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













3












$begingroup$


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










share|cite|improve this question









New contributor



Martin Mansilla is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$











  • $begingroup$
    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_ n 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















3












$begingroup$


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










share|cite|improve this question









New contributor



Martin Mansilla is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$











  • $begingroup$
    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_ n 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













3












3








3





$begingroup$


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










share|cite|improve this question









New contributor



Martin Mansilla is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




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






share|cite|improve this question









New contributor



Martin Mansilla is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










share|cite|improve this question









New contributor



Martin Mansilla is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share|cite|improve this question




share|cite|improve this question








edited 8 hours ago









GH from MO

60.4k5152231




60.4k5152231






New contributor



Martin Mansilla is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








asked 8 hours ago









Martin MansillaMartin Mansilla

161




161




New contributor



Martin Mansilla is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




New contributor




Martin Mansilla is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.













  • $begingroup$
    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_ n 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_ n 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_ n 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_ n 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










2 Answers
2






active

oldest

votes


















7












$begingroup$

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.






share|cite|improve this answer











$endgroup$




















    2












    $begingroup$

    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).






    share|cite|improve this answer









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      2 Answers
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      active

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$

















        7












        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$















          7












          7








          7





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 7 hours ago

























          answered 8 hours ago









          GH from MOGH from MO

          60.4k5152231




          60.4k5152231





















              2












              $begingroup$

              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).






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                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).






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  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).






                  share|cite|improve this answer









                  $endgroup$



                  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).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Greg MartinGreg Martin

                  9,09213862




                  9,09213862




















                      Martin Mansilla is a new contributor. Be nice, and check out our Code of Conduct.









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