx grow?! (Canonical quote about loglogx going to infinity with dignity.) What intuition...
$endgroup$
– alpoge
3 hours ago
add a comment |
$begingroup$
I need to know, or at least have a good bound for, the asymptotic behaviour on $x$ of amount of integers less o equal than $x$ that are square free and with exactly $k$ primes on its decomposition. That is the cardinal of the following set
$$
mathcalJ_T(x,k) = n in mathbbN : n le x, Omega(n)=k , n mbox is square free .$$
Other way to describe this cardinal, using the Möbius function, is
$$ |mathcalJ_T(x,k) | = sum_Omega(n) = k, n le x |mu(n)|.$$
I am looking for the asymptotic behaviour on $x$, but this will depend also on $k$ in some way. The bound given using
$$
sum_Omega(n) = k, n le x |mu(n)| le sum_n le x |mu(n)| le frac6xpi^2 + O(sqrtx) ll x,
$$
is not good enought for my purposes.
Thanks in advanced, any reference or idea is helpful
nt.number-theory analytic-number-theory prime-numbers asymptotics
New contributor
$endgroup$
I need to know, or at least have a good bound for, the asymptotic behaviour on $x$ of amount of integers less o equal than $x$ that are square free and with exactly $k$ primes on its decomposition. That is the cardinal of the following set
$$
mathcalJ_T(x,k) = n in mathbbN : n le x, Omega(n)=k , n mbox is square free .$$
Other way to describe this cardinal, using the Möbius function, is
$$ |mathcalJ_T(x,k) | = sum_Omega(n) = k, n le x |mu(n)|.$$
I am looking for the asymptotic behaviour on $x$, but this will depend also on $k$ in some way. The bound given using
$$
sum_Omega(n) = k, n le x |mu(n)| le sum_n le x |mu(n)| le frac6xpi^2 + O(sqrtx) ll x,
$$
is not good enought for my purposes.
Thanks in advanced, any reference or idea is helpful
nt.number-theory analytic-number-theory prime-numbers asymptotics
nt.number-theory analytic-number-theory prime-numbers asymptotics
New contributor
New contributor
edited 8 hours ago
GH from MO
60.4k5152231
60.4k5152231
New contributor
asked 9 hours ago
Martin MansillaMartin Mansilla
161
161
New contributor
New contributor
$begingroup$
Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using sum_d^2 mu(d)) the number of such squarefree integers leq x with omega(n) = k as an alternating sum of, for each dleq x^1/2, the number of integers leq x/d^2 with exactly k - omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below:
$endgroup$
– alpoge
3 hours ago
$begingroup$
mathoverflow.net/questions/35927/…
$endgroup$
– alpoge
3 hours ago
$begingroup$
Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers leq x with two prime factors seems to grow like x loglogx / logx. Who among us, even with our computers, has seen loglogx grow?! (Canonical quote about loglogx going to infinity with dignity.) What intuition...
$endgroup$
– alpoge
3 hours ago
add a comment |
$begingroup$
Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using sum_d^2 mu(d)) the number of such squarefree integers leq x with omega(n) = k as an alternating sum of, for each dleq x^1/2, the number of integers leq x/d^2 with exactly k - omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below:
$endgroup$
– alpoge
3 hours ago
$begingroup$
mathoverflow.net/questions/35927/…
$endgroup$
– alpoge
3 hours ago
$begingroup$
Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers leq x with two prime factors seems to grow like x loglogx / logx. Who among us, even with our computers, has seen loglogx grow?! (Canonical quote about loglogx going to infinity with dignity.) What intuition...
