Sets that are both Sum-free and Product-free












24












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Let $P_o$ be the primes excluding $2$. $P_o subset mathbb{N}$ has the following property $Q$:




  • For any $a,b in P_o$, $a + b notin P_o$.

  • For any $a,b in P_o$, $ab notin P_o$.


So both addition and multiplication necessarily leave the set $P_o$.



$P_o$ has natural density $0$.




Q1. Is there a set $S subset mathbb{N}$ with positive density
that satisfies property $Q$?




Answered quickly by @JoséCarlosSantos: Yes.
Permit me then to add a new question:




Q2. What is largest density $S subset mathbb{N}$
that satisfies property $Q$?




Santos's example has density $frac{1}{3}$.










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  • 8




    $begingroup$
    Instead of adding a question after receiving an answer to your original question, you should ask that other question separately
    $endgroup$
    – Hagen von Eitzen
    17 hours ago










  • $begingroup$
    Could you edit your question to specify that you are (apparently) more interested in upper natural density rather than just natural density?
    $endgroup$
    – Erick Wong
    13 secs ago
















24












$begingroup$


Let $P_o$ be the primes excluding $2$. $P_o subset mathbb{N}$ has the following property $Q$:




  • For any $a,b in P_o$, $a + b notin P_o$.

  • For any $a,b in P_o$, $ab notin P_o$.


So both addition and multiplication necessarily leave the set $P_o$.



$P_o$ has natural density $0$.




Q1. Is there a set $S subset mathbb{N}$ with positive density
that satisfies property $Q$?




Answered quickly by @JoséCarlosSantos: Yes.
Permit me then to add a new question:




Q2. What is largest density $S subset mathbb{N}$
that satisfies property $Q$?




Santos's example has density $frac{1}{3}$.










share|cite|improve this question











$endgroup$








  • 8




    $begingroup$
    Instead of adding a question after receiving an answer to your original question, you should ask that other question separately
    $endgroup$
    – Hagen von Eitzen
    17 hours ago










  • $begingroup$
    Could you edit your question to specify that you are (apparently) more interested in upper natural density rather than just natural density?
    $endgroup$
    – Erick Wong
    13 secs ago














24












24








24


7



$begingroup$


Let $P_o$ be the primes excluding $2$. $P_o subset mathbb{N}$ has the following property $Q$:




  • For any $a,b in P_o$, $a + b notin P_o$.

  • For any $a,b in P_o$, $ab notin P_o$.


So both addition and multiplication necessarily leave the set $P_o$.



$P_o$ has natural density $0$.




Q1. Is there a set $S subset mathbb{N}$ with positive density
that satisfies property $Q$?




Answered quickly by @JoséCarlosSantos: Yes.
Permit me then to add a new question:




Q2. What is largest density $S subset mathbb{N}$
that satisfies property $Q$?




Santos's example has density $frac{1}{3}$.










share|cite|improve this question











$endgroup$




Let $P_o$ be the primes excluding $2$. $P_o subset mathbb{N}$ has the following property $Q$:




  • For any $a,b in P_o$, $a + b notin P_o$.

  • For any $a,b in P_o$, $ab notin P_o$.


So both addition and multiplication necessarily leave the set $P_o$.



$P_o$ has natural density $0$.




Q1. Is there a set $S subset mathbb{N}$ with positive density
that satisfies property $Q$?




Answered quickly by @JoséCarlosSantos: Yes.
Permit me then to add a new question:




Q2. What is largest density $S subset mathbb{N}$
that satisfies property $Q$?




Santos's example has density $frac{1}{3}$.







number-theory elementary-number-theory prime-numbers infinity






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share|cite|improve this question













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edited 14 mins ago







Joseph O'Rourke

















asked yesterday









Joseph O'RourkeJoseph O'Rourke

18.1k349111




18.1k349111








  • 8




    $begingroup$
    Instead of adding a question after receiving an answer to your original question, you should ask that other question separately
    $endgroup$
    – Hagen von Eitzen
    17 hours ago










  • $begingroup$
    Could you edit your question to specify that you are (apparently) more interested in upper natural density rather than just natural density?
    $endgroup$
    – Erick Wong
    13 secs ago














  • 8




    $begingroup$
    Instead of adding a question after receiving an answer to your original question, you should ask that other question separately
    $endgroup$
    – Hagen von Eitzen
    17 hours ago










  • $begingroup$
    Could you edit your question to specify that you are (apparently) more interested in upper natural density rather than just natural density?
    $endgroup$
    – Erick Wong
    13 secs ago








8




8




$begingroup$
Instead of adding a question after receiving an answer to your original question, you should ask that other question separately
$endgroup$
– Hagen von Eitzen
17 hours ago




$begingroup$
Instead of adding a question after receiving an answer to your original question, you should ask that other question separately
$endgroup$
– Hagen von Eitzen
17 hours ago












$begingroup$
Could you edit your question to specify that you are (apparently) more interested in upper natural density rather than just natural density?
$endgroup$
– Erick Wong
13 secs ago




$begingroup$
Could you edit your question to specify that you are (apparently) more interested in upper natural density rather than just natural density?
$endgroup$
– Erick Wong
13 secs ago










5 Answers
5






active

oldest

votes


















18












$begingroup$

New answer:



Kurlberg, Lagarias and Pomerance, On sets of integers which are both sum-free and product-free (arXiv:1201.1317) answers the question. The upper density of any such set is strictly less than 1/2, but can be arbitrarily close to 1/2. I don't see that they state this explicitly in the paper, but it follows pretty quickly from Theorem 1.3.



Explicitly: Theorem 1.3 implies that for any $varepsilon>0$ there is some $n$ and some subset $Ssubsetmathbb{Z}/nmathbb{Z}$ of residue classes that is sum-free and product-free consisting of at least $(frac{1}{2}-varepsilon)n$ classes. Then taking all integers in those residue classes gives a product-free sum-free set of integers of density at least $(frac{1}{2}-varepsilon)$.



