Probability for composite $n$ to have prime factor $geq sqrt n$












1












$begingroup$


Let $operatorname{GPF}(n)$ denote the largest prime factor of
$ninmathbb N_{>1}$. My computer tests for intervalls $[m,n]$, where $n<10,000,000$, suggests that the probability
$operatorname{P}(operatorname{GPF}(n)geqsqrt n)$ is asymptotically equivalent with a constant $approxfrac{8}{11}$.



Can this be proved, heuristically understood or falsified?






probability divisibility prime-factorization conjectures







share|cite|improve this question









$endgroup$












  • $begingroup$
    The probability over what? What you are saying seems to more closely resemble the idea that the density of naturals with this property is $approx frac 8{11}$.
    $endgroup$
    – DreamConspiracy
    Dec 29 '18 at 14:56






  • 2




    $begingroup$
    The precise limit is $ln 2$. See also this.
    $endgroup$
    – metamorphy
    Dec 29 '18 at 14:57










  • $begingroup$
    @DreamConspiracy: the meaning is $lim_{ntoinfty}p_n$, where $p_n$ is the probability that a random integer $in[1,n]$ has a factor $>n^{1/2}$.
    $endgroup$
    – metamorphy
    Dec 29 '18 at 14:59












  • $begingroup$
    @metamorphy yes, that is the definition of what I said above
    $endgroup$
    – DreamConspiracy
    Dec 29 '18 at 15:00






  • 3




    $begingroup$
    See also math.stackexchange.com/questions/375270/…
    $endgroup$
    – lhf
    Dec 29 '18 at 16:46
















1












$begingroup$


Let $operatorname{GPF}(n)$ denote the largest prime factor of
$ninmathbb N_{>1}$. My computer tests for intervalls $[m,n]$, where $n<10,000,000$, suggests that the probability
$operatorname{P}(operatorname{GPF}(n)geqsqrt n)$ is asymptotically equivalent with a constant $approxfrac{8}{11}$.



Can this be proved, heuristically understood or falsified?






probability divisibility prime-factorization conjectures







share|cite|improve this question









$endgroup$












  • $begingroup$
    The probability over what? What you are saying seems to more closely resemble the idea that the density of naturals with this property is $approx frac 8{11}$.
    $endgroup$
    – DreamConspiracy
    Dec 29 '18 at 14:56






  • 2




    $begingroup$
    The precise limit is $ln 2$. See also this.
    $endgroup$
    – metamorphy
    Dec 29 '18 at 14:57










  • $begingroup$
    @DreamConspiracy: the meaning is $lim_{ntoinfty}p_n$, where $p_n$ is the probability that a random integer $in[1,n]$ has a factor $>n^{1/2}$.
    $endgroup$
    – metamorphy
    Dec 29 '18 at 14:59












  • $begingroup$
    @metamorphy yes, that is the definition of what I said above
    $endgroup$
    – DreamConspiracy
    Dec 29 '18 at 15:00






  • 3




    $begingroup$
    See also math.stackexchange.com/questions/375270/…
    $endgroup$
    – lhf
    Dec 29 '18 at 16:46














1












1








1





$begingroup$


Let $operatorname{GPF}(n)$ denote the largest prime factor of
$ninmathbb N_{>1}$. My computer tests for intervalls $[m,n]$, where $n<10,000,000$, suggests that the probability
$operatorname{P}(operatorname{GPF}(n)geqsqrt n)$ is asymptotically equivalent with a constant $approxfrac{8}{11}$.



Can this be proved, heuristically understood or falsified?






probability divisibility prime-factorization conjectures







share|cite|improve this question









$endgroup$




Let $operatorname{GPF}(n)$ denote the largest prime factor of
$ninmathbb N_{>1}$. My computer tests for intervalls $[m,n]$, where $n<10,000,000$, suggests that the probability
$operatorname{P}(operatorname{GPF}(n)geqsqrt n)$ is asymptotically equivalent with a constant $approxfrac{8}{11}$.



Can this be proved, heuristically understood or falsified?






probability divisibility prime-factorization conjectures




probability divisibility prime-factorization conjectures






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 14:46









LehsLehs

7,02931664




7,02931664












  • $begingroup$
    The probability over what? What you are saying seems to more closely resemble the idea that the density of naturals with this property is $approx frac 8{11}$.
    $endgroup$
    – DreamConspiracy
    Dec 29 '18 at 14:56






  • 2




    $begingroup$
    The precise limit is $ln 2$. See also this.
    $endgroup$
    – metamorphy
    Dec 29 '18 at 14:57










  • $begingroup$
    @DreamConspiracy: the meaning is $lim_{ntoinfty}p_n$, where $p_n$ is the probability that a random integer $in[1,n]$ has a factor $>n^{1/2}$.
    $endgroup$
    – metamorphy
    Dec 29 '18 at 14:59












  • $begingroup$
    @metamorphy yes, that is the definition of what I said above
    $endgroup$
    – DreamConspiracy
    Dec 29 '18 at 15:00






  • 3




    $begingroup$
    See also math.stackexchange.com/questions/375270/…
    $endgroup$
    – lhf
    Dec 29 '18 at 16:46


















  • $begingroup$
    The probability over what? What you are saying seems to more closely resemble the idea that the density of naturals with this property is $approx frac 8{11}$.
    $endgroup$
    – DreamConspiracy
    Dec 29 '18 at 14:56






  • 2




    $begingroup$
    The precise limit is $ln 2$. See also this.
    $endgroup$
    – metamorphy
    Dec 29 '18 at 14:57










