Does this property characterize the odd primes?












6












$begingroup$


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 $frac{1}{3}$.






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







share|cite|improve this question











$endgroup$

















    6












    $begingroup$


    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 $frac{1}{3}$.






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







    share|cite|improve this question











    $endgroup$















      6












      6








      6


      3



      $begingroup$


      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 $frac{1}{3}$.






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







      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




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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 4 hours ago







      Joseph O'Rourke

















      asked 5 hours ago









      Joseph O'RourkeJoseph O'Rourke

      18.1k349109




      18.1k349109






















          3 Answers
          3






          active

          oldest

          votes


















          6












          $begingroup$

          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
            3 hours ago



















          4












          $begingroup$

          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
            3 hours ago






          • 1




            $begingroup$
            @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
            3 hours ago








          • 4




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






          • 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
            1 hour ago










          • $begingroup$
            @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
            1 min ago



















          3












          $begingroup$

          This is not an answer, but too long for a comment. 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.






          share|cite|improve this answer









          $endgroup$













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











          Your Answer





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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          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
            3 hours ago
















          6












          $begingroup$

          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
            3 hours ago














          6












          6








          6





          $begingroup$

          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 5 hours ago









          José Carlos SantosJosé Carlos Santos

          163k22130233




          163k22130233












          • $begingroup$
            Very nice! ${}$
            $endgroup$
            – Joseph O'Rourke
            3 hours ago


















          • $begingroup$
            Very nice! ${}$
            $endgroup$
            – Joseph O'Rourke
            3 hours ago
















          $begingroup$
          Very nice! ${}$
          $endgroup$
          – Joseph O'Rourke
          3 hours ago




          $begingroup$
          Very nice! ${}$
          $endgroup$
          – Joseph O'Rourke
          3 hours ago











          4












          $begingroup$

          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
            3 hours ago






          • 1




            $begingroup$
            @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
            3 hours ago








          • 4




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






          • 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
            1 hour ago










          • $begingroup$
            @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
            1 min ago
















          4












          $begingroup$

          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
            3 hours ago






          • 1




            $begingroup$
            @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
            3 hours ago








          • 4




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






          • 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
            1 hour ago










          • $begingroup$
            @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
            1 min ago














          4












          4








          4





          $begingroup$

          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 4 hours ago









          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
            3 hours ago






          • 1




            $begingroup$
            @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
            3 hours ago








          • 4




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






          • 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
            1 hour ago










          • $begingroup$
            @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
            1 min ago


















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






          • 1




            $begingroup$
            @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
            3 hours ago








          • 4




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






          • 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
            1 hour ago










          • $begingroup$
            @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
            1 min ago
















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




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




          1




          1




          $begingroup$
          @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
          3 hours ago






          $begingroup$
          @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
          3 hours ago






          4




          4




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




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




          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
          1 hour ago




          $begingroup$
          @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
          1 hour ago












          $begingroup$
          @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
          1 min ago




          $begingroup$
          @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
          1 min ago











          3












          $begingroup$

          This is not an answer, but too long for a comment. 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.






          share|cite|improve this answer









          $endgroup$













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
















          3












          $begingroup$

          This is not an answer, but too long for a comment. 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.






          share|cite|improve this answer









          $endgroup$













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














          3












          3








          3





          $begingroup$

          This is not an answer, but too long for a comment. 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.






          share|cite|improve this answer









          $endgroup$



          This is not an answer, but too long for a comment. 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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 47 mins ago









          vadim123vadim123

          76.1k897190




          76.1k897190












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


















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
















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




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


















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