Old answer:



Andrew Treglow's talk On sum-free and solution-free sets of integers cites the following result of Deshouillers, Freiman, Sós and Temkin (1999):




If $Ssubseteq[n]$ is sum-free then at least one of the following holds:




  1. $lvert Srvertle2n/5+1$


  2. $S$ consists of odds


  3. $lvert Srvertlemin(S)$.




Therefore, if the density of a sum-free product-free set $P$ of integers is greater than 2/5, then $Pcap[n]$ must fall in the second case for sufficiently large $n$. (We can't be in the third case because $min(P)<2n/5$ for sufficiently large $n$.)



So, the only way we could hope to do better than 2/5 is to use only odd numbers, and as a corollary the highest density we could hope for is 1/2.



In fact, the proof of Remark 2.7 of Kurlberg, Lagarias and Pomerance, Product-free sets with high density carries over to the case of only odd numbers, showing that we cannot attain a density of 1/2. For completeness, we repeat the argument here with the appropriate modifications: Let $a$ denote the least element of $P$, and let $P(x):=Pcap[1,x]$. Since $P(x)setminus{P(x/a)}$ lies in $(x/a,x]$, $lvert P(x)rvertle lvert P(x/a)rvert+frac{x-lfloor x/arfloor}{2}+1$. Also, multiplying each member of $P(x/a)$ creates products in $[1,x]$ which cannot lie in $P$, so we have $lvert P(x)rvertle frac{x}{2}+1-lvert P(x/a)rvert$. Adding these two inequalities and dividing both sides by 2 gives $lvert P(x)rvertle
frac{x}{2}-frac{lfloor x/arfloor}{2}+2$
, which implies that the upper density of $P$ is at most $frac{1}{2}-frac{1}{2a}$.






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    27












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    What about $S={3n-1,|,ninmathbb N}$? Its natural density is $frac13$.






    share|cite|improve this answer









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    • $begingroup$
      Very nice! ${}$
      $endgroup$
      – Joseph O'Rourke
      yesterday



















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    This did not begin as an answer, but see edit below.



    See this talk by Carl Pomerance, on sum-free sets, and product-free sets. One way to answer the OP (and this is the approach of the other answers) is to choose an $n$, and a subset $S$ of $mathbb{Z}/nmathbb{Z}$, such that $S$ is both sum-free (if $a,bin S$, then $a+bnotin S$) and product-free (if $a,bin S$, then $abnotin S$) . Then, we take all integers that are congruent to an element of $S$, modulo $n$. The desired asymptotic density is seen to be $frac{|S|}{n}$.



    This might not be a simple question at all. Looking just at the sum-free property , we can easily get asymptotic density $0.5$ by taking the odd numbers. The product-free property is quite subtle: the linked talk gives a construction of a very large $n$ (with over 100 million digits) such that there is an $S$ satisfying $frac{|S|}{n}>0.5003$. In fact, we could raise $0.5003$ to be arbitrarily close to $1$ (although no construction is given in the linked talk). The general approach is to make $n$ have many small primes, to large powers, as factors.



    One would not expect that this product-free set is also sum-free, but we can always remove some elements from it, until we have a subset of $S$ that is both sum-free and product-free. I have no idea how big that resulting subset would be.






    EDIT: Following the methods of the linked talk, choose $n$ (assumed even) and product-free $S$, so that $frac{|S|}{n}ge 1-epsilon$. Hence $|S|ge n(1-epsilon)$. $S$ contains at most $frac{n}{2}$ even numbers (since $Ssubseteq {0,1,ldots, n-1}$, half of which are even). Take $T$ to be the set of all the odd numbers in $S$. We have $|T|ge |S|-frac{n}{2}=n(frac{1}{2}-epsilon)$. Since $Tsubseteq S$, $T$ is product-free. $T$ is also sum-free, since the sum of two elements of $T$ are even (and hence not in $T$). The asymptotic density of all naturals congruent to an element of $T$ modulo $n$ is $frac{|T|}{n}ge frac{1}{2}-epsilon$.

    Note that an asymptotic density of $frac{1}{2}$ is best-possible for sum-free sets (as proved in Pomerance's slides), much less sum-free product-free sets. The above construction gives a subset of $mathbb{N}$ arbitrarily close to this bound.






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    • 1




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      I am glad to know the terms "sum-free" and "product-free." Much more memorable than "property $Q$"!
      $endgroup$
      – Joseph O'Rourke
      yesterday



















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    Let $S = {n : n equiv 2 rm{or} 3 pmod 5}$. This has density $2/5$, which beats $1/3$.



    Incidentally, this sequence can be generated with a greedy algorithm, starting with $S = {2}$ and progressively adding every larger number that maintains the requirement.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This feels like it might be the max, because of the greedy property you mentioned.
      $endgroup$
      – Joseph O'Rourke
      yesterday






    • 2




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      @JosephO'Rourke Not necessarily: using a greedy algorithm starting at $1$ instead of $2$ leads to a different sequence that is not as dense: ${1, 3, 5, 7, 11, 13, 17, 19, 23, 27, 29, 31, 37, 41, 43, 45, 47, 53, 59, 61, ldots}$. (I'm surprised not to find this on OEIS. You might be interested in adding it there.) Other starting points are similarly bad. But it's interesting that the greedy method starting at $2$ gives a consistent structure, and I agree that it feels like it could be the best result.
      $endgroup$
      – Théophile
      yesterday








    • 6




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      @Théophile Doesn't 1*3=3 violate the constraint?
      $endgroup$
      – alphacapture
      yesterday






    • 1




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      @Théophile What it looks like you are doing is starting with 3 and running the greedy algorithm while restricting yourself to odd numbers; this will result in the set of odd numbers with an odd number of prime factors
      $endgroup$
      – alphacapture
      yesterday






    • 3




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      @alphacapture, indeed, dropping the $1$ and checking OEIS for the rest gives oeis.org/A067019 . (It's generally a good idea to drop a few early terms when checking OEIS, especially $1$'s and $0$'s.)
      $endgroup$
      – Barry Cipra
      yesterday



