  • $begingroup$
    @DreamConspiracy: the meaning is $lim_{ntoinfty}p_n$, where $p_n$ is the probability that a random integer $in[1,n]$ has a factor $>n^{1/2}$.
    $endgroup$
    – metamorphy
    Dec 29 '18 at 14:59












  • $begingroup$
    @metamorphy yes, that is the definition of what I said above
    $endgroup$
    – DreamConspiracy
    Dec 29 '18 at 15:00






  • 3




    $begingroup$
    See also math.stackexchange.com/questions/375270/…
    $endgroup$
    – lhf
    Dec 29 '18 at 16:46
















$begingroup$
The probability over what? What you are saying seems to more closely resemble the idea that the density of naturals with this property is $approx frac 8{11}$.
$endgroup$
– DreamConspiracy
Dec 29 '18 at 14:56




$begingroup$
The probability over what? What you are saying seems to more closely resemble the idea that the density of naturals with this property is $approx frac 8{11}$.
$endgroup$
– DreamConspiracy
Dec 29 '18 at 14:56




2




2




$begingroup$
The precise limit is $ln 2$. See also this.
$endgroup$
– metamorphy
Dec 29 '18 at 14:57




$begingroup$
The precise limit is $ln 2$. See also this.
$endgroup$
– metamorphy
Dec 29 '18 at 14:57












$begingroup$
@DreamConspiracy: the meaning is $lim_{ntoinfty}p_n$, where $p_n$ is the probability that a random integer $in[1,n]$ has a factor $>n^{1/2}$.
$endgroup$
– metamorphy
Dec 29 '18 at 14:59






$begingroup$
@DreamConspiracy: the meaning is $lim_{ntoinfty}p_n$, where $p_n$ is the probability that a random integer $in[1,n]$ has a factor $>n^{1/2}$.
$endgroup$
– metamorphy
Dec 29 '18 at 14:59














$begingroup$
@metamorphy yes, that is the definition of what I said above
$endgroup$
– DreamConspiracy
Dec 29 '18 at 15:00




$begingroup$
@metamorphy yes, that is the definition of what I said above
$endgroup$
– DreamConspiracy
Dec 29 '18 at 15:00




3




3




$begingroup$
See also math.stackexchange.com/questions/375270/…
$endgroup$
– lhf
Dec 29 '18 at 16:46




$begingroup$
See also math.stackexchange.com/questions/375270/…
$endgroup$
– lhf
Dec 29 '18 at 16:46










1 Answer
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$begingroup$

Begin by noting that if a natural $n$ has a prime factor $p geq sqrt{n}$, then clearly $p$ is unique for $n$. So let's fix some prime $p$ and consider all naturals for which $p$ satisfies this property. Clearly, these are just all naturals of the form $ap$, where $ain Bbb N, a leq p$. From this, we get that if we fix some natural $N$, then the number of naturals less than $N$ satisfying this property is precisely
$$sum_{p in Bbb P, p leq sqrt{N}}p + sum_{p in Bbb P, sqrt{N} < p leq N}lfloor frac np rfloor$$
where $Bbb P$ is of course the set of primes.






share|cite|improve this answer









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    $begingroup$

    Begin by noting that if a natural $n$ has a prime factor $p geq sqrt{n}$, then clearly $p$ is unique for $n$. So let's fix some prime $p$ and consider all naturals for which $p$ satisfies this property. Clearly, these are just all naturals of the form $ap$, where $ain Bbb N, a leq p$. From this, we get that if we fix some natural $N$, then the number of naturals less than $N$ satisfying this property is precisely
    $$sum_{p in Bbb P, p leq sqrt{N}}p + sum_{p in Bbb P, sqrt{N} < p leq N}lfloor frac np rfloor$$
    where $Bbb P$ is of course the set of primes.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Begin by noting that if a natural $n$ has a prime factor $p geq sqrt{n}$, then clearly $p$ is unique for $n$. So let's fix some prime $p$ and consider all naturals for which $p$ satisfies this property. Clearly, these are just all naturals of the form $ap$, where $ain Bbb N, a leq p$. From this, we get that if we fix some natural $N$, then the number of naturals less than $N$ satisfying this property is precisely
      $$sum_{p in Bbb P, p leq sqrt{N}}p + sum_{p in Bbb P, sqrt{N} < p leq N}lfloor frac np rfloor$$
      where $Bbb P$ is of course the set of primes.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Begin by noting that if a natural $n$ has a prime factor $p geq sqrt{n}$, then clearly $p$ is unique for $n$. So let's fix some prime $p$ and consider all naturals for which $p$ satisfies this property. Clearly, these are just all naturals of the form $ap$, where $ain Bbb N, a leq p$. From this, we get that if we fix some natural $N$, then the number of naturals less than $N$ satisfying this property is precisely
        $$sum_{p in Bbb P, p leq sqrt{N}}p + sum_{p in Bbb P, sqrt{N} < p leq N}lfloor frac np rfloor$$
        where $Bbb P$ is of course the set of primes.






        share|cite|improve this answer









        $endgroup$



        Begin by noting that if a natural $n$ has a prime factor $p geq sqrt{n}$, then clearly $p$ is unique for $n$. So let's fix some prime $p$ and consider all naturals for which $p$ satisfies this property. Clearly, these are just all naturals of the form $ap$, where $ain Bbb N, a leq p$. From this, we get that if we fix some natural $N$, then the number of naturals less than $N$ satisfying this property is precisely
        $$sum_{p in Bbb P, p leq sqrt{N}}p + sum_{p in Bbb P, sqrt{N} < p leq N}lfloor frac np rfloor$$
        where $Bbb P$ is of course the set of primes.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 16:45









        DreamConspiracyDreamConspiracy

        9391216




        9391216






























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