$endgroup$
– alpoge
3 hours ago
$begingroup$
Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using sum_d^2 mu(d)) the number of such squarefree integers leq x with omega(n) = k as an alternating sum of, for each dleq x^1/2, the number of integers leq x/d^2 with exactly k - omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below:
$endgroup$
– alpoge
3 hours ago
$begingroup$
Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using sum_d^2 mu(d)) the number of such squarefree integers leq x with omega(n) = k as an alternating sum of, for each dleq x^1/2, the number of integers leq x/d^2 with exactly k - omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below:
$endgroup$
– alpoge
3 hours ago
$begingroup$
mathoverflow.net/questions/35927/…
$endgroup$
– alpoge
3 hours ago
$begingroup$
mathoverflow.net/questions/35927/…
$endgroup$
– alpoge
3 hours ago
$begingroup$
Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers leq x with two prime factors seems to grow like x loglogx / logx. Who among us, even with our computers, has seen loglogx grow?! (Canonical quote about loglogx going to infinity with dignity.) What intuition...
$endgroup$
– alpoge
3 hours ago
$begingroup$
Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers leq x with two prime factors seems to grow like x loglogx / logx. Who among us, even with our computers, has seen loglogx grow?! (Canonical quote about loglogx going to infinity with dignity.) What intuition...
$endgroup$
– alpoge
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method.
I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula
$$sum_text$n$ square-freez^omega(n)n^-s=prod_pleft(1+fraczp^sright).$$
Then you will need to "factor out" $zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.
$endgroup$
add a comment |
$begingroup$
It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas
beginalign*
# nle xcolon omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)! \
# nle xcolon Omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)!.
endalign*
(Here $omega$ and $Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)
Note that any nonsquarefree integer $n$ with $Omega(n) = k$ satisfies $omega(n) = j$ for some $1le jle k-1$. Therefore the bounds
beginalign*
# nle xcolon Omega(n) = k,, ntext is squarefree &le # nle xcolon Omega(n) = k \
# nle xcolon Omega(n) = k,, ntext is squarefree &ge # nle xcolon Omega(n) = k - sum_j=1^k-1 # nle xcolon omega(n) = j ,
endalign*
together with the Selberg–Sathe asymptotic formulas above, immediately imply that
$$
# nle xcolon Omega(n) = k,, ntext is squarefree sim frac xlog x frac(loglog x)^k-1(k-1)!.
$$
This argument can be made to work with some uniformity in $k$ as well (I think $k=o(loglog x)$ is what is needed).
$endgroup$
add a comment |
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2 Answers
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votes
$begingroup$
You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method.
I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula
$$sum_text$n$ square-freez^omega(n)n^-s=prod_pleft(1+fraczp^sright).$$
Then you will need to "factor out" $zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.
$endgroup$
add a comment |
$begingroup$
You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method.
I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula
$$sum_text$n$ square-freez^omega(n)n^-s=prod_pleft(1+fraczp^sright).$$
Then you will need to "factor out" $zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.
$endgroup$
add a comment |
$begingroup$
You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method.
I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula
$$sum_text$n$ square-freez^omega(n)n^-s=prod_pleft(1+fraczp^sright).$$
Then you will need to "factor out" $zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.
$endgroup$
You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method.
I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula
$$sum_text$n$ square-freez^omega(n)n^-s=prod_pleft(1+fraczp^sright).$$
Then you will need to "factor out" $zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.
edited 7 hours ago
answered 8 hours ago
GH from MOGH from MO
60.4k5152231
60.4k5152231
add a comment |
add a comment |
$begingroup$
It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas
beginalign*
# nle xcolon omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)! \
# nle xcolon Omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)!.
endalign*
(Here $omega$ and $Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)
Note that any nonsquarefree integer $n$ with $Omega(n) = k$ satisfies $omega(n) = j$ for some $1le jle k-1$. Therefore the bounds
beginalign*
# nle xcolon Omega(n) = k,, ntext is squarefree &le # nle xcolon Omega(n) = k \
# nle xcolon Omega(n) = k,, ntext is squarefree &ge # nle xcolon Omega(n) = k - sum_j=1^k-1 # nle xcolon omega(n) = j ,
endalign*
together with the Selberg–Sathe asymptotic formulas above, immediately imply that
$$
# nle xcolon Omega(n) = k,, ntext is squarefree sim frac xlog x frac(loglog x)^k-1(k-1)!.
$$
This argument can be made to work with some uniformity in $k$ as well (I think $k=o(loglog x)$ is what is needed).