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    The answer to the title question is no. $Q$ doesn't characterize the odd primes, since, for example, ${2,3,15}vDash Q$. Any subset of the odd primes satisfies $Q$, as does any set formed by taking the odd primes along with an even integer $k$ and deleting one of every pair of primes that differ by $k$. Or forget about the primes altogether and take any (finite or infinite) set ${a_1, a_2, dots}$ such that $2leq a_1 < a_2$ and, for all $i>2$, $a_i > a_{i-1}a_{i-2}$.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      I especially like your last example: $2,3,7,22,155,3411,ldots$.
      $endgroup$
      – Joseph O'Rourke
      yesterday











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    5 Answers
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    5 Answers
    5






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    active

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    active

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    18












    $begingroup$

    New answer:



    Kurlberg, Lagarias and Pomerance, On sets of integers which are both sum-free and product-free (arXiv:1201.1317) answers the question. The upper density of any such set is strictly less than 1/2, but can be arbitrarily close to 1/2. I don't see that they state this explicitly in the paper, but it follows pretty quickly from Theorem 1.3.



    Explicitly: Theorem 1.3 implies that for any $varepsilon>0$ there is some $n$ and some subset $Ssubsetmathbb{Z}/nmathbb{Z}$ of residue classes that is sum-free and product-free consisting of at least $(frac{1}{2}-varepsilon)n$ classes. Then taking all integers in those residue classes gives a product-free sum-free set of integers of density at least $(frac{1}{2}-varepsilon)$.



    Old answer:



    Andrew Treglow's talk On sum-free and solution-free sets of integers cites the following result of Deshouillers, Freiman, Sós and Temkin (1999):




    If $Ssubseteq[n]$ is sum-free then at least one of the following holds:




    1. $lvert Srvertle2n/5+1$


    2. $S$ consists of odds


    3. $lvert Srvertlemin(S)$.




    Therefore, if the density of a sum-free product-free set $P$ of integers is greater than 2/5, then $Pcap[n]$ must fall in the second case for sufficiently large $n$. (We can't be in the third case because $min(P)<2n/5$ for sufficiently large $n$.)



    So, the only way we could hope to do better than 2/5 is to use only odd numbers, and as a corollary the highest density we could hope for is 1/2.



    In fact, the proof of Remark 2.7 of Kurlberg, Lagarias and Pomerance, Product-free sets with high density carries over to the case of only odd numbers, showing that we cannot attain a density of 1/2. For completeness, we repeat the argument here with the appropriate modifications: Let $a$ denote the least element of $P$, and let $P(x):=Pcap[1,x]$. Since $P(x)setminus{P(x/a)}$ lies in $(x/a,x]$, $lvert P(x)rvertle lvert P(x/a)rvert+frac{x-lfloor x/arfloor}{2}+1$. Also, multiplying each member of $P(x/a)$ creates products in $[1,x]$ which cannot lie in $P$, so we have $lvert P(x)rvertle frac{x}{2}+1-lvert P(x/a)rvert$. Adding these two inequalities and dividing both sides by 2 gives $lvert P(x)rvertle
    frac{x}{2}-frac{lfloor x/arfloor}{2}+2$
    , which implies that the upper density of $P$ is at most $frac{1}{2}-frac{1}{2a}$.






    share|cite|improve this answer











    $endgroup$


















      18












      $begingroup$

      New answer:



      Kurlberg, Lagarias and Pomerance, On sets of integers which are both sum-free and product-free (arXiv:1201.1317) answers the question. The upper density of any such set is strictly less than 1/2, but can be arbitrarily close to 1/2. I don't see that they state this explicitly in the paper, but it follows pretty quickly from Theorem 1.3.



      Explicitly: Theorem 1.3 implies that for any $varepsilon>0$ there is some $n$ and some subset $Ssubsetmathbb{Z}/nmathbb{Z}$ of residue classes that is sum-free and product-free consisting of at least $(frac{1}{2}-varepsilon)n$ classes. Then taking all integers in those residue classes gives a product-free sum-free set of integers of density at least $(frac{1}{2}-varepsilon)$.



      Old answer:



      Andrew Treglow's talk On sum-free and solution-free sets of integers cites the following result of Deshouillers, Freiman, Sós and Temkin (1999):




      If $Ssubseteq[n]$ is sum-free then at least one of the following holds:




      1. $lvert Srvertle2n/5+1$


      2. $S$ consists of odds


      3. $lvert Srvertlemin(S)$.




      Therefore, if the density of a sum-free product-free set $P$ of integers is greater than 2/5, then $Pcap[n]$ must fall in the second case for sufficiently large $n$. (We can't be in the third case because $min(P)<2n/5$ for sufficiently large $n$.)



      So, the only way we could hope to do better than 2/5 is to use only odd numbers, and as a corollary the highest density we could hope for is 1/2.



      In fact, the proof of Remark 2.7 of Kurlberg, Lagarias and Pomerance, Product-free sets with high density carries over to the case of only odd numbers, showing that we cannot attain a density of 1/2. For completeness, we repeat the argument here with the appropriate modifications: Let $a$ denote the least element of $P$, and let $P(x):=Pcap[1,x]$. Since $P(x)setminus{P(x/a)}$ lies in $(x/a,x]$, $lvert P(x)rvertle lvert P(x/a)rvert+frac{x-lfloor x/arfloor}{2}+1$. Also, multiplying each member of $P(x/a)$ creates products in $[1,x]$ which cannot lie in $P$, so we have $lvert P(x)rvertle frac{x}{2}+1-lvert P(x/a)rvert$. Adding these two inequalities and dividing both sides by 2 gives $lvert P(x)rvertle
      frac{x}{2}-frac{lfloor x/arfloor}{2}+2$
      , which implies that the upper density of $P$ is at most $frac{1}{2}-frac{1}{2a}$.






      share|cite|improve this answer











      $endgroup$
















        18












        18








        18





        $begingroup$

        New answer:



        Kurlberg, Lagarias and Pomerance, On sets of integers which are both sum-free and product-free (arXiv:1201.1317) answers the question. The upper density of any such set is strictly less than 1/2, but can be arbitrarily close to 1/2. I don't see that they state this explicitly in the paper, but it follows pretty quickly from Theorem 1.3.



        Explicitly: Theorem 1.3 implies that for any $varepsilon>0$ there is some $n$ and some subset $Ssubsetmathbb{Z}/nmathbb{Z}$ of residue classes that is sum-free and product-free consisting of at least $(frac{1}{2}-varepsilon)n$ classes. Then taking all integers in those residue classes gives a product-free sum-free set of integers of density at least $(frac{1}{2}-varepsilon)$.



        Old answer:



        Andrew Treglow's talk On sum-free and solution-free sets of integers cites the following result of Deshouillers, Freiman, Sós and Temkin (1999):




        If $Ssubseteq[n]$ is sum-free then at least one of the following holds:




        1. $lvert Srvertle2n/5+1$


        2. $S$ consists of odds


        3. $lvert Srvertlemin(S)$.




        Therefore, if the density of a sum-free product-free set $P$ of integers is greater than 2/5, then $Pcap[n]$ must fall in the second case for sufficiently large $n$. (We can't be in the third case because $min(P)<2n/5$ for sufficiently large $n$.)



        So, the only way we could hope to do better than 2/5 is to use only odd numbers, and as a corollary the highest density we could hope for is 1/2.



        In fact, the proof of Remark 2.7 of Kurlberg, Lagarias and Pomerance, Product-free sets with high density carries over to the case of only odd numbers, showing that we cannot attain a density of 1/2. For completeness, we repeat the argument here with the appropriate modifications: Let $a$ denote the least element of $P$, and let $P(x):=Pcap[1,x]$. Since $P(x)setminus{P(x/a)}$ lies in $(x/a,x]$, $lvert P(x)rvertle lvert P(x/a)rvert+frac{x-lfloor x/arfloor}{2}+1$. Also, multiplying each member of $P(x/a)$ creates products in $[1,x]$ which cannot lie in $P$, so we have $lvert P(x)rvertle frac{x}{2}+1-lvert P(x/a)rvert$. Adding these two inequalities and dividing both sides by 2 gives $lvert P(x)rvertle
        frac{x}{2}-frac{lfloor x/arfloor}{2}+2$
        , which implies that the upper density of $P$ is at most $frac{1}{2}-frac{1}{2a}$.






        share|cite|improve this answer











        $endgroup$



        New answer:



        Kurlberg, Lagarias and Pomerance, On sets of integers which are both sum-free and product-free (arXiv:1201.1317) answers the question. The upper density of any such set is strictly less than 1/2, but can be arbitrarily close to 1/2. I don't see that they state this explicitly in the paper, but it follows pretty quickly from Theorem 1.3.



        Explicitly: Theorem 1.3 implies that for any $varepsilon>0$ there is some $n$ and some subset $Ssubsetmathbb{Z}/nmathbb{Z}$ of residue classes that is sum-free and product-free consisting of at least $(frac{1}{2}-varepsilon)n$ classes. Then taking all integers in those residue classes gives a product-free sum-free set of integers of density at least $(frac{1}{2}-varepsilon)$.



        Old answer:



        Andrew Treglow's talk On sum-free and solution-free sets of integers cites the following result of Deshouillers, Freiman, Sós and Temkin (1999):




        If $Ssubseteq[n]$ is sum-free then at least one of the following holds:




        1. $lvert Srvertle2n/5+1$


        2. $S$ consists of odds


        3. $lvert Srvertlemin(S)$.




        Therefore, if the density of a sum-free product-free set $P$ of integers is greater than 2/5, then $Pcap[n]$ must fall in the second case for sufficiently large $n$. (We can't be in the third case because $min(P)<2n/5$ for sufficiently large $n$.)



        So, the only way we could hope to do better than 2/5 is to use only odd numbers, and as a corollary the highest density we could hope for is 1/2.



        In fact, the proof of Remark 2.7 of Kurlberg, Lagarias and Pomerance, Product-free sets with high density carries over to the case of only odd numbers, showing that we cannot attain a density of 1/2. For completeness, we repeat the argument here with the appropriate modifications: Let $a$ denote the least element of $P$, and let $P(x):=Pcap[1,x]$. Since $P(x)setminus{P(x/a)}$ lies in $(x/a,x]$, $lvert P(x)rvertle lvert P(x/a)rvert+frac{x-lfloor x/arfloor}{2}+1$. Also, multiplying each member of $P(x/a)$ creates products in $[1,x]$ which cannot lie in $P$, so we have $lvert P(x)rvertle frac{x}{2}+1-lvert P(x/a)rvert$. Adding these two inequalities and dividing both sides by 2 gives $lvert P(x)rvertle
        frac{x}{2}-frac{lfloor x/arfloor}{2}+2$
        , which implies that the upper density of $P$ is at most $frac{1}{2}-frac{1}{2a}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 10 hours ago









        David Richerby

        2,18011324




        2,18011324










        answered yesterday









        alphacapturealphacapture

        2,165425




        2,165425























            27












            $begingroup$

            What about $S={3n-1,|,ninmathbb N}$? Its natural density is $frac13$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Very nice! ${}$
              $endgroup$
              – Joseph O'Rourke
              yesterday
















            27












            $begingroup$

            What about $S={3n-1,|,ninmathbb N}$? Its natural density is $frac13$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Very nice! ${}$
              $endgroup$
              – Joseph O'Rourke
              yesterday














            27












            27








            27





            $begingroup$

            What about $S={3n-1,|,ninmathbb N}$? Its natural density is $frac13$.






            share|cite|improve this answer









            $endgroup$



            What about $S={3n-1,|,ninmathbb N}$? Its natural density is $frac13$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            José Carlos SantosJosé Carlos Santos

            163k22131234




            163k22131234












            • $begingroup$
              Very nice! ${}$
              $endgroup$
              – Joseph O'Rourke
              yesterday


















            • $begingroup$
              Very nice! ${}$
              $endgroup$
              – Joseph O'Rourke
              yesterday
















            $begingroup$
            Very nice! ${}$
            $endgroup$
            – Joseph O'Rourke
            yesterday




            $begingroup$
            Very nice! ${}$
            $endgroup$
            – Joseph O'Rourke
            yesterday











            12












            $begingroup$

            This did not begin as an answer, but see edit below.