$endgroup$
add a comment |
$begingroup$
It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas
beginalign*
# nle xcolon omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)! \
# nle xcolon Omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)!.
endalign*
(Here $omega$ and $Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)
Note that any nonsquarefree integer $n$ with $Omega(n) = k$ satisfies $omega(n) = j$ for some $1le jle k-1$. Therefore the bounds
beginalign*
# nle xcolon Omega(n) = k,, ntext is squarefree &le # nle xcolon Omega(n) = k \
# nle xcolon Omega(n) = k,, ntext is squarefree &ge # nle xcolon Omega(n) = k - sum_j=1^k-1 # nle xcolon omega(n) = j ,
endalign*
together with the Selberg–Sathe asymptotic formulas above, immediately imply that
$$
# nle xcolon Omega(n) = k,, ntext is squarefree sim frac xlog x frac(loglog x)^k-1(k-1)!.
$$
This argument can be made to work with some uniformity in $k$ as well (I think $k=o(loglog x)$ is what is needed).
$endgroup$
add a comment |
$begingroup$
It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas
beginalign*
# nle xcolon omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)! \
# nle xcolon Omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)!.
endalign*
(Here $omega$ and $Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)
Note that any nonsquarefree integer $n$ with $Omega(n) = k$ satisfies $omega(n) = j$ for some $1le jle k-1$. Therefore the bounds
beginalign*
# nle xcolon Omega(n) = k,, ntext is squarefree &le # nle xcolon Omega(n) = k \
# nle xcolon Omega(n) = k,, ntext is squarefree &ge # nle xcolon Omega(n) = k - sum_j=1^k-1 # nle xcolon omega(n) = j ,
endalign*
together with the Selberg–Sathe asymptotic formulas above, immediately imply that
$$
# nle xcolon Omega(n) = k,, ntext is squarefree sim frac xlog x frac(loglog x)^k-1(k-1)!.
$$
This argument can be made to work with some uniformity in $k$ as well (I think $k=o(loglog x)$ is what is needed).
$endgroup$
It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas
beginalign*
# nle xcolon omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)! \
# nle xcolon Omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)!.
endalign*
(Here $omega$ and $Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)
Note that any nonsquarefree integer $n$ with $Omega(n) = k$ satisfies $omega(n) = j$ for some $1le jle k-1$. Therefore the bounds
beginalign*
# nle xcolon Omega(n) = k,, ntext is squarefree &le # nle xcolon Omega(n) = k \
# nle xcolon Omega(n) = k,, ntext is squarefree &ge # nle xcolon Omega(n) = k - sum_j=1^k-1 # nle xcolon omega(n) = j ,
endalign*
together with the Selberg–Sathe asymptotic formulas above, immediately imply that
$$
# nle xcolon Omega(n) = k,, ntext is squarefree sim frac xlog x frac(loglog x)^k-1(k-1)!.
$$
This argument can be made to work with some uniformity in $k$ as well (I think $k=o(loglog x)$ is what is needed).
answered 2 hours ago
Greg MartinGreg Martin
9,09213862
9,09213862
add a comment |
add a comment |
Martin Mansilla is a new contributor. Be nice, and check out our Code of Conduct.
Martin Mansilla is a new contributor. Be nice, and check out our Code of Conduct.
Martin Mansilla is a new contributor. Be nice, and check out our Code of Conduct.
Martin Mansilla is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using sum_d^2 mu(d)) the number of such squarefree integers leq x with omega(n) = k as an alternating sum of, for each dleq x^1/2, the number of integers leq x/d^2 with exactly k - omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below:
$endgroup$
– alpoge
3 hours ago
$begingroup$
mathoverflow.net/questions/35927/…
$endgroup$
– alpoge
3 hours ago
$begingroup$
Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers leq x with two prime factors seems to grow like x loglogx / logx. Who among us, even with our computers, has seen loglogx grow?! (Canonical quote about loglogx going to infinity with dignity.) What intuition...
$endgroup$
– alpoge
3 hours ago