            See this talk by Carl Pomerance, on sum-free sets, and product-free sets. One way to answer the OP (and this is the approach of the other answers) is to choose an $n$, and a subset $S$ of $mathbb{Z}/nmathbb{Z}$, such that $S$ is both sum-free (if $a,bin S$, then $a+bnotin S$) and product-free (if $a,bin S$, then $abnotin S$) . Then, we take all integers that are congruent to an element of $S$, modulo $n$. The desired asymptotic density is seen to be $frac{|S|}{n}$.



            This might not be a simple question at all. Looking just at the sum-free property , we can easily get asymptotic density $0.5$ by taking the odd numbers. The product-free property is quite subtle: the linked talk gives a construction of a very large $n$ (with over 100 million digits) such that there is an $S$ satisfying $frac{|S|}{n}>0.5003$. In fact, we could raise $0.5003$ to be arbitrarily close to $1$ (although no construction is given in the linked talk). The general approach is to make $n$ have many small primes, to large powers, as factors.



            One would not expect that this product-free set is also sum-free, but we can always remove some elements from it, until we have a subset of $S$ that is both sum-free and product-free. I have no idea how big that resulting subset would be.






            EDIT: Following the methods of the linked talk, choose $n$ (assumed even) and product-free $S$, so that $frac{|S|}{n}ge 1-epsilon$. Hence $|S|ge n(1-epsilon)$. $S$ contains at most $frac{n}{2}$ even numbers (since $Ssubseteq {0,1,ldots, n-1}$, half of which are even). Take $T$ to be the set of all the odd numbers in $S$. We have $|T|ge |S|-frac{n}{2}=n(frac{1}{2}-epsilon)$. Since $Tsubseteq S$, $T$ is product-free. $T$ is also sum-free, since the sum of two elements of $T$ are even (and hence not in $T$). The asymptotic density of all naturals congruent to an element of $T$ modulo $n$ is $frac{|T|}{n}ge frac{1}{2}-epsilon$.

            Note that an asymptotic density of $frac{1}{2}$ is best-possible for sum-free sets (as proved in Pomerance's slides), much less sum-free product-free sets. The above construction gives a subset of $mathbb{N}$ arbitrarily close to this bound.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              I am glad to know the terms "sum-free" and "product-free." Much more memorable than "property $Q$"!
              $endgroup$
              – Joseph O'Rourke
              yesterday
















            12












            $begingroup$

            This did not begin as an answer, but see edit below.



            See this talk by Carl Pomerance, on sum-free sets, and product-free sets. One way to answer the OP (and this is the approach of the other answers) is to choose an $n$, and a subset $S$ of $mathbb{Z}/nmathbb{Z}$, such that $S$ is both sum-free (if $a,bin S$, then $a+bnotin S$) and product-free (if $a,bin S$, then $abnotin S$) . Then, we take all integers that are congruent to an element of $S$, modulo $n$. The desired asymptotic density is seen to be $frac{|S|}{n}$.



            This might not be a simple question at all. Looking just at the sum-free property , we can easily get asymptotic density $0.5$ by taking the odd numbers. The product-free property is quite subtle: the linked talk gives a construction of a very large $n$ (with over 100 million digits) such that there is an $S$ satisfying $frac{|S|}{n}>0.5003$. In fact, we could raise $0.5003$ to be arbitrarily close to $1$ (although no construction is given in the linked talk). The general approach is to make $n$ have many small primes, to large powers, as factors.



            One would not expect that this product-free set is also sum-free, but we can always remove some elements from it, until we have a subset of $S$ that is both sum-free and product-free. I have no idea how big that resulting subset would be.






            EDIT: Following the methods of the linked talk, choose $n$ (assumed even) and product-free $S$, so that $frac{|S|}{n}ge 1-epsilon$. Hence $|S|ge n(1-epsilon)$. $S$ contains at most $frac{n}{2}$ even numbers (since $Ssubseteq {0,1,ldots, n-1}$, half of which are even). Take $T$ to be the set of all the odd numbers in $S$. We have $|T|ge |S|-frac{n}{2}=n(frac{1}{2}-epsilon)$. Since $Tsubseteq S$, $T$ is product-free. $T$ is also sum-free, since the sum of two elements of $T$ are even (and hence not in $T$). The asymptotic density of all naturals congruent to an element of $T$ modulo $n$ is $frac{|T|}{n}ge frac{1}{2}-epsilon$.

            Note that an asymptotic density of $frac{1}{2}$ is best-possible for sum-free sets (as proved in Pomerance's slides), much less sum-free product-free sets. The above construction gives a subset of $mathbb{N}$ arbitrarily close to this bound.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              I am glad to know the terms "sum-free" and "product-free." Much more memorable than "property $Q$"!
              $endgroup$
              – Joseph O'Rourke
              yesterday














            12












            12








            12





            $begingroup$

            This did not begin as an answer, but see edit below.



            See this talk by Carl Pomerance, on sum-free sets, and product-free sets. One way to answer the OP (and this is the approach of the other answers) is to choose an $n$, and a subset $S$ of $mathbb{Z}/nmathbb{Z}$, such that $S$ is both sum-free (if $a,bin S$, then $a+bnotin S$) and product-free (if $a,bin S$, then $abnotin S$) . Then, we take all integers that are congruent to an element of $S$, modulo $n$. The desired asymptotic density is seen to be $frac{|S|}{n}$.



            This might not be a simple question at all. Looking just at the sum-free property , we can easily get asymptotic density $0.5$ by taking the odd numbers. The product-free property is quite subtle: the linked talk gives a construction of a very large $n$ (with over 100 million digits) such that there is an $S$ satisfying $frac{|S|}{n}>0.5003$. In fact, we could raise $0.5003$ to be arbitrarily close to $1$ (although no construction is given in the linked talk). The general approach is to make $n$ have many small primes, to large powers, as factors.



            One would not expect that this product-free set is also sum-free, but we can always remove some elements from it, until we have a subset of $S$ that is both sum-free and product-free. I have no idea how big that resulting subset would be.






            EDIT: Following the methods of the linked talk, choose $n$ (assumed even) and product-free $S$, so that $frac{|S|}{n}ge 1-epsilon$. Hence $|S|ge n(1-epsilon)$. $S$ contains at most $frac{n}{2}$ even numbers (since $Ssubseteq {0,1,ldots, n-1}$, half of which are even). Take $T$ to be the set of all the odd numbers in $S$. We have $|T|ge |S|-frac{n}{2}=n(frac{1}{2}-epsilon)$. Since $Tsubseteq S$, $T$ is product-free. $T$ is also sum-free, since the sum of two elements of $T$ are even (and hence not in $T$). The asymptotic density of all naturals congruent to an element of $T$ modulo $n$ is $frac{|T|}{n}ge frac{1}{2}-epsilon$.

            Note that an asymptotic density of $frac{1}{2}$ is best-possible for sum-free sets (as proved in Pomerance's slides), much less sum-free product-free sets. The above construction gives a subset of $mathbb{N}$ arbitrarily close to this bound.






            share|cite|improve this answer











            $endgroup$



            This did not begin as an answer, but see edit below.



            See this talk by Carl Pomerance, on sum-free sets, and product-free sets. One way to answer the OP (and this is the approach of the other answers) is to choose an $n$, and a subset $S$ of $mathbb{Z}/nmathbb{Z}$, such that $S$ is both sum-free (if $a,bin S$, then $a+bnotin S$) and product-free (if $a,bin S$, then $abnotin S$) . Then, we take all integers that are congruent to an element of $S$, modulo $n$. The desired asymptotic density is seen to be $frac{|S|}{n}$.



            This might not be a simple question at all. Looking just at the sum-free property , we can easily get asymptotic density $0.5$ by taking the odd numbers. The product-free property is quite subtle: the linked talk gives a construction of a very large $n$ (with over 100 million digits) such that there is an $S$ satisfying $frac{|S|}{n}>0.5003$. In fact, we could raise $0.5003$ to be arbitrarily close to $1$ (although no construction is given in the linked talk). The general approach is to make $n$ have many small primes, to large powers, as factors.



            One would not expect that this product-free set is also sum-free, but we can always remove some elements from it, until we have a subset of $S$ that is both sum-free and product-free. I have no idea how big that resulting subset would be.






            EDIT: Following the methods of the linked talk, choose $n$ (assumed even) and product-free $S$, so that $frac{|S|}{n}ge 1-epsilon$. Hence $|S|ge n(1-epsilon)$. $S$ contains at most $frac{n}{2}$ even numbers (since $Ssubseteq {0,1,ldots, n-1}$, half of which are even). Take $T$ to be the set of all the odd numbers in $S$. We have $|T|ge |S|-frac{n}{2}=n(frac{1}{2}-epsilon)$. Since $Tsubseteq S$, $T$ is product-free. $T$ is also sum-free, since the sum of two elements of $T$ are even (and hence not in $T$). The asymptotic density of all naturals congruent to an element of $T$ modulo $n$ is $frac{|T|}{n}ge frac{1}{2}-epsilon$.

            Note that an asymptotic density of $frac{1}{2}$ is best-possible for sum-free sets (as proved in Pomerance's slides), much less sum-free product-free sets. The above construction gives a subset of $mathbb{N}$ arbitrarily close to this bound.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited yesterday

























            answered yesterday









            vadim123vadim123

            76.2k897191




            76.2k897191








            • 1




              $begingroup$
              I am glad to know the terms "sum-free" and "product-free." Much more memorable than "property $Q$"!
              $endgroup$
              – Joseph O'Rourke
              yesterday














            • 1




              $begingroup$
              I am glad to know the terms "sum-free" and "product-free." Much more memorable than "property $Q$"!
              $endgroup$
              – Joseph O'Rourke
              yesterday








            1




            1




            $begingroup$
            I am glad to know the terms "sum-free" and "product-free." Much more memorable than "property $Q$"!
            $endgroup$
            – Joseph O'Rourke
            yesterday




            $begingroup$
            I am glad to know the terms "sum-free" and "product-free." Much more memorable than "property $Q$"!
            $endgroup$
            – Joseph O'Rourke
            yesterday











            9












            $begingroup$

            Let $S = {n : n equiv 2 rm{or} 3 pmod 5}$. This has density $2/5$, which beats $1/3$.



            Incidentally, this sequence can be generated with a greedy algorithm, starting with $S = {2}$ and progressively adding every larger number that maintains the requirement.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This feels like it might be the max, because of the greedy property you mentioned.
              $endgroup$
              – Joseph O'Rourke
              yesterday






            • 2




              $begingroup$
              @JosephO'Rourke Not necessarily: using a greedy algorithm starting at $1$ instead of $2$ leads to a different sequence that is not as dense: ${1, 3, 5, 7, 11, 13, 17, 19, 23, 27, 29, 31, 37, 41, 43, 45, 47, 53, 59, 61, ldots}$. (I'm surprised not to find this on OEIS. You might be interested in adding it there.) Other starting points are similarly bad. But it's interesting that the greedy method starting at $2$ gives a consistent structure, and I agree that it feels like it could be the best result.
              $endgroup$
              – Théophile
              yesterday








            • 6




              $begingroup$
              @Théophile Doesn't 1*3=3 violate the constraint?
              $endgroup$
              – alphacapture
              yesterday






            • 1




              $begingroup$
              @Théophile What it looks like you are doing is starting with 3 and running the greedy algorithm while restricting yourself to odd numbers; this will result in the set of odd numbers with an odd number of prime factors
              $endgroup$
              – alphacapture
              yesterday






            • 3




              $begingroup$
              @alphacapture, indeed, dropping the $1$ and checking OEIS for the rest gives oeis.org/A067019 . (It's generally a good idea to drop a few early terms when checking OEIS, especially $1$'s and $0$'s.)
              $endgroup$
              – Barry Cipra
              yesterday
















            9












            $begingroup$

            Let $S = {n : n equiv 2 rm{or} 3 pmod 5}$. This has density $2/5$, which beats $1/3$.



            Incidentally, this sequence can be generated with a greedy algorithm, starting with $S = {2}$ and progressively adding every larger number that maintains the requirement.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This feels like it might be the max, because of the greedy property you mentioned.
              $endgroup$
              – Joseph O'Rourke
              yesterday






            • 2




              $begingroup$
              @JosephO'Rourke Not necessarily: using a greedy algorithm starting at $1$ instead of $2$ leads to a different sequence that is not as dense: ${1, 3, 5, 7, 11, 13, 17, 19, 23, 27, 29, 31, 37, 41, 43, 45, 47, 53, 59, 61, ldots}$. (I'm surprised not to find this on OEIS. You might be interested in adding it there.) Other starting points are similarly bad. But it's interesting that the greedy method starting at $2$ gives a consistent structure, and I agree that it feels like it could be the best result.
              $endgroup$
              – Théophile
              yesterday








            • 6




              $begingroup$
              @Théophile Doesn't 1*3=3 violate the constraint?
              $endgroup$
              – alphacapture
              yesterday






            • 1




              $begingroup$
              @Théophile What it looks like you are doing is starting with 3 and running the greedy algorithm while restricting yourself to odd numbers; this will result in the set of odd numbers with an odd number of prime factors
              $endgroup$
              – alphacapture
              yesterday






            • 3




              $begingroup$
              @alphacapture, indeed, dropping the $1$ and checking OEIS for the rest gives oeis.org/A067019 . (It's generally a good idea to drop a few early terms when checking OEIS, especially $1$'s and $0$'s.)
              $endgroup$
              – Barry Cipra
              yesterday














            9












            9








            9





            $begingroup$

            Let $S = {n : n equiv 2 rm{or} 3 pmod 5}$. This has density $2/5$, which beats $1/3$.



            Incidentally, this sequence can be generated with a greedy algorithm, starting with $S = {2}$ and progressively adding every larger number that maintains the requirement.






            share|cite|improve this answer









            $endgroup$



            Let $S = {n : n equiv 2 rm{or} 3 pmod 5}$. This has density $2/5$, which beats $1/3$.



            Incidentally, this sequence can be generated with a greedy algorithm, starting with $S = {2}$ and progressively adding every larger number that maintains the requirement.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            ThéophileThéophile

            19.9k12946




            19.9k12946












            • $begingroup$
              This feels like it might be the max, because of the greedy property you mentioned.
              $endgroup$
              – Joseph O'Rourke
              yesterday






            • 2




              $begingroup$
              @JosephO'Rourke Not necessarily: using a greedy algorithm starting at $1$ instead of $2$ leads to a different sequence that is not as dense: ${1, 3, 5, 7, 11, 13, 17, 19, 23, 27, 29, 31, 37, 41, 43, 45, 47, 53, 59, 61, ldots}$. (I'm surprised not to find this on OEIS. You might be interested in adding it there.) Other starting points are similarly bad. But it's interesting that the greedy method starting at $2$ gives a consistent structure, and I agree that it feels like it could be the best result.
              $endgroup$
              – Théophile
              yesterday








            • 6




              $begingroup$
              @Théophile Doesn't 1*3=3 violate the constraint?
              $endgroup$
              – alphacapture
              yesterday






            • 1




              $begingroup$
              @Théophile What it looks like you are doing is starting with 3 and running the greedy algorithm while restricting yourself to odd numbers; this will result in the set of odd numbers with an odd number of prime factors
              $endgroup$
              – alphacapture
              yesterday






            • 3




              $begingroup$
              @alphacapture, indeed, dropping the $1$ and checking OEIS for the rest gives oeis.org/A067019 . (It's generally a good idea to drop a few early terms when checking OEIS, especially $1$'s and $0$'s.)
              $endgroup$
              – Barry Cipra
              yesterday


















            • $begingroup$
              This feels like it might be the max, because of the greedy property you mentioned.
              $endgroup$
              – Joseph O'Rourke
              yesterday






            • 2




              $begingroup$
              @JosephO'Rourke Not necessarily: using a greedy algorithm starting at $1$ instead of $2$ leads to a different sequence that is not as dense: ${1, 3, 5, 7, 11, 13, 17, 19, 23, 27, 29, 31, 37, 41, 43, 45, 47, 53, 59, 61, ldots}$. (I'm surprised not to find this on OEIS. You might be interested in adding it there.) Other starting points are similarly bad. But it's interesting that the greedy method starting at $2$ gives a consistent structure, and I agree that it feels like it could be the best result.
              $endgroup$
              – Théophile
              yesterday








            • 6




              $begingroup$
              @Théophile Doesn't 1*3=3 violate the constraint?
              $endgroup$
              – alphacapture
              yesterday






            • 1




              $begingroup$
              @Théophile What it looks like you are doing is starting with 3 and running the greedy algorithm while restricting yourself to odd numbers; this will result in the set of odd numbers with an odd number of prime factors
              $endgroup$
              – alphacapture
              yesterday






            • 3




              $begingroup$
              @alphacapture, indeed, dropping the $1$ and checking OEIS for the rest gives oeis.org/A067019 . (It's generally a good idea to drop a few early terms when checking OEIS, especially $1$'s and $0$'s.)
              $endgroup$
              – Barry Cipra
              yesterday
















            $begingroup$
            This feels like it might be the max, because of the greedy property you mentioned.
            $endgroup$
            – Joseph O'Rourke
            yesterday




            $begingroup$
            This feels like it might be the max, because of the greedy property you mentioned.
            $endgroup$
            – Joseph O'Rourke
            yesterday




            2




            2




            $begingroup$
            @JosephO'Rourke Not necessarily: using a greedy algorithm starting at $1$ instead of $2$ leads to a different sequence that is not as dense: ${1, 3, 5, 7, 11, 13, 17, 19, 23, 27, 29, 31, 37, 41, 43, 45, 47, 53, 59, 61, ldots}$. (I'm surprised not to find this on OEIS. You might be interested in adding it there.) Other starting points are similarly bad. But it's interesting that the greedy method starting at $2$ gives a consistent structure, and I agree that it feels like it could be the best result.
            $endgroup$
            – Théophile
            yesterday






            $begingroup$
            @JosephO'Rourke Not necessarily: using a greedy algorithm starting at $1$ instead of $2$ leads to a different sequence that is not as dense: ${1, 3, 5, 7, 11, 13, 17, 19, 23, 27, 29, 31, 37, 41, 43, 45, 47, 53, 59, 61, ldots}$. (I'm surprised not to find this on OEIS. You might be interested in adding it there.) Other starting points are similarly bad. But it's interesting that the greedy method starting at $2$ gives a consistent structure, and I agree that it feels like it could be the best result.
            $endgroup$
            – Théophile
            yesterday






            6




            6




            $begingroup$
            @Théophile Doesn't 1*3=3 violate the constraint?
            $endgroup$
            – alphacapture
            yesterday




            $begingroup$
            @Théophile Doesn't 1*3=3 violate the constraint?
            $endgroup$
            – alphacapture
            yesterday




            1




            1




            $begingroup$
            @Théophile What it looks like you are doing is starting with 3 and running the greedy algorithm while restricting yourself to odd numbers; this will result in the set of odd numbers with an odd number of prime factors
            $endgroup$
            – alphacapture
            yesterday




            $begingroup$
            @Théophile What it looks like you are doing is starting with 3 and running the greedy algorithm while restricting yourself to odd numbers; this will result in the set of odd numbers with an odd number of prime factors
            $endgroup$
            – alphacapture
            yesterday




            3




            3




            $begingroup$
            @alphacapture, indeed, dropping the $1$ and checking OEIS for the rest gives oeis.org/A067019 . (It's generally a good idea to drop a few early terms when checking OEIS, especially $1$'s and $0$'s.)
            $endgroup$
            – Barry Cipra
            yesterday




            $begingroup$
            @alphacapture, indeed, dropping the $1$ and checking OEIS for the rest gives oeis.org/A067019 . (It's generally a good idea to drop a few early terms when checking OEIS, especially $1$'s and $0$'s.)
            $endgroup$
            – Barry Cipra
            yesterday











            4












            $begingroup$

            The answer to the title question is no. $Q$ doesn't characterize the odd primes, since, for example, ${2,3,15}vDash Q$. Any subset of the odd primes satisfies $Q$, as does any set formed by taking the odd primes along with an even integer $k$ and deleting one of every pair of primes that differ by $k$. Or forget about the primes altogether and take any (finite or infinite) set ${a_1, a_2, dots}$ such that $2leq a_1 < a_2$ and, for all $i>2$, $a_i > a_{i-1}a_{i-2}$.






            share|cite|improve this answer









            $endgroup$









            • 2




              $begingroup$
              I especially like your last example: $2,3,7,22,155,3411,ldots$.
              $endgroup$
              – Joseph O'Rourke
              yesterday
















            4












            $begingroup$

            The answer to the title question is no. $Q$ doesn't characterize the odd primes, since, for example, ${2,3,15}vDash Q$. Any subset of the odd primes satisfies $Q$, as does any set formed by taking the odd primes along with an even integer $k$ and deleting one of every pair of primes that differ by $k$. Or forget about the primes altogether and take any (finite or infinite) set ${a_1, a_2, dots}$ such that $2leq a_1 < a_2$ and, for all $i>2$, $a_i > a_{i-1}a_{i-2}$.






            share|cite|improve this answer









            $endgroup$









            • 2




              $begingroup$
              I especially like your last example: $2,3,7,22,155,3411,ldots$.
              $endgroup$
              – Joseph O'Rourke
              yesterday














            4












            4








            4





            $begingroup$

            The answer to the title question is no. $Q$ doesn't characterize the odd primes, since, for example, ${2,3,15}vDash Q$. Any subset of the odd primes satisfies $Q$, as does any set formed by taking the odd primes along with an even integer $k$ and deleting one of every pair of primes that differ by $k$. Or forget about the primes altogether and take any (finite or infinite) set ${a_1, a_2, dots}$ such that $2leq a_1 < a_2$ and, for all $i>2$, $a_i > a_{i-1}a_{i-2}$.






            share|cite|improve this answer









            $endgroup$



            The answer to the title question is no. $Q$ doesn't characterize the odd primes, since, for example, ${2,3,15}vDash Q$. Any subset of the odd primes satisfies $Q$, as does any set formed by taking the odd primes along with an even integer $k$ and deleting one of every pair of primes that differ by $k$. Or forget about the primes altogether and take any (finite or infinite) set ${a_1, a_2, dots}$ such that $2leq a_1 < a_2$ and, for all $i>2$, $a_i > a_{i-1}a_{i-2}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            David RicherbyDavid Richerby

            2,18011324




            2,18011324








            • 2




              $begingroup$
              I especially like your last example: $2,3,7,22,155,3411,ldots$.
              $endgroup$
              – Joseph O'Rourke
              yesterday














            • 2




              $begingroup$
              I especially like your last example: $2,3,7,22,155,3411,ldots$.
              $endgroup$
              – Joseph O'Rourke
              yesterday








            2




            2




            $begingroup$
            I especially like your last example: $2,3,7,22,155,3411,ldots$.
            $endgroup$
            – Joseph O'Rourke
            yesterday




            $begingroup$
            I especially like your last example: $2,3,7,22,155,3411,ldots$.
            $endgroup$
            – Joseph O'Rourke
            yesterday


















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